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Karnataka 1st PUC Chemistry Question Bank Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

1st PUC Chemistry Organic Chemistry: Some Basic Principles and Techniques One Mark Questions and Answers

Question 1.
Which compound is formed when an organic nitrogen compound is fused with sodium metal?
Answer:
NaCN

Question 2.
Name two elements which are detected by Lassaigne’s test.
Answer:
Nitrogen, Chlorine, Bromine, Iodine and Sulphur.

Question 3.
Name the blood red coloured compound formed in Lassaigne’s test conducted on an organic compound containing both nitrogen and sulphur.
Answer:
Ferric Thiocynate [Fe(CNS)₃]

Question 4.
Name the compound used to absorb water in the estimation of hydrogen in an organic compound by Liebig’s method.
Answer:
Anhydrous calcium chloride.

Question 5.
Name the compound formed when an organic compound containing. nitrogen is heated with concentration sulphuric acid and potassium
sulphate.
Answer:
Ammonium Sulphate.

Question 6.
During estimation of nitrogen by Kieldahl’s method, copper sulphate is added to sulphuric acid. Why?
Answer:
Copper sulphate acts as catalyst.

Question 7.
Which method is employed for the estimation of carbon and hydrogen organic compounds?
Answer:
Liebig’s method.

Question 8.
Which type of organic compounds cannot be Kjeldahlised?
Answer:
Nitro compounds (R-NO₂) and azo (-N = N-) compounds.

Question 9.
Name the type of isomerism shown by the following compounds CH₃COOH and HCOOCH₃.
Answer:
Functional isomerism.

Question 10.
What is the difference in the molecular formulae of any two successive members in a homologous series?
Answer:
-CH2 –

Question 11.
Is Isobutene a straight chain or branched chain hydrocarbon?
Answer:
Branched chain hydrocarbon.

Question 12.
Ring compounds containing more than one type of atoms in the ring are
called
Answer:
Heterocyclic compounds.

Question 13.
The molecular formulae of four hydrocarbons belonging to a homologous series are CH₄, C₂H₆, C₅H₈ and C₄H₁₀. Write the general formula of this
series.
Answer:
C_nH_2n+2

Question 14.
Why do organic compounds of a given homologous series show similar properties?
Answer:
They possess the same functional group.

Question 15.
What is the acronym for International Union of Pure and Applied Chemistry?
Answer:
IUPAC

Question 16.
Write resonance hybrid structure of benzene.
Answer:

Question 17.
Define resonance (mesomeric) effect.
Answer:
The permanent polarity is produced by the interation of lone pair & π electrons in conjugate a system of an organic molecule.

Question 18.
Indicate the type of electron pair involved in M effect.
Answer:
π-electron pair or lone pair electrons.

Question 19.
What is +R effect ?
Answer:
Shift of electron pair away from substituent in congjuagte system.

Question 20.
Define inductive effect.
Answer:
The permanent polarity in an saturated organic molecule due to small displacement of sigma bond pair of electrons towards more electronegative atom.

Question 21.
Which electron displacement effect involves displacement of s-electrons?
Answer:
Sigma bond pair of electrons.

Question 22.
What is +1 effect ?
Answer:
Small displacement of sigma bond pair of electrons away from substituent.

Question 23.
Give an example of a group which exerts +1 effect.
Answer:
Alkyl group (methyl group)

Question 24.
Out of (CH₃)₂ CH – and CH₃ – group, which is a better electron releasing group?
Answer:
(CH₃)₂CH-

Question 25.
What is -1 effect ?
Answer:
Small displacement of sigma bond pair of electrons towards substituent.

Question 26.
Give an example of electron withdrawing group.
Answer:
Nitro group (-NO₂)

Question 27.
Identify the stronger electron withdrawing substituent from each of the following pairs.

1. -NO₂ and -Cl
2. -C ≡ N and -Br
3. -Cl and -OCH₃
4. -C₆H₅ and -F

Answer:

1. -N0₂
2. -C ≡ N
3. -Cl
4. -F

Question 28.
What is R-effect?
Answer:
Shift of electron pair towards the substituent in a conjugate system.

Question 29.
What is Inductive effect?
Answer:
The polarisation of one σ bond caused by polarization of adjacent a bond due to difference in electro-negativity.

Question 30.
Sketch the resonance structure of Benzene.
Answer:

Question 31.
What is the suitable adsorbent in the process of column chromography?
Answer:
Al₃O₃ (alumina)

Question 32.
Which process is used to purify impure sample or camphor, contaminated with sand?
Answer:
Sublimation

Question 33.
Which gas is liberated in Kjeldahl’s method?
Answer:
Ammonia gas (NH₃)

Question 34.
What is Lassaigne’s extract ?
Answer:
When organic compound is fused with sodium metal and then extracted by water, it is called Lassaigne’s extract.

Question 35.
Which gas is liberated in Dumas method?
Answer:
N₂

Question 36.
Define Crystallization.
Answer:
The process of getting crystals when hot saturated solution of a compound is cooled is called crystallization.

Question 37.
What is basic principle of chromotography ?
Answer:

• Paper,
• Fhinlayer,
• Column

Question 38.
What is steam distillation ?
Answer:
The distillation which is carried out with the help of steam is called steam distillation.

Question 39.
Suggest the method to purify:
(a) Camphor containing salt impurity
(b) Kerosene Oil containing water
(c) Mixture of Benzene and Toulene
(d) Sugar and Naphtalene
Answer:
(a) sublimation
(b) by solvent extraction
(c) fractional Distillation
(d) treat with water and filter

Question 40.
Why is fusion of an organic compound with sodium required ?
Answer:
It is done so as to convert organic compound into inorganic compound.

Question 41.
Name two classes of compounds in which Kjeldahl’s method cannot be used for estimation of nitrogen.
Answer:
Nitro compounds, Azo compounds and compounds containing nitrogen in ring, e.g., pyridine.

Question 42.
Which class of compounds are tested with the help of Beilstein test.
Answer:
Halogen containing organic compounds but some other compounds like urea is respond to this test.

Question 43.
What are formulae of

• Sodium nitroprusside
• Ferric ferrocyanide ?

Answer:

• Na₂[Fe(CN)₅NO]
• Fe₂[Fe(CN)₆]₃

Question 44.
Which elements are estimated by Liebig’s Method?
Answer:
Carbon and Hydrogen

Question 45.
What is the relationship between molecular mass and equivalent mass of an acid?
Answer:

Question 46.
What is relationship between molecular formula and empirical formula of a substance ?
Answer:

Question 47.
Which effect involves the displacement of electron pair under the infulence of an attacking reagent ?
Answer:
Electromeric effect.

Question 48.
What is an electromeric effect ?
Answer:
The shift of π electron pair of a multiple bond to one of the bonded atoms during the attack of electrophile or nucleophile.

Question 49.
Which type of E effect Operates during the attack of proton on ethane molecule ?
Answer:
+E effect.

1st PUC Chemistry Organic Chemistry: Some Basic Principles and Techniques Two Marks Questions and Answers

Question 1.
What is Electrophile ? Give example.
Answer:
The electron deficient molecules or positively charged ions which are capable of accepting on electron from substrate molecule are called electrophiles.
Positive electrophiles : H⁺, Cl⁺, Br⁺, NO₂, N⁺ = O, R⁺ (carbocations) etc.
Neutral electrophiles : SO₃, BF₃, AICl₃, Cl

Question 2.
What are Nucleophiles ?
Answer:
The molecules or negatively charged ions which are capable of donating an electron pair to electron deficient centre of the substrate are called nucleophiles.
Negative nucleophiles:

Question 3.
What is the difference between organic and inorganic compounds ?
Answer:

Property	Organic Compounds	Inorganic Compounds
Chemical nature	Compounds of carbon	Compounds of elements other than carbon
Bonding	Covalent	Ionic
Melting . and boiling points	Low, generally Volatile	High, generally non-­volatile
Solubility	Soluble in organic solvents, insoluble in water.	Soluble in water, insoluble in organic solvents.
Electrical conduction	Non – conductors ; solutions and liquids – non – electrolytes	Conductors ; solutions and liquids – conductors

Question 4.
What is the difference between Electrophilic Reagents and Nucleophilic Reagents.
Answer:

Electrophilic Reagents	Nucleophilic Reagents
1. These are electron loving species.	1. These are nucleus loving species
2. These are electron deficient species.	2. These are electron rich species.
3. They attack the site of high electron density in the substrate.	3. They attack the site of low electron density in the substarte.
4. They may be +vely charged ions (cations) or neutral molecules.	4. They may be negatively charged ions (anions) or neutral molecules.
5. They possess an atom with incomplete octet.	5. They possess an atom withlone pair of electrons.
6. They are Lewis acids.	6. They are Lewis bases.
7. They accept an electron pair from the substrate to form a covalent bond.	7. They done an electron pair to the substrate to form a covalent bond.

Question 5.
What is Homolytic fission (Homolysis) ? Give reason.
Answer:
Definition: The fission of a covalent bond, in which each of the two species produced, retains one electron of shared electron pair, is called hemolytic fission or homolysis,
Eg: \(\mathrm{Cl}-\mathrm{Cl} \stackrel{\text { Homolysis }}{\longrightarrow} 2 \mathrm{Cl}^{-}\) (Chloroine free radicals)

Question 6.
What is Heterolytic Fission (Heterolysis)?
Answer:
The fission of a covalent bond, in which ionic species are produced due to unequal disrtribution of bonded electron pair, is called heterolytic fission or heterolysis.

Question 7.
What is Homolytic Fission and Heterolytic Fission?
Answer:

Homolytic Reagent	Heterolytic Fission
1. In this case, the covalent bond breaks symmetrically.	1. In this case, the covalent bond breaks unsymmetrically.
2. Each species, obtained, retains one electron from shared electron pair.	2. The more electronegative otom retains the shred electron pair.
3. Electrically neutral free radicals are formed which carry an old certain.	3. Electrically charged ions (cations and anions) are formed.
4. It takes place in the presence of sunlight, U.V. light or by pyrolysis.	4. It takes place in the presence of polar solvent.

Question 8.
What are Free Radical ? Give example.
Answer:
A free radical can be defined as an atom or group of atoms having an odd or impaired electron. Putting a dot (•) against the symbol of atom or group of atoms.
Example :
(Chlorine free radicals)

Question 9.
What are the characteristics of Free Radicals.
Answer:
Characteristics of Free Radicals

• Free radicals are generally electrically neutral.
• They carry on odd (unpaired) electron.
• They are highly unstable.
• They are short-lived.
• They are highly reactive species due to the presence of odd electrons.
• They have a tendency to pair up with the odd electrons to form a covalent bond.
• They are generally formed either in presence of U.V. or visible light or in the presence of peroxides.

Question 10.
What is Alicylic compounds? Give Example.
Answer:
These are saturated hydrocarbons joined by covalent bond to form ring structure.

Question 11.
What are Aromatic Compounds ? Give example.
Answer:
These are the compounds containing atleast one benzene ring.

Question 12.
What are Heterocyclic compound? Give example.
Answer:
These are the compounds containing ring structure in which one or more carbon atoms are replaced by hetero atoms such as N, S, O etc.,

Question 13.
What is functional group? Give example.
Answer:
A functional group is defined as “an atom or group of atoms present in a molecule which determines most of the chemical properties of the particular class of organic compounds”.
Example : CH₃ – OH methyl alcohol.

Question 14.
What are carbanions ? Give example.
Answer:
A reaction intermediate formed by heterolysic fission of a covalent bond which contains one negatively charged carbon with eight electrons in its valence shell is called carbanion.
The heterolytic cleavage of a covalent bond as indicated in the following reactions gives carbanions.
Example : \(\mathrm{CH}_{3}-\mathrm{MgI} \longrightarrow \mathrm{CH}_{3}^{-}+\mathrm{M}^{+} \mathrm{gI}\)

Question 15.
What are carbocations ? Give Example.
Answer:
A reaction intermediate formed by heterolytic fission of a covalent bond which contains one positively charged carbon with three bond pair electrons (sextext of electrons) is called carbocation.
The heterolytic fission of bromomethane yields methyl carbocation and bromide ion as shown below.
Example :

Question 16.
What is the difference between Carbocation & Carbanion.
Answer:

Carbocation	Carbanion
1. The central carbon atom is +vely charged.	1. The central carbon atom is -vely charged.
2. It is an electron deficient species.	2. It is an electron rich species.
3. The central carbon atom possesses six electrons in its outermost shell.	3. The central carbon atom possesses eight electrons in its outermost shell.
4. The central carbon atom is in sp2 state of hybridisation.	4. The central carbon atom is in sp3 state of hybridisation and carries a lone pair of electrons.
5. Its shape is triangular planner.	5. Its shape is pyramidal.
6. It can accept an electron pair from a nucleophile to form a covalent bond.	6. It can donate an electron pair to an electrophile to form a covalent bond.

Question 17.
What is positive inductive effect (+1 effect) ? Give example.
Answer:
In this effect the substituent (Y) releases electron pair away from itself. In other words a bond pair of electrons are displaced away from the substituent.
Eg : All alkyl groups.

Question 18.
What is negative inductive effect (-1 effect) ? Give example.
Answer:
In this effect, the sigma bond pair of electrons are displaced towards electron withdrawing substituent (X).
The order of electron withdrawing ability (intensity of -1 effect) of a few substituent is given below.

Question 19.
What is electromeric effect ? Give example.
Answer:
It is the complete transfer of shared pair of electrons of a multiple bond to one of the atom in the presence of attacking reagent.
Eg.: H+, -CN^–, etc.,

Question 20.
What is +E effect ? Give example.
Answer:
When the transfer of electrons takes place towards the attacking reagent, the effect is called +E effect.
For example, the addition of an acid to alkenes.

Question 21.
What is -E Effect ? Give example.
Answer:
When the transfer of electron takes place way from the attacking reagent, the effect is called -E effect.
For example, the addition of cyanide ion (CN-) to carbonyl group (>C = 0)

Question 22.
What is mesomeric effect ? Give example.
Answer:
The polarity developed in a molecule, as a result of interaction between two π-bond or a π-bond and a lone pair of electrons is referred to as mesomeric effect.
Groups with +M or +R effect.
—Cl, —Br, —I, -NH₂, -NHR, -NR₂, -OH, -OR, -SH, -OCH₃ etc.,
Groups with -M or -R effect.
-NO₂, -C ≡ N, -C-, -CHO, -COOH, -COOR etc.,

Question 23.
What is the difference between Inductive effect & Mesomeric effect?
Answer:

Inductive Effect	Mesomeric Effect
1. It operates in saturated compounds.	1. It operates in unsaturated compounds especially having conjugated systems.
2. It involves electrons of a – bonds.	2. It involves electrons of π – bonds or lone pair of electrons.
3. The electron pair is slightly displace from its position and hence partial charges are developed.	3. The electron pair is completely transferred and hence unit positive and negative charges are developed.
4. It is transmitted over a short distance it becomes negligible after second carbon atom in the chain.	4. It is transmitted through the entire chain provided conjugation is present.

Question 24.
What is the difference between Inductive Effect and Electromeric Effect.
Answer:

Inductive Effect	Electromeric Effect
1. It is permanent in nature.	1. It is temporary in nature.
2. It is due to electronegative atom present in the molecule itself.	2. It is due to approach of the attacking reagent.
3. It is the mobility of electrons along C-C single bond.	3. It is the mobility of electrons in a multiple bond (double or triple bond).

Question 25.
What is resonance energy ? Mention the resonance energy of Benzene.
Answer:
The phonomenon in which two or more structures can be written for a compound but none of them represents it correctly is called resonance. The actual structure of the compound is said to be a resonance hybrid. 36 kcal mol^-1 is the resonance energy of benzene.

Question 26.
Explain hyperconjucgation effect.
Answer:
The electron release of alkyl group bonded to unsaturated system in which delocalization of electrons takes place through overlap between C – H sigma (σ) orbital and pi (π) bond orbited or vacant p-orbital is known as hyperconjugation.

Question 27.
What is substitution reaction ? Give example.
Answer:
The reaction in which an atom of group atoms attached to carbon atom in a substrate molecule, is replaced another atom is called substitution reaction.
Ex: \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\mathrm{UV}}{\longrightarrow} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{HCl}\)

Question 28.
What is addition reaction ? Give example.
Answer:
The reaction in which the attacking reagent adds up to the substrate molecule without elimination of any molecule are called addition reaction.
Ex :
\(\mathrm{CH} \equiv \mathrm{CH}+\mathrm{H}_{2} \frac{\mathrm{Ni}}{14: \mathrm{C}} \cdot \mathrm{CH}_{2}=\mathrm{CH}_{2}\)

Question 29.
What is the test for sulphur ?
Answer:
To a small amount of sodium extract, freshly prepared solution of sodium nitro prusside is added, a deep violet colour indicates the presence of sulphur.
Ex:

1st PUC Chemistry Organic Chemistry: Some Basic Principles and Techniques Three/Four Marks Questions and Answers

Question 1.
How is the detection of carbon and hydrogen by copper oxide test ?
Answer:
Organic compounds undergo oxidation in the presence of a suitable oxidizing agent. In this process, carbon is oxidized to CO₂ and hydrogen is oxidized to water.

Procedure : The compound is initially lime water mixed with dry cupric oxide. The mixture is strongly heated in a hard glass test, tube fitted with a cork and a delivery tube. The liberated gases are passed into lime water. Carbon present in compound is oxidised by cupic oxide to carbon
dioxide, which turns lime water milky. The hydrogen present in the compound is converted into water which turns anhydrous copper sulphate to blue hydrated salt.

Question 2.
How is the prepartion on of Lassaigne’s is filtrate ?
Answer:
Procedure : A piece of dry sodium is introduced into a fusion tube and heated till it melts. A drop of few crystals of the organic compound is added to the fusion tube. The mixture is fused gently on a Bunsen flame initially. The tube is then heated until red hot and plunged into in a mortar containing distilled water. The contents are ground throughly and filtered. The filtrate is known as sodium fusin extract, stock solution or Lassaigne’s extract. The filtrate is divided into three parts, which are used for the detection of nitrogen, sulphur and halogen in organic compound.

Question 3.
How is detection of nitrogen by Lassaigne’s filtrate ?
Answer:
A few crystal of ferrous sulphate are added to the first part of the filtrate. The mixture is boiled and cooled. It is acidified with hydrochloric acid and a few drops of ferric chloride solution are added. Sodium cyanide in the filtrte reacts with ferrous sulphate to give sodium ferrocyanide. It further reacts with ferric chloride to give a blue coloured solution of ferric ferrocyanide.

Question 4.
How is the detection of sulphur by Lassaigne’s is filtrate ?
Answer:
Lead acetate test: second part of the filtrate is treated with excess of acetic acid and lead acetate solution. A black precipitate of lead sulphide is formed.

Question 5.
Explain the test for detection of halogens by Lassaigne’s is filtrate
Answer:
Silver nitrate test : A portion of the stock solution is boiled with dil. HN03, cooled and silver nitrate is added. A white precipitate soluble in ammonium hydroxide shows the presence of chlorine. A pale yellow precipitate slightly in ammonium hydroxide shows the presence of bromine. A yellow precipitate insoluble in ammonium hydroxide ” shows the presence of iodine.

Question 6.
Explain the test for detection of phs phosphorus in an organic compound.
Answer:
Organic compound containing phosphorous is fused with sodium peroxide. The
phosphorus of the compound is oxidised to phosphate. The fused mass is extracted with water and filtered. The filtrate containing sodium phosphate is boiled with nitric acid and then treated with ammonium molybdate. A yellow solution of precipitate indicates the presence of phosphorus.

Question 7.
Explain the estimation of phosphorus in organic compound by carius method.
Answer:
It is also estimated by carius method the known mass of organic compound containing phosphorus is heated with fuming HNO₃ when phosphorus is oxidised to phosphoric acid (H₃PO₄). To this, magnesia mixture (MgSO₄ + NH₄OH + NH₃CI) is added when phosphoric acid precipitates as magnesium ammonium phosphate (MgNH₄PO₄). This precipitate is filtered, washed, dried and ignited when it is converted to magnesium pyrophosphate (Mg₂P₂O₃).
\(2 \mathrm{MgNH}_{4} \mathrm{PO}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}+2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}\)
The weight of Mg₂P₂CO₇ is determied from which the percentage of phosphorus is the compound can be calculated.
Observations and Calculations:

1. Mass of organic compound taken = w₁, g.
2. Mass of Mg₂P₂O₇ obtained = W₂g.

We have \(\underset{222 \mathrm{g}}{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \equiv \underset{62 \mathrm{g}}{2 \mathrm{P}} \text { Now, } 222 \mathrm{g} \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\)contains 62g of phosphorus.
W₂g of Mg₂p₂O₇ Will contain \(\frac{62 \times w_{2}}{222} g\) of phosphorus
This amount of phosphorus was present in wig of the compound
\(\therefore \% \mathrm{P}=\frac{62 \times \mathrm{w}_{2}}{222} \times \frac{100}{\mathrm{w}_{1}}\)

Question 8.
0.189g of an organic substance containing chlorine gave carius method
0. 287g of silver chloride. Calculate the percentage of chlorine in the substance.
Answer:

1. Mass of organic compound taken (W1) = 0.189 g
2. Mass of AgCli ppt (W2) = 0.287 g

Now , %\(\mathrm{Cl} 2=\frac{35.5 \times \mathrm{w}_{2}}{143.5} \times \frac{100}{\mathrm{w}_{1}}=\frac{35.5 \times 0.287 \times 100}{143.5 \times 0.189}=37.57\)

Question 9.
0.2632g of silver bromide is obtained from 0.2562g of an organic compound. Find the percentage of bromine in the compound.
Answer:

1. Mass of organic compound (W1) = 0.2562g
2. Mass of AgBr obtained (W2) = 0.2632g.

\(\text { Now } \% \mathrm{Br}_{2}=\frac{80 \times \mathrm{w}_{2}}{188} \times \frac{100}{\mathrm{w}_{1}}=\frac{80 \times 0.2632 \times 100}{188 \times 0.2562}=43.71\)

Question 10.
In Leibig’s method. 0.24 g of organic compound on combustion with dry oxygen produced of 0.62 g of CO2 and 0.11 g of II2O. Determine the percentage composition of the compound.
Answer:
Mass of organic compound = m = 0.24 g .
Mass of carbon dioxide formed = 0.62 g
Mass of water formed = 0.11 g
Percentage of carbon = \(\frac{12}{4} \times \frac{0.62}{0.24} \times 100=70.45\)
Percentage of hydrogen = \(\frac{2}{4} \times \frac{0.11}{0.24} \times 100=5.09\)
Percentage of oxygen = [100 – (70.5 + 5.0)] = 24.46

Question 11.
In Carius method of estimation of halogen, 0.20 g of organic compound gave 0.15 g of silver bromide. Calculate the percentage of bromine in the compound.
Answer:
Mass of organic compound (m1) = 0.20 g
Mass of silver bromide formed (m2) = 0.15 g
188 (108 + 80) g of AgBr contains 80 g of bromine
∴ 0.20 g of AgBr contains = \(\frac{80 \times 0.15}{188} \mathrm{g}=0.0638 \mathrm{g} \text { of } \mathrm{Br}\)
Percentage of bromine = \(\frac{80 \times 15 \times 100}{188 \times 20}=31.92\)
The percentage of bromine in the given compound = 31.92

1st PUC Chemistry Organic Chemistry: Some Basic Principles and Techniques Five Marks Questions and Answers

Question 1.
Describe with a neat diagram the estimation of carbon and hydrogen by Leibig’s method.
Answer:
Principle : A known mass of an organic compound is strongly heated with dry cupric oxide (CuO), when carbon and hydrogen are quantitatively oxidized to CO₂ and H₂O respectively. The masses of CO₂ and H₂O thus formed are determined. From this, the percentages of carbon and hydrogen can be calculated.
Procedure : Pure and dry oxygen is passed through the entire assembly of the apparatus (Fig) till the CO₂ and moisture is completely removed.

A boat containing weighed organic substances is introduced inside from one end of the combustion tube by opening it for a while. The tube is now strongly heated till the whole of the organic compound is burnt up. The flow of oxygen is continued to drive CO₂ and water vapours completely to the U-tubes. The apparatus is cooled and the U-tubes are weighed separately.
Observed and Calculations.

1. Mass of organic compound taken = w.g.
2. Mass of water produced = x g (Increase in mass of CaCk tube)
3. Mass of carbon dioxide produced = y g. (Increase in mass of KOH tube)

To determine % of carbon
Molar mass of CO₂ = 44g mol^-1
Now, 44g of CO₂ = contains 12 g of C.
∴ y g of CO₂ will contain of \(\frac { 12y }{ 44 }\) g of C.
This amount of carbon was present in w. g. of the substance
\(\therefore \% \mathrm{C}=\frac{12 \mathrm{y}}{44} \times \frac{100}{\mathrm{w}}\)
To determine % of Hydrogen
Molar mass of water = 18 g mol^-1
Now 18g of H₂O contains 2 g of H₂
∴ x g of H₂O will contain \(\frac { 2x }{ 18 }\) g of H₂
This amount of hydrogen was present in weight of substance.
\(\therefore \% \mathrm{H}_{2}=\frac{2 \mathrm{x}}{18} \times \frac{100}{\mathrm{w}}\)

Question 2.
How is the estimation of Nitrogen in ogranic compound by Dumas method.
Answer:
Principle : The organic compound containing nitrogen when heated with excess of copper oxide in the atmosphere of carbon dioxide, yields nitrogen in addition to carbon dioxide and water.

Traces of nitrogen oxides formed during combustion of organic compound are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The percentage of nitrogen present in a given organic compound is calculated from the volume of nitrogen collected over potassium hydroxide solution from a known mass of organic compound.

Procedure : The apparatus used for the estimation of nitrogen by this method is shown in the figure.

A known mass of organic compound is mixed with copper oxide and placed in the combustion tube. The carbon dioxide gas is passed through the combustion tube to displace air present in the tube. The combustion tube is now heated in the furnace, the nitrogen evolved collects in the nitrometer. The volume of the nitrogen collected is recorded after adjusting the levels of potassium hydroxide solution in the two limbs are equal. Room temperature and atmosphere pressure are recorded.
Calculation:
Mass of organic compound = mg
Volume of nitrogen in nitrometer = V cm³
Room temperature = t° C = (273 + t) K
Atmosphere pressure = P1 mm
Aqueous tension at room temperature = P’ mm
Pressure of dry nitrogen gas formed = P = (P – P’) mm
Volume of nitrogen at STP (V0) = \(\frac{\mathrm{PV} \times 273}{760 \times(273+\mathrm{t})} \mathrm{cm}^{3}\)
22,400 cm³ of nitrogen of STP = 28 g of nitrogen
Mass of V0 cm³ of nitrogen = \(\frac{28 \times V}{22,400} g\)
Percentage of nitrogen = \(\frac{28 \times V_{0} \times 100}{22,400 \times m}\)

Question 3.
Describe an experiment to determine the percentage of nitrogen in an organic compound by Kjeldahl’s method.
Answer:
Principle : When a nitrogenous organic compound is heated with cone. H₂SO₄ using (CUSO₄K₂SO₄) as a catalyst, the nitrogen from the compound is quantitatively converted to ammonium sulphate.

This ammonium sulphate is decomposed by heating with excess of alkali and the ammonia evolved is absorbed in known excess of a standard solution of H₂SO₄. Part of acid is neutralized by ammonia. The excess of acid left behind after neutralization with ammonia is estimated by back titration with standard alkali. From this, the amount of acid actually consumed by ammonia can be obtained which can be used to determine the percentage of nitrogen in the compound.

Procedure : A known exact mass of organic compound (about 0.5g) ix mixed with lOg K₂SO₄, Ig CuSO₄ and 25 ml of cone. H₂SO₄. The mixture is heated strongly in a Kjeldahl’s flask. Till the contents become clear. This step is known as digestion.

The Kjeldahl’s flask is now cooled and the liquid is heated in a round-bottomed flask with excess of caustic soda solution The ammonia evolved is absorbed in a known volume of a standard acid.
The amount of unreacted acid is determined by titrating it against a standard alkali.
(NH₄ )2S0₄ + 2NaOH → 2NH₃ ↑+ Na₂S0₄ + 2H₂0
Observation :

1. Mass of organic compound taken = Weight
2. Normality of standard acid = N
3. Volume of standard acid taken = V1 ml
4. Volume of alkali (Normality = N) required for back titration = V2 ml

Calculation: Volume of acid used up by ammonia = Volume of ammonia produced
= (V2 – V1) = V ml of normality N.
Now,
1000 ml of 1 normal NH₃ = 17g NH₃ = 14g N₂
∴ Vml of N- normal ammonia will contain \(\frac{14 \times \mathrm{N} \times \mathrm{V}}{1000} \mathrm{gN}_{2}\)
This amount of nitrogen was present in w g of the compound
\(\therefore \% \mathrm{N}_{2}=\frac{14 \times \mathrm{N} \times \mathrm{V}}{1000} \times \frac{100}{\mathrm{w}} \text { or } \% \mathrm{N}_{2}=\frac{1.4 \mathrm{NV}}{\mathrm{w}}\)
Where, N and V are the normality and volume respectively of the acid used up by ammonia.

Question 4.
How is the estimation of halogens by Carius method ?
Answer:
When an organic compound containing halogen (Cl, Br or 1) is heated in a sealed tube with fuming nitric acid and excess of silver chloride, silver halide is formed from the mass of silver halide obtained, the percentage of the halogen can be calculated.
Procedure : In a hard glass tube (Carius tube), 5ml of fuming HNO₃ and 2 to 2.5 g AgNO₃ are taken. A small narrow weighing tube, containing a small amount (nearly 0.15-0.2g) of accurately weighed organic compound, is introduced in the Carius tube in – such a way that nitric acid does not enter the weighing tube. The Carius tube is now sealed and heated in a furnace at 300°C for about six hours.

The tube is then cooled and its narrow end is cut off and the contents are completely transferred to a beaker by washing with water. The precipitate of silver halide formed is filtered through a weighed sintered glass crucible. It is washed, dried and weighed. Observation and calculation :

1. Mass of organic compound taken = W1g
2. Mass of silver halide obtained = W2 g

(a) For chlorine :
\(\underset{143.5 \mathrm{g}}{\mathrm{AgCl}} \equiv \underset{35.5 \mathrm{g}}{\mathrm{Cl}}\)
143.5g of AgCl contains 35.5 g of chlorine
w2g of Agcl Will contain \(\frac{35.5 \times w_{2}}{143.5} g\) of chlorine
This amount of chlorine was present in wig of the compound.
\(\therefore \quad \% \mathrm{Cl}_{2}=\frac{35.5 \times \mathrm{w}_{2}}{143.5} \times \frac{100}{\mathrm{w}_{1}}\)

(b) For bromine :
\(\underset{188 \mathrm{g}}{\mathrm{AgBr}} \equiv \underset{80 \mathrm{g}}{\mathrm{Br}}\)
188g of AgBr contains 80g of bromine
W2 g of AgBr will contain \(\frac{980 \times w_{2}}{188} g\) of bromine.
\(\therefore \% \mathrm{Br}_{2}=\frac{80 \times \mathrm{w}_{2}}{188} \times \frac{100}{\mathrm{w}_{1}}\)

(c) For Iodine :
\(\underset{235 \mathrm{g}}{\mathrm{Agl}} \equiv \underset{127 \mathrm{g}}{\mathrm{I}}\)
235g of Agl contains 127g of iodine
W2 g of Agl will contain \(\frac{127 \times w_{2}}{235} g\) of iodine
\(\therefore \quad \% \mathrm{I}_{2}=\frac{127 \times \mathrm{w}_{2}}{235} \times \frac{100}{\mathrm{w}_{1}}\)

You can Download Chapter 3 Motion in a Straight Line Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line

1st PUC Physics Motion in a Straight Line TextBook Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. a railway carriage moving without jerks between two stations.
2. a monkey sitting on top of a man cycling smoothly on a circular track.
3. a spinning cricket ball that turns sharply on hitting the ground.
4. a tumbling beaker that has slipped off the edge of a table.

Answer:

1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.
2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.
3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.
4. Different parts of the beaker experience different magnitude of forces during its motion. Hence it cannot be treated as a point object.

Question 2.
The position-time (x-i) graphs for two children A and B returning from their school! O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/ different) time.
5. (A/B) overtakes (B/A) on the road (once/twice).

Answer:
1. The distance between P and O (the school) is less then the distance between Q and O. (from the graph)
⇒ A lives closer to the school than B.
2. From the graph, we see that the line for A starts before the line for B.
⇒ A starts from the school earlier than B.
3. Velocity = Slope of the x-t graph. Clearly, the slope for B > slope for A.
⇒ B walks faster than A
4. From the graph, we see that both lines end at the same time.
⇒ A and B reach home at the same time.
5. From the graph, we see that the lines intersect once. Also, B walks faster.
⇒ B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Time for the woman to reach her office = \(\frac{2.5 \mathrm{km}}{5 \mathrm{km} \mathrm{h}^{-1}}\)
= 30 min.
∴ She reaches her office at 9.30 am. Time for the woman to return home = \(\frac{2.5 \mathrm{km}}{25 \mathrm{km} \mathrm{h}^{-1}}\)
= 0.1 h = 6 min.
∴ She reaches her home at 5.06 pm

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:

From the above table, see that every 8 s the drunkard moves 2 m forward.

By looking at the shape of the graph, we can see that the graph for entire motion will be like this:-

From the graph, we see that the drunkard will fall into the pit at 137 s.
Analytically:-
The drunkard covers 2 m every 8 s.
∴ He will cover 8 m in 32 s. When he takes 5 steps forwards, he will fall into the pit.
∴ Total time = 32 + 5 = 37 s.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Speed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
= (1500 – 500) km h^-1
= 1000 km h^-1

Question 6.
A car moving along a straight highway with a speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer: v₀ = 126 km h^-1
= 126 × \(\frac{5}{18}\) m s^-1
= 35 m s^-1
Stopping distance = (x – x₀) = 200 m
Since the car is stopped , v = 0.
v₂ = v₀² + 2 a (x – x₀)
0 = 35² + 2a (200)
⇒ a = – 3.06 m s^-2
Retardation = 3.06 m s^-2
v = v₀ + at
⇒ 0 = 35 – 3.06 t
⇒ t = 11.4 s
Time for car to stop = 11.4 s.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B Just brushes past the driver of A, what was the original distance between them?
Answer:
Initially,

Relative Acceleration a = 1 ms^-2
v₀ =72 km h^-1 = 72 × \(\frac{5}{18}\) m s^-1
= 20 ms^-1
Relative displacement = x
Relative velocity initially = V₀ – V₀ = 0
X = 1/2 at²
= 1/2 × 1 × 50² = 1250 m
Original distance between the trains = 1250 m.

Question 8.
On a two-lane road, car A Is travelling with a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

V_A= 36 km h^-1 = 10 m s^-1
VB = Vc = 54 km h^-1 = 15 m s^-1
Velocity Of B with respect to A
= V_BA = V_B – V_A
= 15m -10 = 5 m s^-1
Velocity of C with respect to A .
V_CA = V_C – V_A
= 15 – ( -10) = 25 m s^-1
S = V_CA t
1000 m = 25 × t
⇒ t = 40 s
C will take 40 s to cover AC. In that time, B must cover AB to overtake and avoid collisions.
s = V_BA t + 1/2 at²
1000 = 5 × 40 + 1/2 × a × 40²
⇒ a = 1 ms^-2
Acceleration of B = 1 ms^-2

Question 9.
Two towns A and B are connected by a regular bus service with a bus leaving In either direction every T minutes. A man cycling with a speed of 20 km h^-1 In the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period Tof the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let the speed of the bus be v km h^-1
Distance covered by one bus before the next one leaves the same town is vT For buses going in the same direction as the cyclist, frequency of the bus is 18 min i.e.,
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 18 min
\(\frac{v T}{v-20}\) = 18……………………… (1)
Similarly, for buses going in the opposite direction, frequency of the bus = 6 min
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 6 min
\(\frac{v T}{v+20}\) = 6 ……………………… (2)
\((1) \div(2)\)
⇒ \(\frac{v+20}{v-20}\)
Substitute for v in (1)
\(\frac{40 \times T}{40-20}\) =18 ⇒ T = 9 min.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.

1. What is the direction of acceleration during the upward motion of the ball?
2. What are the velocity and acceleration of the ball at the highest point of its motion?
3. Choose the X = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of X-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
4. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s^-2 and neglect air resistance).

Answer:
1. The direction of acceleration is vertically downwards (towards the Earth)
2. At the highest point, velocity = 0 m s^-1 acceleration = g = 9.8 m s^-2 vertically downwards.
3. During upward motion:

• position – positive
• velocity – negative
• acceleration – positive

During downward motion:

• position – positive
• velocity – positive
• acceleration – positive

4. v = 0 at the highest point
v₀ = 29.4 m s^-1
g = -9.8 m s^-2
v² = v₀² + 2 g h
0 = 29.4² = 2 × (-9.8) × 4
⇒ h = 44.1m
Height of ascension = 44.1 m
v = v₀ + g t
0 = 29.4 – 9.8 t
t = 3 s
Time of ascension = Time of descension
∴ Time for the ball to return = 6 s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity,
3. with constant speed must have zero acceleration,
4. with positive value of acceleration must be speeding up.

Answer:
1. True:
Consider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero       ( 9.8ms^-2 vertically downwards). Acceleration does not depend on instantaneous speed.

2. False:
Since the magnitude of velocity is speed, a body with zero speed must have zero velocity.

3. True:
In the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.

4. False:
This depends on the chosen positive direction. The statement is true when the direction of motion and acceleration are along the chosen positive direction.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Velocity of the ball when it hits the ground = v
v₀ = 0, h = 90 m, g = 9.8m s^-2
v² = v₀² + 2 g h
v² = 0² + 2 × 9.8 × 90
⇒ v = 42 ms^-1
Time for descent = t
v = v₀ + g t
42 = 0 + 9.8 t
t = 4.3 s
During the rebound, initial speed of the ball = 9/10 of v
= 9/10 × 42
= 37.8 ms^-1
Total time for ascent and descent = t₁
Net displacement = 0
0 = 37.8 t – 1/2 gt²
37.8 t = 1/2 × 9.8 × t² ⇒ t = 7.7 s
Time at which maximum height is achieved = 4.3 + 1/2 7.7
= 8.15 s

Question 13.
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show In both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Answer:
(a) Consider a person who leaves home for work in the morning and returns home in the evening.
The magnitude of the displacement of the person during the interval is zero but the total length of path covered = 2 × distance between the home and the workplace.
Magnitude of displacement is the magnitude of the shortest length between the two points. The second quantity is clearly always greater than or equal to the first quantity. The equality holds if the body has not moved at all (path length = magnitude of displacement = 0)

(b) Average velocity = \(\frac{\Delta x}{\Delta t}\). If Δ X is zero (like in the example explained in (a)) then average velocity = 0.
Average speed = \(\frac{\text { Total path length }}{\text { Time taken }}\)
2 × \(\frac{\text { Distance between the home and the work place }}{\text { Time taken to traverse this distance }}\)
= Clearly average speed ≥ magnitude of average velocity. Equality exists when body does not undergo any motion.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the

1. magnitude of average velocity, and
2. average speed of the man over the interval of time

(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 tp 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:
v₁ = 5 km h^-1, v₂ = 7.5 km h^-1, d = 2.5 km
Time for the man to reach the market
= t₁ = d/v₁= 2.5/5 = 0.5 h = 30 min
Time for the man to return from the market
= t₂= d/v₂ = 2.5/7.5 = 1/3 h= 20 min

1. To find the average velocity,
Average velocity = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min,
  Displacement = 2.5 km
  time = 30 minutes
  ∴ Average velocity = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• interval between 0 and 50 minutes,
  Displacement = 0
  ∴ Average velocity = 0
• During between 0 to 40 minutes,
  Displacement = 2.5 km – 7.5 kmh^-1 × 10 min = 1.25 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{1.25 \mathrm{km}}{40 \mathrm{min}}\) = 1.875 km h^-1
  

2. To find the average speed:
Average speed = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min
  Path length = 2.5 km
  time = 30 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• During the interval between 0 and 50 minutes,
  Path length = 2.5 km + 2.5 km
  Time taken = 50 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}+2.5 \mathrm{km}}{50 \mathrm{min}}\) = 6 km h^-1
• During 0 to 40 minutes
  Path length = 5 kmh^-1 × 30 min + 7.5 kmh^-1 × 10 min
  2.5 km + 1.25 km = 3.75 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{3.75 \mathrm{km}}{40 \mathrm{min}}\) = 5.625 km h^-1

Question 15.
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.

Question 16.
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:
(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.

(d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent one-dimensional motion of a particle.

Question 17.
Figure 3.21 shows the X- t plot of onedimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, it is wrong to make such statements about the trajectory of the particle because an x-t graph does not show the particle’s trajectory.
Context:
Consider a body dropped from the top of a tower at time t = 0. If the vertically downward direction is chosen as the positive direction, then the body’s x -t graph would resemble the one given in the question.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away In the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m s^-1, with what speed does the bullet hit the thief’s car?
Answer:
Speed of the police van = v_p
= 30 km h^-1 = 25/3 ms^-1.
Speed of the thief’s car is
v_t = 192 km h^-1 = 160/3 ms^-1
Speed of the bullet = v_b = 150 ms^-1
Speed of the bullet with respect to the police car = v_b + v_p = 150 + 25/3
= 475/3 msv^-1
Speed with which the bullet hits the thief’s car = Speed of the bullet with respect to the thief’s car
= 475/3 – v_t = 475/3 – 160/3 =105 ms^-1

Question 19.
Suggest a suitable physical situation for each of the following graphs (Fig 3.22):


Answer:
(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.

(b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented the given v -t graph in this context.

(c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a – t graph for the ball may be similar to the one given above.

Question 20.
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.

Answer:
Let the maximum amplitude of the sine wave be A, where A is positive. We can see that the particle obeys the
equation x = – A sin \(\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right)\)
Where T = 2 s = period of the sine wave
∴ position = x = – A sin (π t)
velocity – v = \(\frac{dx}{dt}\) = – A_π cos (π t)
acceleration = a = \(\frac{dv}{dt}\) = A_π² sin(π t)
At t = 0.3 s
x = – A sin (0.3_π) = negative
v = – A cos (0.3_π) = negative
a = A _π² sin (0.3_π) = positive
Since sin(0.3_π) > 0 and cos(0.3_π) > 0
At t = 1.2 s
x = – A sin (1.2_π) = – A sin (1.2_π) = positive
v = – A _π cos (1.2_π) = – A cos (1.2_π) = positive
a = A_π² sin (1.2_π) = negative
since sin(1.2_π)<0 and cos(1.2_π) < 0
At t = – 1.2 s
x = – A sin (- 1.2_π) = A sin (1.2_π)= negative
v = – A it cos(- 1.2_π) = – A_π cos (1.2_π) = positive.
a = A_π² sin (- 1.2_π) = – A_π 2sin(1.2_π) = positive
Since sin(-θ) = – sinθt cos(-θ) = cosθ sin(1.2_π) <0 and cos(1.2_π) < 0

Question 21.
Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each Interval.

Answer:
The magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
v > 0 for intervals 1 and 2
v < 0 for intervals 3.

Question 22.
Figure 3.25 gives a speed-time graph of a particle In motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and an in the three Intervals. What are the accelerations at the points A, B, C and D?

Answer:
The magnitude of the slope of the speed-time graph is greatest for interval 2. Hence average acceleration is greatest in magnitude in interval 2. The value of speeds in interval 3 is the highest. Hence the average speed in interval 3 is highest. Sign of a = sign of the slope of the speed-time graph, a > 0 for 1 and 3 a < 0 for 2. At points, A, B, C and D, the tangent to the curve will be parallel to the time axis. Hence slope at the points is zero,
∴ a = 0 at points A, B, C and D.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:
Initial velocity u = 0.
Acceleration a = 1 ms^-2.
Acceleration time = 10 s
Let s_n be the total displacement (or distance in this case) in time ‘n’ seconds.
s_n = ut + 1/2 at²
s_n = 0(n) + 1/2 a(n²) = 1/2 an² for n ≤ 10
Distance covered in ‘n’^th second d_n = s_n – s_n – 1
= 1/2 an² – 1/2 a(n-1)²
= an – 1/2 a (for n ≤ 10)
= (n – 1/2) m
At the end of 10 seconds, velocity acquired v = u + at = 0 + a (10)
10 a = 10 ms^-1
Distance covered in the ‘n’^th second = 10m for n >10
∴ d_n = {(n-1/2)m n ≤ 10,
10m n > 10
The plot is a straight line inclined to the time axis for uniformly accelerated motion. (It is a straight line parallel to the time axis for uniform motion).

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^-1. How much time does the bail take to return to his hands? if the lift starts moving up with a uniform speed of 5 m s^-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:
Initial velocity n = 49 ms^-1
Acceleration g = -9.8 ms²
Displacement s = 0
s = ut + 1/2 at²
0 = 49 t + 1/2 (-9.8) t²
⇒ t = 0 or t = 10 s
t = 0 represents initial time. The required ‘t’ here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms^-1. The relative displacement is also 0. Hence the time required in this case is also 10 s.

Question 25.
On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

1. speed of the child running in the direction of motion of the belt?.
2. speed of the child running opposite to the direction of motion of the belt?
3. time taken by the child in (1) and (2)?

Which of the answers alter If motion is viewed by one of the parents?

Answer:
1. Speed of the child running in the direction of motion of the belt = Speed of the child with respect to the belt + Speed of the belt with respect to the ground
= 9 + 4 = 13 km h^-1.
2. Speed of the child running opposite to the direction of motion of the belt = speed of the child with respect to the belt speed of the belt with respect to the ground
= 9 – 4 = 5 km h^-1.
3. Time taken by the child in both cases is the same because the speed of the child with respect to the belt does not change.
Speed v = 9 km h^-1 = 2.5 ms^-1
Distance d = 50 m
Time t = \(\frac{\text { Distance }}{\text { Speed }}\) = \(\frac{50}{2.5}\) = 20 s
The answer to (a) and (b) when the motion is viewed by one of the parents is 9 km h^-1. This is because the parents are also on the belt and are moving with respect to the ground. They only see the motion of the child with respect to the belt. In (c), the answer remains unaltered because the speed of the child with respect to the belt does not depend on the speed of the parents.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with Initial speeds of 15 m s^-1 and 30 m s^-1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take
g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:
Initial velocity of stone 1 is u₁ = 15 ms^-1
Initial velocity of stone 2 = u₂ = 30 ms^-1
Acceleration = g = -10ms^-2 (vertically upwards direction is chosen as positive)
x₁ = 200 + u₁t + 1/2 at²
= 200 + 15 t – 5 t²
x₂ = 200 + u₂t + 1/2 at²
= 200 + 30 t – 5 t².
Also x₁ = 0 for t = 8 s and
x₂ = 0 for t = 10s.
∴ For 0 ≤ t ≤ 8 s
x₂ – x₁ = 15 t
At t = 8 s,
x₂ – x₁ = 120 m
For 8 < t; t ≤ 10 s
x₂ – x₁ = 200 + 30 t – 5 t² – 0
= 200 + 30 t – 5 t².
For t > 10 s,
x₂ – x₁ = 0
∴ The given graph is correct and because it matches with the equations obtained.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown In Fig. 3.28. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s.
(b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?.
Answer:
(a) Distance covered = Area under the speed – time graph
= 1/2 × 10 × 12 = 60 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{60 m}{10 s}\) = 6 ms^-1

(b) For 0 ≤ t ≤ 5 s,
Speed v = \(\frac{12}{5}\) t ms^-1
For 5 < t ≤ 10 s,
Speed v = 12 – \(\frac{12}{5}\) (t – 5) ms^-1
∴ Speed at t = 2 s is 4.8 ms^-1 and speed at t = 6 s is 9.6 ms^-1
Distance covered = Area of trapezium ABEF + Area of trapezium BCDE

= 1/2 (4.8 + 12) × 3 + 1/2 (12 + 9.6) × 1
= 36 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{36}{6-2}\) = 9 ms^-1

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂

1. x(t₂) = x(t₁ + v (t₁) (t₂ – t₁ + (1/2) a (t₂– t₁)²
2. v(t₂) = v(t₁) + a(t₂ – t₁)
3. v_average = (x(t₂) – x(t₁)/(t₂-t₁)
4. average = (v(t₂ – v(t₁))/(t₂ – t₁)
5. x(t₂) = x(t₁) + v_average (t₂ – t₁) + (1/2) a_average (t₂ – t₁)²
6. x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:

1. Wrong, because it is not known whether the acceleration ‘a’ is constant.
2. Wrong, because it is not known whether the acceleration ‘a’ is constant.
3. Correct. By definition.
4. Correct. By definition
5. Wrong. The formula should contain v₁ (t) instead of v_average.
6. Correct By definition.

1st PUC Physics Motion in a Straight Line One Mark Questions and Answers

Question 1.
When do you say that a body is in motion?
Answer:
A body is said to be in motion when it changes its position with respect to time and surrounding.

Question 2.
What is a particle?
Answer:
A particle is a geometrical mass point.

Question 3.
Define displacement.
Answer:
The change of position of a body in a particular direction is called displacement.

Question 4.
Can the displacement of a moving body be zero?
Answer:
Yes

Question 5.
Define speed.
Answer:
The rate of change of position of a body in any direction is known as speed.

Question 6.
Define velocity.
Answer:
The rate of change of position of a body in a particular direction is called velocity or Rate of displacement of a body is called velocity.

Question 7.
Define uniform velocity.
Answer:
If a body covers equal distances in equal intervals of time in a given direction, however small the intervals maybe then its velocity is said to be uniform.

Question 8.
Define variable velocity.
Answer:
If the velocity of a body either changes in magnitude or in direction or both, then its velocity is said to be variable.

Question 9.
Define average velocity.
Answer:
The average velocity of a body is defined as the ratio of its total displacement to the total time.

Question 10.
Define acceleration.
Answer:
Rate of change of velocity of a body is called acceleration.

Question 11.
Define uniform acceleration.
Answer:
If the velocity of a body changes by an equal amount in equal interval of time, however, small the interval maybe then its acceleration is said to be uniform.

Question 12.
What is a position-time graph?
Answer:
A graph drawn by taking time along the x-axis and displacement along the y-axis is called position-time graph.

Question 13.
What does the slope of the position-time graph indicate?
Answer:
The slope of the position-time graph indicates velocity.

Question 14.
Draw the position-time graph of a particle which is at rest.
Answer:

Question 15.
Draw the position-time graph of an object starting from rest and moving with uniform velocity.
Answer:

Question 16.
What is a velocity-time graph?
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph.

Question 17.
What does the slope of the velocity-time graph indicate?
Answer:
The slope of the velocity-time graph indicates the acceleration.

Question 18.
What does the area under the velocity-time graph indicate?
Answer:
Area under velocity-time graph indicates the displacement.

Question 19.
An object is moving with uniform velocity. What Is its acceleration?
Answer:
Zero

Question 20.
How do you determine the instantaneous velocity of a particle from the position-time graph?
Answer:
Instantaneous velocity at any point is given by the slope of the tangent drawn to the position-time graph at that point.

Question 21.
Draw the velocity-time graph of a particle moving with

1. uniform velocity
2. variable velocity.

Answer:
1)

2)

Question 22.
Draw the velocity-time graph of an object starting from rest and moving with uniform acceleration.
Answer:

Question 23.
Draw the velocity-time graph of an object moving with constant retardation.
Answer:

Question 24.
Draw the position time of particle which is initially at rest and moving with uniform acceleration.
Answer:

Question 25.
What is the acceleration time graph?
Answer:
A graph drawn by taking time along x-axis and acceleration along the y-axis is called velocity-time graph.

Question 26.
Write the expression for the distance travelled by a body during the nth second of its motion.
Answer:
Distance travelled by a body during the n^th second of motion is given by
s_n = u + a(n – 1/2)
where u is the initial velocity and a is the uniform acceleration.

Question 27.
Write the expression for the acceleration in terms of distance travelled in two consecutive intervals of time.
Answer:
Acceleration is given by,
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
where s₁ and s₂ are the distances travelled in two consecutive intervals of time ‘t’ seconds each.

Question 28.
What is relative velocity?
Answer:
The velocity of a particle in motion relative to another particle is called relative velocity.

Question 29.
When is the relative velocity of two bodies maximum?
Answer:
Relative velocity of two bodies is maximum when they are moving in opposite directions.

Question 30.
What does the area of a-t graph indicate?
Answer:
Area of a-t graph represents the change in velocity of the body in a given time interval.

Question 31.
Draw the a-t graph of a body moving with uniform acceleration.
Answer:

Question 32.
Why is it not necessary for a body following another, to stop, to avoid collision?
Answer:
If the relative velocity is zero, no collision will occur.

Question 33.
If in case df a motion, displacement is directly proportional to the square of the tune elapsed, what do you think about its acceleration, le, constant or variable? Explain why?
Answer:
x ∝ t²
⇒ x = kt² Where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{d}{d t}\) (2 k t)
= 2k = constant
∴ Acceleration is constant.

Question 34.
Why does the earth Impact the same acceleration to all bodies?
Answer:
Since acceleration is the force on unit mass, the acceleration due to gravity is constant.

Question 35.
What will be the nature of the velocity-time graph for a uniform motion?
Answer:
The v -t graph will be a line parallel to the time axis.

Question 36.
The position coordinate of a moving particle is given by x = 6 + 18t + 9t2(x is in metres and t is in seconds). What is its velocity at t s 2s?
Answer:
x = 6 + 18 t + 9t²
v= \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = 18 + 18 t
v at t = 2 s = 18 + 18 (2)
= 54 ms^-1

Question 37.
A ball is thrown straight up. What is its velocity and acceleration at the top?
Answer:
At the top, v = 0, a = g.

1st PUC Physics Motion in a Straight Line Two Mark Questions and Answers

Question 1.
Distinguish between speed and velocity.
Answer:

1. Speed is the rate of change of position of a body in any direction while velocity is the rate of change of position of a body in a particular direction.
2. Speed is a scalar quantity whereas velocity is a vector quantity.

Question 2.
What is a velocity-time graph? What Is Its importance?
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. The velocity-time graph can be used

1. To find the nature of the motion of a body.
2. To determine the velocity of a body at any instant.
3. To calculate the displacement of a body in a given time.

Question 3.
If V₁ and v₂ are the velocities of two bodies then what is their relative velocity when they move in the

1. same and
2. opposite direction?

Answer:

1. When the bodies move in the same direction, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ – v₂
2. When the bodies move in the opposite directions, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ + v₂.

Question 4.
A car moving with a uniform velocity 54kmph is brought to rest in travelling a distance of 5m What is the retardation produced by brake?
Answer:
Initial velocity of the car.,
u = 54km/hr
= \(\frac{54 \times 1000}{3600}\) = 15 ms^-1
Final velocity of the car, v =0.
Let a be the retardation of the car. Then using the relation.
v² = u² + 2as
0² = 15² + 2a × 5
a = \( -\frac{15 \times 15}{2 \times 5}\) = \(\frac{225}{10}\) = – 22.5 ms^-2.

Question 5.
An automobile moving with a uniform velocity of 500ms^-1 is brought to rest in travelling a distance of 5m. What is the acceleration produced by the brakes?
Answer:
Initial velocity of the car, u = 500ms^-1
Final velocity of the car v = 0
Let a be the acceleration of the car.
Then using the relation.
v² = u² + 2aS
0² = 500² + 2a × 5
a = \( -\frac{500^{2}}{10}\) = – 25 × 10³mS^-2.

Question 6.
The equation of motion of a body is given by S = 2t + 2t², where S is in metre and time in second. What is the acceleration of the body?
Answer:
S = 2t + 2t². This is of the standard form S = ut + 1/2 at²
By comparison, 1/2 a = 2 or a = 4m/s².

Question 7.
Draw a velocity-time graph for a particle in the following situations:

1. Starts from rest and moving with uniform acceleration
2. Moving with uniform retardation.

Answer:
1. Body moving the uniform acceleration.

2. Body moving with uniform retardation.

Question 8.
Define velocity and acceleration.
Answer:
Velocity is defined as the rate of displacement. Acceleration is known as the rate of change of velocity.

Question 9.
Two parallel nail tracks run North-South. Train A moves due north with a speed of 54 km h^-1 and train B moves due south with a speed of 90 km h^-1. What is the relative speed of B with respect to A in ms^-1?
Answer:
V_A = 54 km h^-1
V_B = – 90 km h^-1
(north direction is chosen as positive)
V_BA = V_B – V_A
= – 90 – 54 = -144 km h^-1
= – 40 ms^-1

Question 10.
Differentiate between average speed and instantaneous speed of an object.
Answer:
The distance covered in unit time is called average speed.
V_avg = \(\frac{x\left(t_{2}\right)-x\left(t_{1}\right)}{t_{2}-t_{1}}\)
The speed at any instant of time is called instantaneous speed.
V_inst= \(\Delta \mathrm{t} \rightarrow 0 \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\)

Question 11.
Velocity time graph of a moving object is shown below. What is the acceleration of the object? Also, draw displacement time graph for the motion of the object.

Answer:
Acceleration = Zero (because velocity is constant) Displacement time graph is shown below

Question 12.
Two straight lines are drawn on the same displacement time graph make angles 30° and 60° with the time axis respectively in the figure. Which line represents greater velocity? What is the ratio of the velocity of line A to that of line B?

Answer:
Velocity = Slope of x -t graph
V_A = tan 30° = \(1_{\sqrt{3}}\)
V_B = tan 60° = Clearly, V_B > V_A.
\(\frac{V_{A}}{V_{B}}\) = \(\frac{1_{\sqrt{3}}}{\sqrt{3}}\) = \(\frac{1}{3}\)

1st PUC Physics Motion in a Straight Line Three Mark Questions and Answers

Question 1.
The distance x travelled by a body In a straight line is directly proportional to t². Decide on the type of motion associated. If x ∝ t³ what change will you observe?
Answer:
x ∝ t² = kt² where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (2 k t )
= 2 k
∴ Acceleration is constant / uniform.
If x ∝ t³ where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (3 k t²)
= 6 k t
∴ Acceleration is non-uniform

Question 2.
Draw the following graph for an object under free-fall:

1. Variation of acceleration with respect to time.
2. Variation of velocity with respect to time
3. Variation of distance with respect to time.

Answer:
1.

2.

Line 2 appears only when the body bounces.
3.

Question 3.
A body starts from rest accelerates uniformly along a straight line at the rate of 10 ms^-2 for 5 seconds. It moves for 2 seconds with a uniform velocity of 50 ms^-1. Then it retards uniformly and comes to rest in 3s. Draw the velocity-time graph of the body and find the total distance travelled by the body.
Answer:

Total distance travelled = Area under the v – t graph
= 1/2 × 50 × (10 + 2)
= 300 m

Question 4.
The displacement (in metre) of a particle moving along x-axis given by x = 18 t + 5t². Calculate:-

1. the instantaneous velocity at t = 2s
2. average velocity between ta 2 s and 3 s
3. instantaneous acceleration

Answer:
x = 18 t + 5 t²
v(t) = \(\frac{d x}{d t}\) = 18 + 10 t,
a(t) = \(\frac{d v}{d t}\) = 10 ms^-2
1. v (2) = 18 + 10 (2) = 38 ms^-1
2. v (2) = 38 ms^-1

v(3) = 18 + 10 (3) = 48 ms^-1
Since acceleration is constant,
V_avg \(\frac{v(3)+v(2)}{2}\) = \(\frac{48+38}{2}\)
= 43 ms^-1.

3. Instantaneous acceleration a = 10 ms^-2.

1st PUC Physics Motion in a Straight Line FourFive Mark Questions and Answers

Question 1.
Derive an expression for velocity of the particle after time ‘t’ or Derive v = u + at using v-t graph.
Answer:
Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its initial velocity and v be its velocity after a time t. The velocity-time graph AB of this particle is shown in the figure.

Acceleration of the particle is given by the slope of the v-t graph.
From the graph,
slope of the line AB = \(\frac{B C}{A C}=\frac{B C}{O D}=\frac{B D-C D}{O D}\)
But, BD = v, the final velocity,
OA = u, the initial velocity,
OD = t, the time.
∴ Slope of the graph AB is,
a = \(\frac{B D-C D}{O D}\)
= \(\frac{v-u}{t}\)
Thus a = \(\frac{v-u}{t}\) or v = u + at.

Question 2.
What is the v-t graph? Derive an expression for distance covered by the particle in time ‘t’ or Derive the equation s = ut + 1/2 at² using v-t graph.
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 3.
Derive an expression for the velocity of the particle after covering distance ‘s’ OR Derive the equation v² = u² + 2as using the velocity-time graph.
Answer:
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or
2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1),
2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 4.
Derive an expression for distance travelled during the n^th second of motion.
OR
Derive s_n= u + a(n – 1/2)
Answer:
Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.
Distance travelled during n^th second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n-1) + 1/2 a(n-1)²]
= un + 1/2 an² – [ un-u + 1/2 a(n²-2n +1)]
= un + 1/2 an² – [ un-u + 1/2 an²-an + 1/2 a]
= un + 1/2 an² – un+u – 1/2 an²+an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)

Question 5.
Derive an expression for acceleration in terms of distance travelled in two successive equal interval of time.
OR
Derive the equation a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
Answer:
Consider a particle moving with uniform acceleration a. Let it start from a point with initial velocity ‘u’ and covers the distances s₁ and s₂ in two successive equal interval of time ‘t’ each. Distance covered in the first interval
s₁ = ut + 1/2 at² ………. (1)
Distance covered at the end of time 2t,
s₁ + s₂ = u(2t) + 1/2 a (2t)²
s₁ + s₂ = 2ut + 2at²……………. (2)
(s₁ + s₂ ) – 2s₁ = (2ut + 2at²) – (2ut + at²)
s₂ – s₁ = at²
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)

Question 6.
Define uniform velocity obtain v² – u² = 2as from the v – t graph, where the symbols have the usual meaning.
Answer:
If a body undergoes equal displacements in equal intervals of time, however small interval of time may be then velocity is said to be uniform velocity.
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t= 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelles by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or 2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1), 2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 7.
A particle is moving with uniform acceleration covers a distance of 45 m in 6^th second and 75 m in 12^th second during its motion. Calculate the displacement of the particle after 20s?
Answer:
S₆ = 45m, S₁₂=75m
using S_n = u + a (n – 1/2 )
i.e., S₆  = u + a or 45 = u + 5.5a ……… (1)
S₁₂= u + a(12 – 1/2) or 75 = u + a(11.5) …….. (2)
Eq(2) – (1) gives 6 a = 30
Or a = 30/6 = 5m/s²
From (1), 45 = u + 5 × 5.5
u = 17.5m/s.
Distance travelled in t = 20s,
S = ut+ 1/2at2
= 17.5 × 20 + 1/2 × 5 × 20²
=1350m.

Question 8.
Draw a velocity-time graph of uniformly accelerated motion in one dimension. From the velocity-time graph of uniformly accelerated motion deduce the equations of motion in distance and time.
Answer:
A graph drawn by taking time along x-axis and velocity along y-axis is called velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 9.
Derive an equation for the distance covered by a uniformly accelerated body in the n^th second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.
Answer:
Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.
Distance travelled during nth second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n – 1) + 1/2 a(n – 1)²]
= un + 1/2 an² – [ un – u + 1/2 a(n² – 2n + 1)]
= un + 1/2 an² – [ un – u + 1/2 an² – an + 1/2 a]
= un + 1/2 an² – un + u – 1/2 an² + an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)
s_n = 4 + a (n – 1/2)
Given u = 0, a = g
s = 1/2 g t² = 1/2 g n²
By the given condition 1/2 s = s_n
⇒ \(\frac{\mathrm{g} n^{2}}{4}\) = g(n – 1/2)
⇒ n² – 4 n + 2 = 0
⇒ n = \(\frac{4 \pm \sqrt{16-8}}{2}\)
⇒ n = 2 ± \(\sqrt{2}\)
∴ Time of the body’s fall = (2 ± \(\sqrt{2}\)) s

1st PUC Physics Motion in a Straight Line Numerical Problems Questions and Answers

Question 1.
A man runs from his home to office at a speed of 2 ms^-1 on a straight road and returns back to home at a speed of 4 ms^-1. Find

1. Average speed and
2. average velocity.

Solution:
1. To find the average speed. Let ‘s’ is the distance between the home and the office.
∴ Time taken to reach the office
t₁ \(\frac{\text { distance }}{\text { velocity }}\) = \(\frac{s}{2}\)
Similarly, time taken to walk back to the home from the office is t₂ = \(\frac{s}{4}\)
∴ Total time taken = t₁ + ₂ = \(\frac{s}{2}\) + \(\frac{s}{4}\) = \(\frac{3 s}{4}\)
Total distance moved = s + s = 2s
Hence. the average speed \(=\frac{\text { toal distance moved }}{\text { total time taken }}\)
= 2.67 ms^-1
2. To find the average velocity Since the man comes back to home, the final and initial position is same.
∴ Total displacement is zero.
Hence average velocity
\(=\frac{\text { toal displacement }}{\text { total time }}\) = 0

Question 2.
A car moving along a straight high-way with a speed of 35 m/s is brought to stop within a distance of 200 m. What Is the retardation, and how long does it take for the car to stop? Solution:
Initial speed of the car, u = 35 ms^-1
Final speed of the car, v = 0
Let a be the retardation of the car.
Then using the relation,
v² = u² + 2as, we have 0 = (35)² + 2a × 200
a = – \(\frac{35 \times 35}{2 \times 200}\) = – 3.06 ms^-2
Let t be the time taken by the car to come to stop. Then,
v = u + at ;
0 = 35 – 3.06 × t or
t = \(\frac{35}{3.06}\) = 11.43 s.

Question 3.
A car starts from rest and accelerates from rest uniformly for 10 s to a velocity of 36 kmhr^-1. It then runs at a constant velocity and is finally brought to rest in 50 m with uniform retardation. If the total distance covered by the car is 500 m, find

1. acceleration and
2. retardation.

Solution:
1. Initial velocity of the car u = 0;
Final velocity v = 36 kmhr^-1
= \(\frac{36 \times 1000}{3600}\) ms^-1
= 10 ms^-1
time taken t=10 s
∴ acceleration a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{\mathrm{10}-\mathrm{0}}{\mathrm{10}}\) 1ms^-2

2. To find the retardation
During the motion with retardation, Initial velocity of the car u = 10 ms^-1
Final velocity v = 0, distance travelled, s = 50 m
Using the relation v² = u² + 2as,
0 = 10² – 2a × 50
a = \(-\frac{100}{2 \times 50}\) = -1 ms^-2

Question 4.
A body travelling with a uniform acceleration travels a distance of 100m in the first 5 second and 200m in the next 5 second, calculate the initial velocity and acceleration of the body.
Solution:
If s₁ and s₂ are the distances travelled by a body in two successive intervals of t seconds, then the acceleration is given
by, a = \(\frac{s_{2}-s_{1}}{t^{2}}=\frac{200-100}{5^{2}}=\frac{100}{25}\) = 4ms^-2
we have, s = ut + 1/2 at²
Here, s = s₁ = 100m, a = 4ms^-2, t = 5 s
100 = u × 5 + 1/2 × 4 × 25 OR
5u = 50 ∴u= 10ms^-1.

Question 5.
A body moving with uniform acceleration along a straight line covers 60m in the 5^th second and 80m in the 8^th second of its motion. Calculate the initial velocity and uniform acceleration.
Solution:
The distance travelled during the n^th second is given by,
s_n = u + a/2(2n – 1), n = 5 and s_n = 60
∴ 60 = u+ a/2 (2 × 5 – 1)
60 = u + 9a/2 ………… (1)
Similarly, for n = 8, s_n =80
∴ 80 = u+ a/2 (2 × 8 – 1)
80 = u + 15a/2 …………….. (2)
(2) – (1), gives
20 = 6a/2 = 3a
∴ a = 20/3 ms^-2
Using this in equation (1), we get
60 = u + \(\frac{9 \times 20}{2 \times 3}\)
u = 60 – 30 = 30 ms^-1.

Question 6.
A stone is thrown vertically upwards with an initial velocity of 19.6 m^-1. After how long will the stone strike the ground? Take g = 9.8 ms^-2.
Solution:
Initial velocity u = +19.6 ms^-1. The stone after reaching the highest point comes back to the initial point
∴ The displacement = 0.
a = g = – 9.8 ms^-2
Let t be the time taken to reach the ground
From s = ut + 1/2at², we have
0=19.6t – 1/2 × 9.8 × t²
4.9t² – 19.6t = 0
Dividing throughout by 4.9t, we have t = 4s. Hence, the stone reaches the ground after 4s.

Question 7.
A stone is thrown vertically upwards with a velocity of 10ms^-1 from the top of a tower 40m tall and it finally falls to the ground,

1. Find the time taken by the stone to reach the ground
2. After how long will it pass through the point of projection
3. Calculate the velocity when it strikes the ground. Take g = 10 ms^-2

Solution:
Let the stone be thrown upward with a velocity u = 10ms^-1 from the tower AB of height 40 m. Let C be the highest point reached.

1. Let t be the total time taken by the stone to move from B to C and back to the ground.
∴ The displacement = BA = – 40 m (directed downward since displacement is measured from the initial to the final position)
a = -10 ms^-2.
From s = ut + 1/2at², we have
– 40= 10 × t – \(\frac{10 t^{2}}{2}\)
– 40 = 10t – 5t²
– 8 = 2t -t²
t² – 2t – 8 = 0
(t – 4) (t + 2) = 0.
OR
t = + 4s or – 2s
As time cannot be negative total time taken to reach the ground = 4s.

2. For the motion from B to C and back to B,
displacement = 0
u = 10ms^-1, a = g = -10ms^-2
Let t₁ be the time taken.
From s = ut + 1/2at², we have
0= 10t₁ – \(\frac{10 t_{1}^{2}}{2}\) or 10₁² – 5t₁² = 0
∴ t₁ = 2s.
Hence, the stone reaches the point of projection after 2 seconds.

3. Let ‘v’ be the velocity of the stone when it strikes the ground. For the motion from B to C and back to A, we have,
u = 10 ms^-1, a = g = -10 ms^-2, t = 4s
From, v = u + at, we have
v = 10 – 10 × 4 = – 30 ms^-1.
The -ve sign indicates that it is directed downward.

Question 8.
A stone is projected vertically upwards from the ground with a velocity of 49 ms^-1. At the same time, another stone is dropped from a height 98m to fall freely along the same path as the first. Find where and when the two stones meet each other.
Solution:

Let P be the stone dropped from B at a height 98m from the ground A. Q is the other stone thrown up vertically upwards with a velocity 49ms^-1 at the same instant. Let the two stones meet at C after a time t seconds.
For the stone P, we have
u = 0, a = + g, s = + h (=BC), t = t
From s = ut + 1/2 at², we have,
+h = 0 × t + 1/2 gt² or h = 1/2 gt² ………. (1)
For the stone Q,
u = 49ms^-1,
a = -g,
s = +AC = +(98 – h),
(98 – h) = 49t – 1/2 gt²
Using eqn. (1) in (2), we have,
(98 – h) = 49t – h or 49t =98
∴ t = 2s.
Using this in (1), we have
h = BC = 1/2 × 9.8 × 4 = 19.6 m
∴ AC =96 – 19.6 = 76.4 m
∴ Two stones meet at a height of 76.4 m from the ground, after 2 seconds.

Question 9.
A train is moving southwards with a speed of 30ms^-1. A man is running on the roof of the train with speed of 5ms^-1 with respect to the train. Find the velocity of the man as observed by the person on the ground if the man is running

1. southwards
2. northwards.

Solution:
If v_A and v_B are the velocity of the two bodies moving along the same direction with respect to the ground, relative velocity of A with respect to B is given by,
V_AB = V_A – V_B …………. (1)
Here, the velocity of the man with respect to the train,
v_AB = 5ms^-1.
velocity of the train with respect to the ground is,
v_B = 30 ms^-1
∴ velocity of the man with respect to the ground,
v_A = v_B + v_AB = 30 + 5 = 35 ms^-1 if the man moves southwards.
On the other hand, if the man moves northwards,
v_AB = – 5 ms^-1
∴ V_A = V_B – V_AB
= 25 ms^-1.

Question 10.
A police van moving on a highway with a speed of 10ms^-1 fires a bullet at a smuggler’s car speeding away in the same direction with a speed of 30ms^-1. If the muzzle velocity of the bullet is 140ms^-1, with what speed the bullet will hit the smuggler’s car?
Solution:
Speed of the police car v_p = 10 ms^-1.
Muzzle velocity of the bullet v_M=140 ms^-1
Net velocity of the bullet,
v_A = v_p + v_M = 10 + 140 = 150 ms^-1
(∵ gun is mounted on the van)
speed of the smuggler’s car v_B = 30 ms^-1
∴ Velocity of the bullet relative to smuggler’s car,
v_AB = V_A – V_B
= 150 – 30 = 120 ms^-1.

Question 11.
Two trains A and B are moving on two parallel tracks with uniform speed of 20ms^-1 in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms^-2. After 50 seconds, they are moving together, find the original distance of separation.
Solution:
Initially, both A and B are moving with the same velocity in the same direction,
∴ The initial relative velocity .
u_AB = u_A – u_B = 0
when t = 50 s,
relative acceleration = a_AB = 1 ms^-2
If s is the distance moved, then
From s = ut + 1/2at²
S = u_AB × t + 1/2 a_ABt²
S = 0 × t + 1/2 × 1 × 50²
= 2500/2 = 1250 m.
Hence, the Initial distance of separation is 1250m.

Question 12.
A body is starting from rest and is subjected to a uniform acceleration of 10ms^-2 determine.

1. Velocity of the body at the end of 5s
2. Displacement of the body In the first 3s
3. velocity of the body after a displacement of 20m.
4. Displacement of the body In the 4^th second of its motion (Tumkur 05)

Solutions:
1. Initial velocity, u = 0
acceleration, a = 10ms^-2
time, t = 5s,
v=?
From, the Equation, v = u + at we have,
v= 0 + 10 × 5
v= 50 ms^-1

2. Initial velocity, u = 0
acceleration a = 10 ms^-2
time, t = 36s
Distance travelled, s =?
From the equation s = ut+ 1/2 at² we have
= 0 + 1/2 × 10 × 9
s = 45 m

3. Initial velocity u= 0,
acceleration a = 10ms^-2
Displacement s = 20m
velocity v=?
From the equation v²= u² + 2as ,
= 0² + 2 × 10 × 20
= 400
v = \(\sqrt{400}\) = 20ms^-1

4. Initial velocity u=0
Acceleration a = 10ms^-1
n = 4 sec
Displacement S₄ = ?
From the Equation, S_n = u+ a/2 (2n-1)
S₄=0 + 10/2 (2 × 4 – 1) = 0 + 5(7)
S₄ = 35m.

Question 13.
A stone dropped from the top of a building travels 24.5 m in the last second of its fall. Find the height of the tower, (g = 9.8 m/s2)
Answer:
Distance travelled in the last second of its fall is S_n = 24.5m
Also, S_n = ut + a (n – 1/2)
But u = 0, a = g
∴ S_n = 0 + g (n – 1/2)

24.5 = 9.8n – 4.9
n =\(\frac{24.5+4.9}{9.8}\) = 3 seconds.
we know that S = ut + 1/2 at²
∴ Height of the tower;
h = 0 × 3+ 1/2 × 9.8(3)²
(∵ s = h, u = 0, a = g = 9.8m/s)
∴ h = 0 + 4.9(3)²
= 44.1m.

Question 14.
A body moving along a straight line with uniform acceleration covers 23m and 35m respectively In the 5^th and 8^th second of its motion calculate the distance travelled by the body in

1. 10^th second
2. complete 10 second and what is its velocity at the end of 6 seconds.

Answer:
S₅ = 23m, S₈ = 35m
W.k.that s_n = u + a(n – 1/2)
i.e., S₅ = u + a(5 – 1/2) or 23 = u + 4.5 a….(1)
S₈ = u + a(8 – 1/2) or 35 = u + a(7.5)……….(2)
Eq(2) – (1) gives 3a = 12
or a = 12/3 = 4 m/s²
From (1), 23 = u + 4.5 × 4
u = 5 m/s
1. Distance traveled in 10^th second
S₁₀=u + a(n – 1/2) = 5 + 4(10 – 1/2) = 43m

2. Distance traveled in 10s is
S= ut + 1/2at² = 5 × 10 + 1/2 × 4 × 10² = 250m
Velocity at the end of 6 seconds is v = u + at = 5 + 4 × 6 =29 m/s.

Question 15.
A point object is thrown vertically upwards at such a speed that it returns the thrown after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed-time graph for the object and use it to find the distance travelled by it in the last second of its journey.
Answer:
Time of rise t₁ = time of fall = t₂
t = 6 s
v = u – g t (upward direction is chosen as positive)
v = – u
– u=u-gt
⇒ – 2u = – 9.8 × 6
⇒ u = 29.4 ms^-1
v² = u² – 2 gh
At the top, v = 0
⇒ h = \(\frac{u^{2}}{2 g}\)
= \(\frac{(29.4)^{2}}{2 \times 9.8}\)
h = 44.1 m
Initial speed = 29.4 ms^-1
Height the object attained = 44.1 m

Distance covered in last second = area under speed – time graph.
= 1/2 × (6 – 5) × (19.6 + 29.4)
= 24.5 m

Question 16.
A race car is moving on a straight road with a speed of 180 km h^-1. If the driver stops the car in 25 s by applying the brakes, calculate the distance covered by the car during the time brakes are applied. Assume acceleration of the car is uniform throughout the retarding motion.
Answer:
Initial speed μ = 180 km h^-1
μ = 50 ms^-1
t = 25 s
v = u – at
v = 0
∴ 0 = 50 – a × 25
⇒ a = 2 ms^-2
v² = u² – 2as
⇒ s = μ²/2 a
\(\frac{50^{2}}{2 \times 2}\)
∴ Stopping distance = 625 m

Question 17.
A body covers 12 m in the 2^nd second and 20 m in the 4^th second. Find what distance the body will cover in the 4 seconds after the 5^th second.
Answer:
s_n= u + a/2 (2n -1)
s₂ = 12
⇒ 12 = u+ a/2 (2 × 2 – 1)
⇒ 12 = u + 3 a/2 ………. (1)
s₄ = 20
⇒ 20 = u+ a/2 (2 × 4 – 1)
⇒ 20 = u + 7a/2 ………. (2)
Subtract equation (1) from (2)
s = 4a/2
⇒ a = 4 ms^-2
12 = u + 3/2 × 4
⇒ u = 6 ms^-1
Distance covered in the 4 seconds after 5^th second,
= S₉ – S₅
= u (9) + 1/2 a (9)² – [u(5) + 1/2 a (5)² ]
= 4u + a/2 (81 – 25)
= 4 × 6 + 4/2 × 56 = 136 m.

Question 18.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a building of height 25 m from the ground.

1. How high will the ball reach?
2. How long will the ball tale to reach the ground ? (g = 10 ms^-2)

Answer:
u = 20 ms^-1
h = 25 m
g = 10ms^-2
1. v² = u² – 2gh₁
At the top, v = 0
⇒ h₁ = \(\frac{\mathrm{u}^{2}}{2 \mathrm{g}}\)
= \(\frac{20^{2}}{2 \times 10}\)
= 20 m
∴ Height the ball reaches = 20 + 25 = 45 m from the ground.
2. – h = u t – 1/2 g t²
– 25 = 20 t – 1/2 × 10 t²
⇒ 5 t² – 20 t – 25 = 0
⇒ t² – 4 t – 5 = 0
⇒ (t – 5 ) (t + 1 ) = 0
= t = 5, -1
∴ t = 5 s (Since t cannot be negative) Total time to reach the ground = 5 s

1st PUC Physics Motion in a Straight Line Hard Questions and Answers

Question 1.
The speed of a train increases at a constant rate a from zero to v and then remains constant for an interval, and finally decreases to zero at a constant rate p. If L be the total distance covered prove that the total time taken
is \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\infty}+\frac{1}{\beta}\right)\)
Answer:
During time t₁, the train accelerates uniformly from 0 to V.
v = α t₁
Distance covered d₁ = 1/2 α t₁²
= 1/2 (∝ t₁) t₁
d₁ = 1/2 v t₁
⇒ \(t_{1}=\frac{2 d_{1}}{v} \alpha\)
During time t₂, the train travels at a uniform speed of v
Distance covered d₂ = vt₂
⇒ t₂ = \(\frac{\mathrm{d}_{2}}{\mathrm{v}}\)
During time t₃, the train decelerates uniformly from v to 0.
0 = v – β t₃
⇒ v = β t₃
Distance covered d₃ = vt₃ – 1/2 β t₃²
= vt₃ – 1/2 (β t₃) t₃
d₃ = vt₃ – 1/2 v t₃
⇒ d₃ = 1/2 v t₃
⇒ t₃ = \(\frac{2 \mathrm{d}_{3}}{\mathrm{v}}\)
Total time T = t₁ + t₂ + t₃

Using v² = u² + 2as
d₁ = \(\frac{v^{2}}{2 \alpha}\) and d₃ = \(\frac{v^{2}}{2 \beta}\)
∴ T = \(\frac{L}{v}+\frac{v}{2 \alpha}+\frac{v}{2 \beta}\)
T = \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
Hence proved.

Question 2.
In a car race, car A takes ‘t’ seconds less than car B and passes the finish line with a velocity ‘V’ more than that of the car B. Of the cars start from rest and travel with constant acceleration a₁ and a₂ respectively, show that
V = \(t \sqrt{a_{1} a_{2}}\).
Answer:
Let the car A take t_A seconds to finish the race.
Car B takes t_B seconds to finish the race. Let the final velocities of the cars be v_A and v_B
By the given conditions,
t_B – t_A = t
v_A = a₁ t_A
v_B = a₂ t_B
v_A – v_B = v
Since the distance d covered is the same,

Now, v_A = a₁ t_A
⇒ v_B + v = a₁ (t_B – t)
⇒ a₂ t_B + v = a₁ (t_B – t)
⇒ v = (a₁ – a₂) t_B – a₁ t
Substitute for t from (1),

Substitute for t_B from (2),

Hence proved.

Question 3.
An object moves with a deceleration \(\alpha \sqrt{\mathbf{v}}\) in a straight line (a is a constant) At time t = 0, the velocity is v₀. What is the distance it traverses before coming to rest? What will be the total time taken?
Answer:
Deceleration = \(\alpha \sqrt{\mathbf{v}}\)
⇒ Acceleration a = – \(\alpha \sqrt{\mathbf{v}}\)
\(\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}\) = – αv^1/2
⇒ \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}^{1 / 2}}\) = – αdt
Integrating,

2v^1/2 = – αt + c
At t = 0, v = v₀
⇒ 2v₀^1/2 = c
∴ 2v^1/2 = – αt + 2v₀^1/2 …………… (1)
When t = T, v = 0
∴ 2(0)^1/2 = – αT + 2v₀^1/2
⇒ T = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
∴ Total time taken = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
From (1), we get

You can Download Chapter 14 Oscillations Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations

1st PUC Physics Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(b) and (c)

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

1. the rotation of earth about its axis.
2. motion of an oscillating mercury column in a U-tube.
3. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
4. general vibrations of a poly-atomic molecule about Its equilibrium position.

Answer:

1. Periodic but not SHM (Simple Harmonic Motion)
2. SHM
3. SHM
4. Periodic but not SHM

Question 3.
Figure below depicts four x

 

 

 

-t plots for linear motion of a particle. Which of the piots represent periodic motion? What is the period of motion (in case of periodic motion)?


Answer:
(b) and (d) are in periodic motion with period of 2 sec.

Question 4.
Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion?
Give period for each case of periodic motion (ω is any positive constant):
(a) sin ω t – cos ω t
(b) sin³ ω t
(c) 3 cos( π/4 – 2 ω t)
(d) cos ω t + cos 3 ω t + cos 5 ω t
(e) exp (- ω² t²)
(f) 1 + ω t+ ω² t²
Answer:
(a) sin ω t – cos ω t = \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right)\)
= \(\sqrt{2} \sin (\omega t-\pi / 4)\) ; SHM , period = \(\frac{2 \pi}{\omega}\)
(b) sin³ ω t = \(\frac{1}{4}\) (3 sin ωt – sin 3ωt) both terms are in SHM hence sin³ ωt is periodic.
period = \(\frac{2 \pi}{\omega}\)
(c) 3 cos (π/4 – 2 ω t) S H M with period = \(\frac{\pi}{\omega}\)
(d) cos ω t + cos 3 ω t + cos 5 ω t Superposition of 3 periodic motion,period = \(\frac{2 \pi}{\omega}\)
(e) exp (- ω² t²) non periodic motion
(f) 1 + ω t+ ω² t² non periodic motion

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration, and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:

Question 6.
Which of the following relationships between the acceleration α and the displacement x of a particle involve simple harmonic motion?

1. α = 0.7x
2. α = -200x²
3. α = -10x
4. α = 100x³

Answer:
α = -10 x (∵ acceleration is proportional and opposite to displacement)

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ω t + Φ). If the initial (t = 0) position of the particle is 1 cm and Its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ω t + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Given: x(t) = A cos(ω t + Φ) ……. (1)
at t = 0, x(0) = 1cm, ω = π s^-1 ,
substituting in (1) we get
1 cm = A cos (0 × π + Φ)
⇒ A cos Φ =1     …… (2)
Differentiate equation (1) w.r.t. t
⇒ \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) – A ω sin (ω t + Φ)    …… (3)
at t = 0, v = ω cm /s, ω = π s^-1
Substituting in 3
⇒ ω = – A ωsin Φ
⇒ 1 = – Asin Φ     ….. (4)
Divide equation (4) by (2)
⇒ 1 = \(\frac{-\sin \phi}{\cos \phi}\)
⇒ tan Φ = -1
⇒ Φ = \(\frac{-\pi}{4}\)
Φ = initial phase.
Substitute Φ value in equation (2) we get
1 = A cos \(\left(\frac{-\pi}{4}\right)\)
⇒ A = \(\sqrt{2} \mathrm{cm}\)
If x (t) = B sin (ω t + α)        ……. (5)
at t = 0,  x = 1 cm, ω = π s^-1
Substituting in (5) we get
1 = B sin(α)                 …… (6)
Differentiate (5) w.r.t. ‘ t’
⇒ v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\)= B ω cos (ω t + α) dt
at t = 0, v = ω = π
⇒ 1 = B cos(α)           …… (7)
Divide(7) by (6)
we get tan Φ = 1
⇒ Φ = π/4 is the initial phase
Substitute Φ in equation (6) we get
1 = B sin (π/4) ⇒ B = \(\sqrt{2} \mathrm{cm}\)

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Maximum length Y_max = 0.2m
Maximum weight M_max = 50 kg
We know that,
F = Mg = ky, at M = M_max, Y = Y_max

⇒ m = 22.36 kg
Weight of the body = 22.36 × 9.8 = 219.1 N

Question 9.
A spring having a spring constant. 1200 N m^-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass Is then pulled sideways to a distance of 2.0 cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass, and
3. the maximum speed of the mass.

Answer:
Given:
K = 1200 Nm^-1, m = 3 kg, a = 2 cm
1.  Frequency of Oscillation
We know that T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}=2 \pi \sqrt{\frac{3}{1200}}\).
⇒ T = 0.3144 s
⇒ f = \(\frac{1}{\mathrm{T}}\) = 3.18/s

2. Maximum acceleration of mass (α_max) maximum acceleration is obtained when y = a
⇒ m α_max = Ka
⇒ α_max = \(\frac{\mathrm{Ka}}{\mathrm{m}}=\frac{1200 \times 0.02}{3}\) = 8 ms^-2

3. Maximum speed of the mass
V_max = a ω = \(a \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}\)
V_max = 0.4 ms^-1

Question 10.
In Exercise 14.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
1. at the mean position,
2. at the maximum stretched position, and
3. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the Initial phase?
Answer:
Given a = 2 cm k = 1200Nm^-1 m = 3 kg
⇒ \(\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=\sqrt{\frac{1200}{3}}=20 \mathrm{s}^{-1}\)
let equation of S H M ⇒ x = A sin ω t
1. time is measured from mean position.
x = A sin ω t
⇒ x = 2 sin 20 t

2.  at maximum structured position phase angle is π/2
⇒ x = a sin (ω t + π/2)
⇒ x = 2 cos (20t)

3. At maximum compressed position phase angle is 3 π/2
x= a sin (ω t + 3 π/2)
⇒ x = – a cos ω t

Question 11.
The figure below correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anticlockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:

(a) Let A be any point on the circle shown in fig (a). Draw AM ⊥ to x-axis. Point M refers to x – projection of the radius vector.
Now, ∠POA = θ = ωt, T = 2s OA = 3cm ⇒ ∠OAM =θ=ωt (∵ Alternate angles)

⇒ x = 3 sin \(\frac{2 \pi}{\mathrm{T}}\)t(cm)
⇒ x = – 3 sin π t (cm)

(b) let A be any point on the circles of fig. (b). From A draw AM ⊥ to x-axis
Now ∠MOA= θ = ωt

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the Initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise In every case: (x Is In cm and t is in s).

1. x = – 2 sin (3t + π/3)
2. x = cos(π/6 – t)
3. x = 3 sin (2 π t + π/4)
4. x = 2 cos π t

Answer:
1.


2.  x = cos(π/6 – t)
radius r = 1 cm
at t = 0, x = cos (π/6) = \(\frac{\sqrt{3}}{2} \mathrm{cm}\),
Φ = – π/6 = – 30°
ω = 1 rad / s

3.

4. x = 2 cos π t
r = 2 cm
ω = π rad / s
at t = 0 , x = 2 cm

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

1. What is the maximum extension of the spring in the two cases?
2. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer:
1. Consider figure (a)
let y be the extension produced in the spring F = ky
consider fig (b) each mass acts as if it is fixed w.r.t the other
⇒ F = ky  ⇒ y = F/k

2. consider fig (a)
F = – ky
ma = – ky   a = \(\frac{-\mathrm{k}}{\mathrm{m}} \mathrm{y}\)
⇒ ω² = \(\frac{\mathrm{k}}{\mathrm{m}}\)
Therefore, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)
Consider fig (b)
let us assume (1) as centre of system and 2 springs each of length 1/2 attached to two masses. So k’ is the spring factor of each spring.
k’ = 2k

⇒ \(\mathrm{T}=\frac{2 \pi}{\omega}\)
\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}\) is the period of oscillation in the case of (b).

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/ min, what is its maximum speed?
Answer:
Stroke = 1 m
Amplitude = \(\frac{\text { stroke }}{2}\) = 1/2 m
ω = 200 rad / min
V_max = ωA
= 200 × 1/2
= 100 m / min
V_max = 1.67 m/s

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s^-2)
Answer:
g_m =1.7 ms^-2  g_e = 9.8 ms^-2  T_e = 3.5 s

Question 16.
Answer the following questions :

1. Time period of a particle in SHM depends on the force constant k and mass m of the particle: T =\(2 \pi \sqrt{\frac{m}{k}}\) . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
2. The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{1}{g}}\) . Think of a qualitative argument to appreciate this result.
3. A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
4. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Answer:

1. In case of a simple pendulum, k is directly proportional to m. Hence ratio of m / k is a constant. Hence time period doesn’t depend on mass.

2. Restoring force that brings body of pendulum back to its mean position F = – mg sin θ
Form small θ, sin θ ≈ θ = \(\frac{\mathrm{y}}{1}\)

If approximation for sin θ = θ is not taken into account then Time period T > \(2 \pi \sqrt{\frac{1}{\mathrm{g}}}\) This happens when θ is not small.

3. Wristwatch work on the principle of spring action. Hence acceleration due to gravity plays no role in the functioning of the wristwatch hence it gives correct time during free fall.

4. During free-fall acceleration, due to gravity is zero hence the pendulum will not vibrate (i.e.: frequency of oscillation is zero).

Question 17.
A simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations In a radial direction about its equilibrium position, what will be its time period?
Answer:
Body of the pendulum is under the action of two accelerations, acceleration due to gravity ‘g’ and centripetal acceleration \(\alpha=\frac{v^{2}}{R}\)
effective acceleration = \(\alpha^{\prime}=\sqrt{\alpha^{2}+g^{2}}\)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density \(\rho_{\mathrm{t}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{1} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Initially in equilibrium,
weight of cork = weight of water displaced

When cork is pushed in, then the restoring force acting on it is :
f = – weight of the portion dipped after pushing

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic, motion.
Answer:
Restoring force acting on the liquid when suction pump is removed is,
f = – mg  ⇒ f= -(A × 2y)ρ × g
where, A = Cross – section area of U tube
ρ = density of liquid
⇒ f = – 2Aρgy
acceleration produced in liquid column
a = f/mass of liquid

hence it is a SHM.

Question 20.
An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Answer:
Volume = V
mass of ball = m
cross-sectional Area = A
initial pressure on either side of the ball = atmospheric pressure = P
Let the charge in volume of air when ball is pressed be ∆V
∆V = Ay (y = displacement)
Now, Bulk modulus of elasticity


The above equation is of the form,

hence it is in SHM.

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of Its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of

1. the spring constant k and
2. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer:
1. Total mass = 3000 kg
mass supported by each wheel = 750 kg
y = 0.15 m
We know that,
mg = kg

2.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the
same period.
Answer:
Let the particle executing SHM starts from mean position.
Displacement can be given by
x = Asin ω t
Velocity v = A ω cosωt
Kinetic energy over one complete cycle

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillators is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J= – α θ, where j is the restoring couple and θ the angle of twist).
Answer:
We know that,

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is

1. 5 cm
2. 3 cm
3. 0 cm.

Answer:
Given,
r = 5 cm
T = 0.2 s
\(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rad} / \mathrm{s}\)
acceleration A = – ω²y
velocity V = \(\omega \sqrt{r^{2}-y^{2}}\)
1. y = 5 cm
A = – (10π)² × 0.05
⇒ A = 0.493 m / s²
V = \(10 \pi \sqrt{(0.05)^{2}-(0.05)^{2}}=0\)

2. y = 3 cm
A = (10π)² × 0.03
⇒ A = – 0.296 m / s²
V= \(10 \pi \sqrt{(0.05)^{2}-(0.03)^{2}}\)
V = 0.4 πm/s

3. y = 0 cm
A = 0
V= \(10 \pi \sqrt{(0.05)^{2}-0}\)
= 0.5 πm/s

Question 25.
A mass attached to a spring is freeto oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and v₀.
[Hint : Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]
Answer:
Let x = a cos (ωt + θ)
∴ v = \(\frac{d x}{d t}\) = – a ω sin (ωt + θ)
When t = 0, x = x₀ and \(\frac{d x}{d t}\) = – v₀
⇒ x_o = a cos θ ….. (1) and
– v₀ = – a ω sin θ
⇒ a sin θ = \(\frac{\mathrm{v}_{0}}{\omega}\) …..(2)
Squarring and adding (1) and (2),
a² = \(=x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}} \Rightarrow a=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}\)

1st PUC Physics Oscillations One Mark Questions and Answers

Question 1.
Mention the relation between period and frequency of periodic motion.
Answer:
Frequency f = \(\frac{1}{T}\)

Question 2.
A particle executes simple harmonic motion. At what point on its path is the acceleration maximum?
Answer:
Acceleration is maximum at a point of maximum displacement from the mean position.

Question 3.
At what position the KE of an oscillating simple pendulum Is maximum?
Answer:
Kinetic energy is maximum at the mean position.

Question 4.
What is the condition for motion of a particle to be SHM?
Answer:
Acceleration should be proportional to displacement and always directed towards mean position.

Question 5.
What is Oscillation?
Answer:
To and for motion in the same path is called Oscillation.

Question 6.
How will the period of a simple pendulum change when its length is doubled?
Answer:
T_new = \(\sqrt{2}\) T_old

Question 7.
Will a pendulum’s time period increases or decreases when taken to the top of the mountain?
Answer:
Increases, g decreases as high altitude and \(\mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{g}}}\)

Question 8.
Two simple pendulum of equal length cross each other at mean position. What is their phase difference?
Answer:
180° (π radians)

Question 9.
How many times in one vibration, KE and PE become maximum?
Answer:
Two.

Question 10.
When is the tension maximum in the spring of a simple pendulum?
Answer:
Mean position.

Question 11.
A spring of spring constant k Is cut Into two equal parts. What is the spring constant of each part?
Answer:
2 K

Question 12.
What Is the phase difference between the displacement and velocity in a SHM?
Answer:
90° (π/2 radians)

Question 13.
State force law for a SHM.
Answer:
Force, F = mω²x
m = mass
ω = angular speed
x = displacement

Question 14.
A pendulum is making one oscilla¬tion in every two seconds. What is the frequency of oscillation?
Answer:
f = \(\frac{1}{T}=\frac{1}{2} s^{-1}\)

Question 15.
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
Frequency is 0 because the acceleration zero during free fall.

Question 16.
A simple pendulum Is inside a spacecraft. What should be its time period vibration?
Answer:
Pendulum does not oscillate.

Question 17.
What are isochronous vibrations?
Answer:
When the time period is independent of amplitude such an oscillation is called isochronous.

1st PUC Physics Oscillations Two Marks Questions and Answers

Question 1.
Define simple harmonic motion (SHM). Give an example.
Answer:
A particle is said to have SHM if the acceleration of the particle is directly proportional to its displacement from the mean position and directed towards the mean position.
E.g.:

1. Vertical oscillations of a loaded spring.
2. Oscillations of bob of a simple pendulum.
3. Vibrations of string of musical instruments.
4. Motion of air particles during the propagation of sound waves.
5. Vibration of a tuning fork.

Question 2.
Mention expression for velocity and acceleration of a particle executing SHM.
Answer:
Velocity of a particle executing SHM is v = \(\omega \sqrt{A^{2}-y^{2}}\)
Acceleration of a particle executing SHM is a = – ω²y
y is the displacement of a particle from its mean position in t seconds. A is the amplitude and ω is the angular frequency.

Question 3.
Mention expression for K.E, P.E and total energy of a particle executing SHM.
Answer:
Potential energy of particle at any instant t second is E_p = 1/2 m ω² y².
Kinetic energy of particle at any instant t second is, E_K = 1/2 m ω² (A² – y²).
Total energy of particle at any instant is constant E = E_K + E_p = 1/2 m ω² A².
A is the amplitude and w is the angular frequency and m is mass of the particle y is the displacement of a particle from its mean position in t seconds.

Question 4.
The acceleration of a particle executing SHM Is 20ms^-1 at a distance of 5 m from the mean position. Calculate its time period and frequency.
Answer:
a = 20m/s², y = 5m.
We know that a = | ω² y |
⇒ \(\omega=\sqrt{\frac{a}{y}}=\sqrt{\frac{20}{5}}\)= 2 rad/s or 2πf = 2
Frequency f = \(\frac{2}{2 \times 3.14}\) = 0.3185 Hz
Period T = \(\frac{1}{f}=\frac{1}{0.3185}\) = 3.14s

Question 5.
Differentiate between forced oscillations and resonance.
Answer:
1. Forced Oscillations:
A body oscillates with the help of external periodic force with a frequency different from natural frequency of body.

2. Resonance:
A body oscillating with its natural frequency with the help of external periodic force whose frequency is equal to natural frequency of body.

Question 6.
Two linear, simple harmonic motion of equal amplitudes and frequencies ω and 2ω are impressed on a particle along the axis of X and Y respectively. If the initial phase difference is π/2, then find the resultant path followed by particle
Answer:
Let
X = Asinωt …… (1)
Y= Acos2ωt …….(2)
(1) and (2) represent SHM with equal amplitude and phase difference π/2 with frequencies ω and 2ω.

Question 7.
Derive an expression for K.E and P.E of a particle SHM.
Answer:
We know that, Potential energy
U = 1/2 kx²
For a particle in SHM, k = mω² and x = Asin ωt
⇒ U = 1/2 mω² A² sin² ωt
Kinetic energy, k= 1/2 mv²
now, v = ωy = ωA cos ωt
⇒ K= 1/2 mω² A²cos²ωt

Question 8.
The amplitude of a oscillating simple pendulum Is doubled. What will be its effect on

1. periodic time
2. total energy
3. maximum velocity

Answer:

1. Periodic time does not change. T is independent of amplitude.
2. Total energy, T.E = 1/2 mω² A²
   A is doubled ⇒ T.E_new = 4 TE_old
3. Maximum velocity, V_max = ωA
   A is doubled ⇒ V_max is doubled.

Question 9.
The frequency of oscillations of a mass m suspended by a spring is V₁. If the length of spring is cut to one half, the same mass oscillates with frequency V₂. Calculate V₂/V₁.
Answer:
V₁ = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
where K is the spring constant and m is the mass
V₂ = \(\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}\)
∴ \(\frac{V_{2}}{V_{1}}=\sqrt{2}\)

Question 10.
Discuss some important characteristics of wave motion.
Answer:

1. Wave transports energy.
2. particles are not transported.
3. Elasticity and inertia determine the motion of particle distribution.

Question 11.
Differentiate between free oscillations and forced oscillations with the help of examples.
Answer:
Consider a pendulum in free space (without air) it oscillates freely. This is free oscillation. Consider another pendulum in a viscous liquid it oscillates only if there is external force this is forced oscillation.

Question 12.
List any two characteristics of SHM.
Answer:

1. SHM is always directed towards mean position.
2. Acceleration is directly proportioned to displacement but opposite in direction.

1st PUC Physics Oscillations Three Marks Questions and Answers

Question 1.
Define the terms

1. amplitude
2. period
3. frequency
4. phase related with a particle executes SHM.

Answer:
1. Amplitude:
Maximum displacement of the particle from the mean position is called amplitude.

2. Period :
Time taken by the particle to complete one oscillation is called time period. (T)

3. Frequency :
The number of oscillations completed by the particle in one second is called frequency (f)

4. Phase:
Phase of a particle is defined as the fraction of the time period that has elapsed since the particle last passed through its mean position in the positive direction.

Question 2.
For an oscillating pendulum, establish the relation \(\frac{d^{2} \theta}{d t^{2}}=-\omega^{2} \theta\) where \(\omega=\sqrt{\frac{g}{1}} \theta\) = small angular displacement.
Answer:

restoring force F_r = – mg sin θ
torque on pendulum \(\tau=\mathrm{I} \alpha=\mathrm{ml}^{2} \alpha\)
restoring torque = torque on pendulum

We know that = \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=\alpha\) and \(\omega=\sqrt{\frac{g}{1}}\)
⇒ \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-\omega^{2} \theta\)

Question 3.
A body of mass 1 kg is suspended from a weightless spring having force constant 600 Nm^-1. Another body of mass 0.5 kg moves vertically upwards lift the suspended body with a velocity of 3 ms^-1 and tets embedded in it. Find the frequency of oscillation and amplitude of motion.
Answer:
Total mass = (1 + 0.5)kg = 1.
k = 600 nm^-1
frequency of oscillations

Let V₁ be velocity of mass after collision
⇒ m₁ V₂ = (m₁ + m_2,) V₁
⇒ 0.5 × 3 = 1.5 V₁
⇒ V₁ = 1 m/s
According to law of conservation of energy
(PE)_max = (KE)_max
\(\frac{1}{2} m v_{1}^{2}=1/2^{K A^{2}}\)
\(\frac{1}{2} 1.5 \times 1^{2}=\frac{1}{2} 600 \mathrm{A}^{2}\)
⇒ A = 5 cm

Question 4.
The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.
Answer:

Let the rikshaw of mass M be at any point P in the dip of radius R. Let O be the centre of this circular path. This case is similar to that of a simple pendulum and assume that e is small.
restoring force
F = -mg sinθ
F = – mg θ
displacement of rickshaw = R θ

Question 5.
Find an expression for body vibrating in SHM.
Answer:
Let the SHM equation be
x = Asin ω t

Potential energy is given by
= 1/2 kx² = 1/2 m ω ² k²
= 1/2 mω² A² sin² ω t
Kinetic energy = 1/2 mv²
= 1/2 m ω ² A² cos² ω t
Total energy    = PE + KE
= 1/2 mω² A² sin² ω t + 1/2 mω² A² cos² ω t
= 1/2 mω ² A²

Question 6.
Show that when a particle is moving in SHM its velocity at a distance \(\frac{\sqrt{3}}{2}\) its amplitude from the central position is half its velocity in central position.
Answer:
For a particle in SHM
\(\mathrm{V}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}\)

Question 7.
A body oscillates with SHM according to the equation.
x(t) = 5cos (2π t + π /4) Where x is in meters and t is in seconds calculate,

1. displacement att = 0
2. Angular frequency
3. V_max

Answer:
1. displacement at t = 0
x(0) = 5cos(2π 0 + π/4)
⇒ x = \(\frac{5}{\sqrt{2}} m\)

2. Angular frequency
ω = 2π radian/s

3. Vmax = aω = 2π × 5 = 10π m/s

Question 8.
What is spring factor? Find its value in case of two springs connected in

1. series
2. Parallel

Answer:
Spring factor (k):
Force acting for unit extension produced is called spring factor.
1. When two springs connected in series,

force experienced by both springs is same.
Total extension (x) is sum of individual extension.
⇒ x = x₁ + x₂
\(\frac{F}{-K_{e q}}=\frac{F}{-K_{1}}+\frac{F}{-K_{2}}\) ⇒ 1/K_eq = 1/K₁ + 1/K₂
K_eq = \(\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{1}+\mathrm{K}_{2}}\)

2. When two springs are in parallel,

Net force experienced is the sum of force experienced in both springs. But the extension will be the same.
F_eq = F₁ + F₂
K_eqx =K₁ X + K₂X
⇒ K_eq = K₁ + K₂

Question 9.
Explain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
Answer:
Let displacement x = Asinωt
Velocity v= ω Acosω t
acceleration a = – ω² Asin ω t = ω² Asin(ω t + π )
Displacement and velocity have phase difference of π/2 radians.
v and α has \(\frac{\pi}{2}\) radians of phase difference. α ans x has π radians of phase difference.
Assume ω >1

Question 10.
For a particle in SHM, the displacement x of the particle at a function of time t is given as x = Asin(2πt) …..(1) where x is in centimeters and t is in seconds. Let the time taken by the particle to travel from x=0 to x = A/2 beT₁ and the time taken to travel from x = A/2 to x = A be T₂. Find T₁/T₂
Answer:
ω =2π = \(\frac{2 \pi}{\mathrm{T}}\) ⇒ T= 1 s
at t = 0 , x = 0
now if x = A/2

now in a SHM time taken from 0 to A is \(\frac{T}{4}\)
⇒ time taken for x = A/2 to A is \(\frac{T}{4}\) -T/12

Question 11.
What is SHM? Show that in SHM acceleration is directly proportional to Its displacement at a given instant.
Answer:
Simple Harmonic Motion is a type of motion in which displacement is always directed towards mean position and acceleration is directly proportional to displacement and opposite in direction.
Let a SHM be represented by
x = Asin ωt
⇒ \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = A ω cos ωt

acceleration is proportional to displacement and opposite in direction.

Question 12.
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretched 7cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer:
Mass added m = 0.02
Length stretched x= 7 cm = 0.07 m

Time period T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)

Question 13.
A particle executes SHM of time period 10 sec. The displacement of particle at any instant is x = 10 sinωt(cm) Find

1. Velocity of body 2 s after it passes through mean position.
2. Acceleration of body 2 s after it passes mean position.

Answer:
Given x = 10 sin ωt T = 10 sec
A = 10 cm   ω = \(\frac{2 \pi}{10}\) rad/s
Let at t = 0 body be at the mean position.
Now at t = 2
1.  V = Aω cos ωt
= 10ω cos(ω2)cm/s
⇒ \( V=10 \times \frac{2 \pi}{10} \cos \left(\frac{2 \pi}{10} 2\right)\)
⇒ V = 1.94 cm/s

2. Acceleration
a = – Aω² sinωt
\(= – 10 \times \frac{4 \pi^{2}}{100} \sin \left(\frac{4 \pi}{10}\right)\)
⇒ a = – 3.75 cm/s²

Question 14.
What is a simple pendulum? Show that the motion of the pendulum is SHM and hence deduce an expression for the time period of pendulum. Also, define Second’s pendulum.
Answer:
Simple pendulum is a point mass body suspended by a weightless thread or string from a rigid support about which it is free to oscillate.

Now restoring force acting on body can be given by
F = – mgsinθ (from figure)
If  θ is small
F= – mgθ    (∵ sin θ ≈ θ)
Let P be any point on the path of pendulum and makes ∠OMP = θ which is small and arc OP be x (displacement) then
\(\dot{\theta}=\frac{\mathrm{OP}}{1}=\frac{\mathrm{x}}{1}\)
⇒ F = \(\frac{-m g x}{1}\) ……(1)
⇒ force ∝ displacement and opposite in direction hence an SHM
Now this of form F = -kx ……(2)
⇒ k = mg/l (from (1) and (2)

Question 15.
A body of mass ‘m’ suspended from a spring executes SHM. Calculate the ratio of kinetic energy and potential energy of the body when it is at half the amplitude far from the mean position.
Answer:
KE of a mass suspend from a spring
KE = mω²(a² – y²)
Potential energy of a mass suspend from a spring
PE = 1/2 mω² y²
\(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right) 2}{\mathrm{y}^{2}}\)
at y = a/2
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{a}^{2} / 4\right)^{2}}{\mathrm{a}^{2} / 4}\)
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=3\)

Question 16.
If x = a cos ωt + b sin ωt, show that it represents SHM.
Answer:
We have,
x = a cos ωt + b sin ωt
Now, \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – aω sinωt + bω cosωt
\(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – aω² cos ωt + bω² sin ωt
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω² (a cos ωt+ b sin ωt)
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω²x
⇒ α = – ω²x
Hence an SHM

Question 17.
Find the expression for the total energy of a particle executing SHM.
Answer:
For a SHM PE = 1/2 kx²
= 1/2 mω² x² = 1/2 mω A² sin² ωt
KE= 1/2 mv²
v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – Aω cos ωt
⇒ KE = 1/2 mA²ω² cos² ωt
Total energy = KE + PE = 1/2 mω² A² cos² ωt + 1/2 mω² sin² ωt
Total energy = 1/2 mω² A²

1st PUC Physics Oscillations Five Marks Questions and Answers

Question 1.

1. Find the total energy of particle executing SHM?
2. Show graphically the variation of PE and KE with time In SHM.
3. What is the frequency of these energies w.r.t the frequency of the particle executing SHM?

Answer:
1. PE at any instant
PE = 1/2 kx²
= 1/2 mω² A² sin² ωt
(∵ k = mω² and x = A sin ωt)
⇒ PE = 1/2 mω² A² sin² ωt
KE at any instant KE = 1/2 mv²
KE = 1/2 mA² ω² cos² ωt
Total energy = KE + PE
⇒ Total energy = 1/2 mω² A²sin² ωt + 1/2 mω² A² cos² ωt
⇒ Total energy = 1/2 mω² A²

2. Graph

3. Frequency = \(\frac{2}{\text { period }}=\frac{2}{\mathrm{T}}\)

1st PUC Physics Oscillations Numerical Problems Questions and Answers

Question 1.
A spring compressed by 0.1m develops a restoring force of 10N. A body of mass 4 kg is placed on it. Deduce the

1. force constant of the spring
2. depression of the spring under the weight of the body
3. period of oscillation, if the body is disturbed.

Answer:
Given
Restoring force, F = 10N
Mass of the body, m = 4kg
displacement, x =0.1m

1. \(\mathrm{k}=\frac{\text { force }}{\text { displacement }}=\frac{10}{0.1}\) = 100 Nm^-1
2. Depression due to weight \(=\frac{\text { force }}{\mathbf{k}}=\frac{4 \times 10}{100}\) = 0.4 m
3. Period of oscillation T = \(2 \pi \sqrt{\frac{m}{k}}\) \(=2 \pi \sqrt{\frac{4}{100}}\) = 0.4 πs
   T = 0.4 πs

Question 2.
A vertical U- tube of uniform crosssection contains water up to a height to 20cm. Calculate the time period of the oscillation of water when it is disturbed.
Answer:
The length of liquid column
L = 2 × 20cm = 40 cm
Time period of oscillation
\(=2 \pi \sqrt{\frac{L}{2 g}}=2 \pi \sqrt{\frac{40}{2 \times 9.80}}\) = 0.9 s

Question 3.
A cylindrical piece of cork of base area ‘A’ and height ‘h’ floats in a liquid of density \(\rho_{\mathrm{e}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period of \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{\mathrm{e}} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let x be the depression created.
Excess upthrust caused is, U_e= g\(\rho_{\mathrm{e}}\) Ax
Restoring force = U = gA\(\rho_{\mathrm{e}}\) x
mass = m = Ahρ
Now F = mα;  α = acceleration
Ahρα = -gA\(\rho_{\mathrm{e}}\) x

Question 4.
If this earth were a homogenous sphere and a straight hole bored in it through its centre. Show that if a body were dropped into the hole it would execute a SHM. Also, find its time period.
Answer:
Let mass of body dropped = m
mass of earth = M
radius of earth = R
mg = \(=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) ⇒  g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
⇒  \(\mathrm{g}=\frac{\mathrm{G}}{\mathrm{R}^{2}} \frac{4}{3} \pi \mathrm{R}^{2} \rho=\frac{4 \pi \mathrm{GR} \rho}{3}\)
where ρ is mean density of earth

Let R be the depth of the body fallen from centre then g at the point is
\(\mathrm{g}^{\prime}=\frac{\mathrm{Gm}^{\prime}}{(\mathrm{R}-\mathrm{d})^{2}}\)
where m’ is mass of sphere of radius (R – d)

Hence acceleration is proportional to displacement and in opposite direction go on SHM.

You can Download Chapter 10 The S-Block Elements Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 10 The S-Block Elements

1st PUC Chemistry The S-Block Elements One Mark Questions and Answers

Question 1.
How does the density of alkali metal change from Li to Cs?
Answer:
Density increases down the group from Li to Cs.

Question 2.
Elements of which group in the periodic table belong to s-block?
Answer:
I and II groups.

Question 3.
In what way the electronic configuration of hydrogen is similar to that of the electronic configuration of alkali metals?
Answer:
Hydrogen and alkali metals both have one electron in the outermost orbital.

Question 4.
What similarity is found in the electronic configurations of hydrogen and halogen?
Answer:
Both hydrogen and halogen are in short of one electron for the completion of the outermost orbital containing electrons.

Question 5.
Name any four alkali metals.
Answer:
Lithium, Sodium, Potassium and Rubidium.

Question 6.
LiCI and MgCk dissolve in alcohol. How do you explain this?
Answer:
Both LiCI and MgCl₂ are covalent compounds and dissolve in alcohol. This is due to high polarizing power of Li⁺ and Mg²⁺ ions.

Question 7.
The alkali metals have no tendency to show variable oxidation states. Give reason.
Answer:
Alkali metals show oxidation state of +1. With the loss of valence electron it attains the stable configuration of nearest inert gas. Its second ionization potential is high. Hence an alkali metal does not show variable oxidation states.

Question 8.
Write the alkali metals in the increasing order of hydration energy.
Answer:
Li ⁺ > Ma⁺ > K⁺> > Rb⁺ > Cs⁺

Question 9.
Why are group 1 elements called alkali metals ?
Answer:
It is because their hydroxides are soluble bases called alkalies. Secondly their ashes are alkaline in nature.

Question 10.
Why do alkali metals have low ionisation energy ?
Answer:
It is due to largest atomic size, they can lose electrons easily.

Question 11.
Alkali and alkaline earth metals cannot be obtained by chemical reduction, why?
Answer:
Alkali and alkaline metals are good reducing agents, they cannot be obtained by chemical reduction.

Question 12.
Why does ionisation energy of alkali metals decrease with the increase in atomic number ?
Answer:
Atomic size increases with increase in atomic number, therefore, nuclear force of attraction between valence electrons and nucleus decreases, hence ionisation energy decreases down the group.

Question 13.
Why group 2 elements (Mg and Ca) are harder and denser than group 1 elements ?
Answer:
They have strong metallic bonds due to smaller size and have more number of valence electrons.

Question 14.
Why is potassium more reactive than sodium ?
Answer:
K has lower ionisation energy than sodium due to bigger atomic size, therefore, it is more reactive.

Question 15.
Why are alkali metals strong reducing agents?
Answer:
It is because of low ionisation energy. They can lose electrons easily, that is why they are strong reducing agents.

Question 16.
Why are alkali metals used in photoelectric cells ?
Answer:
They have low ionisation energy and can lose electrons when light falls on them, that is why they are used in photoelectric cells.

Question 17.
Write electronic configuration of Na (11) and K (19).
Answer:
Na(11): 1s²2s²2p⁶3s¹ K(19): ls²2s²2p⁶3s¹3p⁶4s¹

Question 18.
Why do alkali metals have low melting and boiling points ?
Answer:
It is due to weak metallic bonds which is due to bigger atomic size that is why they how low melting and boiling points.

Question 19.
How will you prepare sodium hydrogen carbonate from sodium chloride ?
Answer:

(NH₃)HCO₃ + NaCl → NaHCO₃ + NH₄Cl

Question 20.
Why do alkali metals not occur in free state ?
Answer:
They are highly reactive, therefore, they occur in combined state and do not occur in free state.

Question 21.
Why is second ionisation energy of alkali metals higher than alkaline earth metals ?
Answer:
Alkali metals acquire, noble gas configuration after losing 1 electron, therefore their second ionization energy is higher than alkaline earth metals.

Question 22.
Which out of K, Mg, Ca and Al from amphoteric oxide ?
Answer:
Al forms amphoteric oxide, i.e., acidic as well as basic in nature.

Question 23.
Which out of Na, K, Al, Mg occur as oxide in nature ?
Answer:
Al occurs as oxide in nature as bauxite Al₂O₃. 2H₂O

Question 24.
Why do alkali metals give characteristic flame colouration?
Answer:
They have low ionization energy and absorb energy from visible region of spectrum and radiate complementary colour.

Question 25.
What happens when K burns in air ? Give chemical equation.
Answer:
K + O₂ → KO₂, pottassium superoxide

Question 26.
What is quick lime ? How is it prepared ?
Answer:
Quick lime is calcium oxide. It is prepared by heating limestone.

Question 27.
Give two uses of plaster of paris. Also give its formula.
Answer:

1. It is used in plastering fracture bones,
2. It is used in preparations of chalks. Its formula is (CaSO₄)₂.H₂O

Question 28.
Arrange the following in order of their increasing covalent character : MCI, MBr, MF, Ml (Where M is alkali metal)
Answer:
MF < MCI < Ml, lesser the difference in electronegativity, more will be covalent character.

Question 29.
One reason on being heated in excess supply of air K, Rb and Cs from superoxide in preference to oxides and peroxides ?
Answer:
K, Rb and Cs are more reactive therefore, they form superoxide in preference to oxides and peroxides K⁺,Rb⁺ and Cs⁺ ions are large cations and superoxide ion \(\mathrm{O}_{2}^{-}\), is also large. Larger cations stabilize larger anions, therefore, they form superoxide.

Question 30.
What happens when KO₂ reacts with water ? Give balanced chemical equation.
Answer:
2KO₂ (Pottasium sup eroxide) + 2H₂O → 2KOH + O₂ + H₂O₂

Question 31.
Complete the reaction : Lil + KF →
Answer:
Lil + KF → LiF + KI; larger cation stabilizes larger anion and smaller cation stabilizes smaller anion.

Question 32.
Name the reagent or one process to distinguish between :

• BeSO₄ and BaSO₄
• Be(OH₂) and Ba(OH)₂

Answer:

• BeSO₄ is soluble in water while BaSO₄ is not.
• Be(OH)₂ dissolves in NaOH while Ba(OH)₂ is insoluble.

Question 33.
Why does Be resemble Al?
Answer:
Be resembles Al because charge over radius ratio is similar, i.e., they have similar polarizing power.

Question 34.
The second ionization enthalpy of Ca is higher than first and yet calcium forms CaCl₂ and not CalCl Why?
Answer:
The hydration energy of Ca²⁺ over comes the second ionization energy of Ca, that is why Ca forms CaCl₂ and not CaCl. Ca⁺ is not stable

Question 35.
Name the alkali metals which form superoxide when heated in air?
Answer:
K, Rb, Cs are alkali metals which form superoxide when heated in air.

Question 36.
Name the metal which floats on water without apparent reaction.
Answer:
Berylium.

Question 37.
Why is BeCl₂ soluble in organic solvents ?
Answer:
BeCl₂ is covalent, therefore, soluble in organic solvents.

Question 38.
Starting from quick lime how slaked lime is prepared ? Is this reaction exothermic or endothermic ?
Answer:
CaO + H₂O → Ca(OH)₂ +heat
When CaO is put in water, it forms calcium hydroxide. It is an exothermic reaction.

Question 39.
Carbon dioxide is passed through a suspension of limestone in water. Write balanced chemical equation for the above reaction.
Answer:
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂

Question 40.
What do we get when crystals of washing soda exposed to air?
Answer:
We get amorphous sodium carbonate because it loses water molecules.

Question 41.
What happens when sodium dissolve in liquid ammonia?
Answer:
It results in the formation of intense blue colour solution which possess conducting power
\(\mathrm{Na}+(\mathrm{x}+\mathrm{y}) \mathrm{NH}_{3} \longrightarrow \mathrm{Na}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}}^{+}+\left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{3}\right]^{-}\)

Question 42.
Name the elements (alkali metals) which form superoxide when heated in excess of air.
Answer:
Potassium, rubidium and caesium

Question 43.
Why is oxidation state of Na and K always + 1?
Answer:
It is due to their high second ionization enthalpy and stability of their ions [Na⁺ K⁺]

Question 44.
What is meant by dead burnt plaster?
Answer:
It is anhydrous calcium suplhate (CasO₄)

Question 45.
What is the reason that sodium reacts with water more vigorously than lithium ?
Answer:
Because sodium is more electro-positive than Li.

Question 46.
Why is sodium thiosulphate used in photography ?
Answer:
Because of its complex forming behaviour.

Question 47.
Why does lithium show anomalous behaviour ?
Answer:
Due to its small size and high charge/size ratio.

1st PUC Chemistry The S-Block Elements Two Marks Questions and Answers

Question 1.
Give reason for diagonal relationship of lithium with magnesium.
Answer:
Both Lithium and magnesium have small size and high charge density. The electronegativities of Li is 1.0 and Mg is 1.2. They are low and almost same. Their ionic radii are similar. Hence they show similarities which is known as diagonal relationship between first element of a group with the second element in the next higher group.

Question 2.
What is photoelectric effect?
Answer:
Alkali metals have the lowest ionization energy in each period of the periodic table. Hence they emit electrons even when exposed to light. This phenomenon is called photoelectric effect. Rubidium and caesium are used in photoelectric cells.

Question 3.
Give two important ores each of Na and K.
Answer:
Rock salt (NaCl), Na₂CO_3, NaHCO₃, 2H₂O (trona) are important ores of Na. Sylvine (kCl); kCl.mgcl₂.6H₂O (carnallite) are of k.

Question 4.
Give one important use of following compounds.

1. NaHCO₃
2. Slaked lime
3. NaOH

Answer:

1. Sodium bicarbonate is used as antacid
2. Slaked lime is used for white washing,
3. NaOH is used in manufacture of soap.

Question 5.
What is effect of heat on the following compounds ? (Give equations for the reactions)
(i) CaCO₃
(ii) CaSO₄2H₂O
Answer:

Question 6.
Name the metals which are found in each of the following minerals :
(a) Chile salt petre
(b) Marble
(c) Epsomite
(d) Bauxite
Answer:
(a) Na
(b) Ca
(c) Mg
(d) Al.

Question 7.
What are the raw materials used in manufacture of Portland cement ? How is it manufactured ?
Answer:
Limestone and clay are raw materials used in manufacture of cement. It is prepared by heating powdery mixture of limestone and clay in dry process. In wet process, fine – powdered mixture is converted into slurry by adding water and then it is heated at a temperature 1500° C to 1600° C, the product formed is called clinkers. It is cooled down and mixed with gypsum (CaSO₄, 2H₂O) and then it is powdered.

Question 8.
What is composition of Portland Cement ? What is average composition of good quality cement ?
Answer:
CaO = 50% ; 60% ; SiO₂ = 20% to 25% ; Al₂O₃ = 5 to 10% ; MgO = 2% to 3% ; Fe₂O = 1 to .2% ; SO₂ = 1 to 2% is composition of Portland cement.
The ratio of SiO₂ (silica) to alumina (Al₂O₃) should be between 2.5 and 4.0 and the ratio of lime (CaO) to total oxides of silicon, aluminium and iron (SiO₂, Al₂O₃ and Fe₂O₃) should be as close to 2 as possible.

Question 9.
Describe in brief the manufacture of caustic soda using the Castner-Kellner cell.
Answer:
The Castner-Kellner cell consists of large rectangular trough divided into three compartments with partition short of reaching the bottom of the tank. Thus mercury in one compartment can flow into another but solution cannot mix. Graphite anodes are used in outer compartments filled with NaCl solution. The middle compartment contains very dilute solution of caustic soda and filled with iron rods as cathode.

On passing electric current CI₂ is liberated in outer compartments and sodium liberated at cathode. Mercury forms amalgam which is passed into middle compartment in which mercury acts as anode (having induced +ve potential).
At anode Na⁺ + e^– → Na
At cathode Na + Hg → Na – Hg
2(Na – Hg) + 2H₂O → 2NaOH + Hg + H₂
The concentration of NaOH goes on increasing in the middle compartment. When the concentration of NaOH reaches 20% the solution is replaced by dilute solution.

Question 10.
Compare four properties of alkali metals and alkaline earth metals.
Answer:

Alkali Metals	Alkaline earth metals
1. The show + 1 oxidation state.	1. They show + 2 oxidation state.
2. They are soft metals	2. They are harder than alkali metals.
3. They do not form complexes except Li.	3. They can form complex compounds
4. Their carbonates are soluble in water except Li2CO3	4. Their carbonates are insoluble water.

Question 11.
What happens when exhaling is made through a tube passing in lime water ? What will happen if continued exhaling is made through it ? If the solution thus obtained is heated, what do we observe ? Explain giving chemical reactions.
Answer:
Lime water turns milky due to formation of CaCO₃
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
Milkiness will disappear if continuously exhaling i.e., CO₂ is passed due to formation of calcium bicarbonate.
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂
Milkiness will reappear if heating is done due to formation. CaCO₃

Question 12.
Complete the following equations :
(i) Ca + N₂ →
(ii) Ca + SO₂ →
(iii) Ca(OH)₂+NH₄Cl →
(iv) Ca + CO₂ →
Answer:
(i) 3 Ca + N₂ → Ca₃N₂
(ii) 2 Ca + SO₂ → 2CaO + S
(iii)

(iv) 2 Ca + CO₂ → CaO + C

Question 13.
What is dead burnt plaster ? How is it obtained from gypsum?
Answer:
CaSO₄ is called dead burnt plaster. It is obtained by heating gypsum at high , temperature.

Question 14.
What is used for drying alcohol and non-acidic gases and why ?
Answer:
Calcium is used for drying alcohol and non-acidic gases because Ca does not react with alcohol.

Question 15.
What is the mixture of CaCN₂ and carbon called ? How is it prepared ? Give its uses.
Answer:
It is called Nitrolim. It is prepared by heating CaC₂ with N₂ at high temperature. It is used as fertilizer.

Question 16.
Convert limestone to calcium carbide.
Answer:

Question 17.
What are isomorphous salts ? Give two examples.
Answer:
Isomorphous salts are those which have same crystalline structure, e.g., MgSO₄, 7H₂O and ZnSO₄,7H₂O are isomorphous.

Question 18.
Which metal is present in chlrophyll ? How does this metal react with N₂ ?
Answer:
Mg is present in chlorophyll. N₂ reacts with Mg to form magnesium nitride.
3Mg + N₂ → Mg,N₂ (magnesium nitride)

Question 19.
Name an alkali metal carbonate which is thermally unstable and why ? Give its decomposition reaction.
Answer:
Li₂CO₃ is thermally unstable because it is covalent. It decomposes to form Li₂O and
CO₂ ; Li₂CO₃ → Li₂O + CO₂

Question 20.
Why are ionic hydrides of only alkali metals and alkaline earth metals are known ? Give two examples.
Answer:
Alkali metals and alkaline earth metals aye most electropositive due to low ionization energy or enthalpy. Therefore, they can form ionic hydrides, e.g., NaH, KH and CaH₂.

Question 21.
Which one of the alkaline earth metal carbonate is thermally most and last stable. Why ?
Answer:
BaCO₃ is thermally most satable due to greater ionic character and high lattice energy whereas BaCO₃ is thermally least stable because it is covalent and has less lattice energy.

Question 22.
Which out of Li, Na, K, Be, Mg, Ca has lowest ionization enthalpy and why ?
Answer:
K has lowest ionisation energy due to larger atomic size among these elements. The force of attraction between valence electron and nucleus is less, therefore, it can lose electron easily.

Question 23.
Which alkali metal ion forms largest hydrate ion in aqueous solution and why?
Answer:
Li⁺ forms largest hydrated cations because it has highest hydration energy. It has smallest size therefore, it is most hydrated.

Question 24.
What is responsible for the blue colour of the solution of alkali metal in liquid ammonia ? Give chemical equation also. [MSE (Chandigarh) 2003]
Answer:
The solvated electron, [e(NH3)]^– or ammoniated electron is responsible for blue colour of alkali metal solution in NH₃. It absorbs light from visible region and radiates complementary colour, (in the equation am = ammoniated)
Na⁺(am) + e^–(am) + NH₂(I) →NaNH₂(am) + \(\frac { 1 }{ 2 }\)H₂(g)

Question 25.
Heat of Hydration of Na+ (size 102 pm) = -397 kJ mol-1 whereas Caz 100 pm) = -1650 kJ mol-1. Explain the difference.
Answer:
Ca²⁺ is smaller in size than Na⁺ and also it has higher charge, therefore, its hydration energy is more than that of Na⁺.

Question 26.
Discuss the diagonal relationship of Be and A1 with regard to
1. action of alkali and
2. the structure of their chloride.
Answer:
1. Be and A1 both react with NaOH to form sodium beryllate and sodium meta aluminate respetively. Be dissolves in excess of NaOH to form [Be(OH)4]² where as
Al forms [ A1 (OH)6 ]^3- in excess of NaOH.
2H₂O + Be + 2NaOH → Na₂ [Be (OH)4 ] (saliun beryllali) + H₂
2A1 + 6NaOH + 6H₂O → 2Na₃ [ A1 (OH)₆ ] (Sodiam meta ilumivali) + 3H₂

2. BeCl₂ is electron deficient, therefore, has polymeric chain structure in solid state.

AlCl₃ is also electron deficient, It exists as dimmer, i.e., Al₂Cl₆

Question 27.
Complete the following:
(i) Li + N₂ →
(ii) \(\text { LiNO }_{3} \stackrel{\text { heat }}{\longrightarrow}\)
(iii) \(\mathrm{NaNO}_{3} \stackrel{\text { heat }}{\longrightarrow}\)
(iv) B₂H₆ →
Answer:

Question 28.
Arrange the (i) hydroxide and (ii) sulphates of alkaline earth metals in order of decreasing solubilities giving a suitable reason for each.
Answer:
Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > Mg(OH)₂ > Be(OH)₂
Solubility of hydroxides goes on increasing down the group because hydration energy dominates over lattice energy.
BaSO₄ < SrSO₄ < CaSO₄ < MgSO₄ < BeSO₄

Question 29.
What makes lithium show properties uncommon to the rest of alkali metals ? Write two points of similarly in properties between lithium and magnesium.
Answer:
Solubility of sulphates goes on decreasing down the group because lattice energy dominates over hydration energy. Lithium has smallest atomic size and highest ionization energy, highest polarizing power that is why it shows uncommon properties to the rest of alkali metals.

• Both Lithium and Magnesium react with N₂ to form nitrides.
• Li and Mg react with O₂ to form monoxides.

Question 30.
Write the chemical equations of the reactions involved in solvay process of preparation of sodium carbonate.
Answer:

Question 31.
Arrange the following in order of the increasing covalent character : MCI, MBr, MF, MI (where M = alkali metal)
Answer:
As the size the anion increases, covalent character increases and hence the order is ‘MF < MCI < MBr < MI.

Question 32.
What is the formula of gypsum? What happens when it is heated?
Answer:
CaSO₄. 2H₂O. When heated to 393 K, it gives plaster of paris (CaSO₄.1/2H₂O ) but at 473 K it gives dead burnt plaster (CaSO₄).

Question 33.
The E° for C1^–/Cl₂ is 1.36, for I^–/I₂ is +0.53, for Ag⁺/Ag is + 0.79, Na⁺ is -2.71 and for Li⁺ / Li is -3.04 V Arrange the following species in decreasing order of reducing strength. I^–, Ag, Cl^– Li, Na
Answer:
The more negative or less positive is the electrode potential, more is the reducing strength of the species. Since the electrode potentials inceases in the order : Li(-3.04V) < Na(-2.7V) < I^–(0.53 V) < Ag(+0.79V) < Cl^– (+1.36V), therefore, the reducing strength decreases in the order Li > Na > I^– > Ag > Cl^–

Question 34.
How do you prepare KO₃ ? Mention Magnetic Behaviour of \(\mathrm{O}_{3}^{-}\) .
Answer:
Potassium ozonide (KO₃) is formed when ozone is passed through KOH.
2KOH + 5O₃ → 2KO₃ + 5O₂ + H₂O
It is an orange coloured solid and contains the paramagnetic O₃ ion.

Question 35.
What are ionic polyhide compounds ?
Answer:
The alkali metals react with halogens and interhalogen compounds forming ionic polyhide compounds.
KI + I₂ → K[I₃] ; KBr + ICI → K[BrICI] ; KF + BrF₃ → K[BrF₄]

1st PUC Chemistry The S-Block Elements Three / Four Marks Questions and Answers

Question 1.
What are alkali metals? Describe their general properties.
Answer:
1st group elements of periodic table i.e., lithium, sodium, potassium, rubidium and caesium are called alkali metals.
General properties:

• Alkali metals have general electronic configuration ns¹E
• Alkali metals exhibit on oxidation state of+1.
• Atomic radius increases down the group from lithium to caesium.
• The metallic property of alkali metals increases from lithium to caesium.
• Alkali metals have low ionization potential.
• Due to low ionization power they are highly electropositive.
• Alkali metals are light metals.

Question 2.
Write the balanced equations from the reaction between
(a) Na₂ O₂ and water
(b) KO₂ and water
(c ) Na₂O and CO₂
Answer:
(a) 2Na₂O₂ + 2H₂O → 4NaOH + O₂
(b) 2KO₂ + 2H2_O → 2KOH + H₂O₂ + O₂
(c) Na₂O + CO₂ → Na₂CO₃

Question 3.
What is the action of heat on the following compound ?
(i) Na₂CO₃ and CaCO₃ ,
(ii) MgCl₂.6H₂O and CaCl₂.6H₂O
(iii) Ca(NO₃)₂ and NaNO₂
Answer:
(i) Na₂CO₃ and CaCO₃
Na₂CO₃ does not decompose on heating while CaCO₃ evolves CO₂
\(\begin{array}{l}{\mathrm{Na}_{2} \mathrm{CO}_{3} \stackrel{\text { heat }}{\longrightarrow} \text { Noaction }} \\ {\mathrm{CaCO}_{3} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}}\end{array}\)

(ii) MgCl₂.6H₂O and CaCl₂.6H₂O
On heating hydrated CaCl₂ is dehydrated while hydrated MgCl₂ changes into MgO

(iii) On heating the two nitrate form different products

Question 4.
Complete the following

1. Ca + H₂0 →
2. Ca(OH)₂ + Cl₂ →
3. BeO + NaOH →

Answer:

1. Ca + H₂O → CaO + H₂O
2. Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
3. BeO + NaOH → Be(OH)₂ + Na₂O

Question 5.
Explain what happens when
(i) Sodium hydrogen carbonate is heated
(ii) Sodium with mercury reacts with water
(iii) Fused sodium metal reacts with ammonia.
Answer:

Question 6.
Why is it that the s-block elements never occur in free state/nature? What are their usual modes of occurrence and how are they generally prepared?
Answer:
The elements belonging to s-block in periodic table (i.e. alkali and alkaline earth metals) are highly reactive because of their low ionization energy. They are highly electropositive forming positive ions. So they are never found in free state.
They are widely distributed in nature in the combined state. They occur in earth’s crust in the form of oxides, chlorides, silicates and carbonates.
Generally group I metals are prepared by the electrolysis of fused solution.
As for example:

These metals are highly reactive and therefore cannot be extracted by the usual methods, because they are strong reducing agents.

Question 7.
What is the effect of heat on the following compounds ?
(a) Magnesium chloride hexahydrate
(b) Gypsum
(c) Magnesium sulphate heptahydrate
Answer:

Question 8.
What is the approximate composition of Portland cement ? What raw materials are used in the manufacture of this cement ? Describe method.
Answer:
Raw materials : The raw materials required for the manufacture of cement are lime stone, stone and clay. Lime stone in calcium carbonate, (CaCO₃) and it provides calcium oxide. (CaO) Clay is hydrated aluminium silicate, (Al2O₃.2SiO₂.2H₂O) and it provides alumina as well as silica. A small amount of gypsum, CaSO₄.2H₂O is also required. It is added in calculated quantity in order to adjust the rate of setting of cement.

Manufacture : Cement is made by strongly heating a mixture of lime stone and clay in a rotatory kiln. Lime stone and clay are finely powdered and a little water is added to get a thick paste called slurry. The slurry is fed into a rotatory kiln from the top through the hopper.

The hot gases produce a temperature of about 1770-1870 K in the kiln. At this high temperature at the lime stone and clay present in slimy combine to form cement in the form of small pieces called clinker. This clinker is mixed with 2-3% by weight of gypsum (CaSO₄.2 H₂O) to regulate the setting time and is then ground to an exceedingly fine powder.
\(\text { Limestone }+\text { Clay } \frac{170-1870 \mathrm{K}}{\text { (Clinker) }} \text { , Cement }+\mathrm{CO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O} \uparrow\)
When mixed with water the cement reacts to form gelatinous mass which sets to a hard mass when three .dimensional cross lines are formed between silica oxygen silica and silica oxygen aluminium as .
…….. Si – O – Si …….. and Si – O – Al ……….. chains
Composition of cement:
CaO = 50 -60%
SiO₂ = 20 – 25%
Al₂O₃ = 5 – 10%
MgO = 2 – 3%
Fe₂G₃ = 1-2%
SO₃ = 1 – 2%
For a good quality cement the ratio of’silica (SiO₂) and alumina (AI₂O₃) should be between 2.5 to 4.0. Similarly the ratio of lime (CaO) to the total oxide mixtures consisting of SiO₂, Al₂O₃ and Fe₂O₃ should be roughly 2:1:1, If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than the required, the cement is weak in strength. Therefore, a proper composition of cement must be maintained to get cement of good quality.

Question 9.
Identify A and B in the following reaction


Answer:
(i) ‘A’ is BeCh and ‘B’ is AlCl₃
(ii) A is CaCO₃, B is CO₂

Question 10.
A white crystalline solid ‘A’ on heating loses the water of crystallization to form a monohydrate ‘B’ above 373K, the monohydrate also becomes completely anhydrous and changes to white powder called soda ash. Identify ‘A’ and ‘B’. Also give two uses of ‘A’
Answer:
‘A’ is Na2CO3.10H₂O (Washing soda)
‘B’ is Na₂CO₃ (Anhydrous sodium carbonate)

Uses of‘A’
(a) Used for softening hard water.
(b) Used in glass and soap industries.

Question 11.
Write the uses, and any two reaction of KO₂.
Answer:
Potassium superoxide (KO₂) is used as a source of oxygen in submarines, space shuttles and in emergency breathing apparatus such as oxygen masks. Such masks are used in rescue work in mines and in other areas where the air is so deficient in oxygen that an artificial atmosphere must be generated.

The moisture of the breath reacts with superoxide to liberate oxygen, and at that same time the potassium hydroxide formed removes carbon dioxides as it is exhaled thereby allowing the atmosphere in the mask to be continuously regenerated.
4KO₂(s) + 2H₂O(g) → 4KOH(aq) + 3₂(g) → KHCO₃ (s)
KO₂ also combines directly with CO₂ forming K₂CO₃ and with CO₂ & moisture forming KHCO₃
4KO₂ + 2CO₂ → 2K₂CO₃ + 3O₂; 4KO₂ + 4CO₂ + 2H₂O → 4KHCO₃ + 3O₂

Question 12.
(a) Write any four uses of Calcium Hydroxide.
(b) Give chemical equation of the reaction of caustic soda with
1. ammonium chloride, and
2. carbon dioxide
Answer:
(a)

• It is used in the building material in the form of mortar.
• It is used in the manufacture of bleaching powder.
• It is used in glass making and the purification of sugar.
• It is used to absorbed acidic gases.

(b)
1. NH₄Cl + NaOH → NaCl + NH₄OH
2. 2NaOH + CO₂ → Na₂CO₃ + H₂O

Question 13.
What is plaster of paris ? How is it prepared ? Give its any two important uses.
Answer:
Plaster of paris is CaSO₄ \(\frac { 1 }{ 2 }\) H₂O. It is prepared by heating gypsum at 373K

Uses:

• It is used for manufacture of statues
• It is used for filling gaps before white washing

Question 14.
Discuss the trends of:
1. Thermal stability of alkaline earth metal carbonates
2. Solubility of sulphates of group 2 elements
3. Basic strength of alkaline earth metal hydroxides
Answer
1. Thermal stability of alkaline earth metal carbonates increases down the group due to increase in ionic character and therefore, increase in lattice energy

2. Solubility of sulphates of group 2 elements decreases down the group because lattice energy dominates over hydration energy.

3. Basic strength of alkaline earth metal hydroxides increases down the group because ionization energy of metal decreases and electropositive character increases down the group.

Question 15.
Explain the different oxides of metals or classify different metal oxides.
Answer:
Classification of oxides on the basis of oxygen content.
On the basis of oxygen content, oxides can be classified into the following types
1. Normal oxides: Those oxides inwhich the oxidation number of the element (M) can be deducted from the empirical formula MxOy by taking the oxidation number of oxygen as – 2 are called normal oxides. For example, H₂O, Na₂O, MgO AI₂O₃, CO₂ etc. All these oxides contain M – O bonds.

2. Polyoxides: These oxides contain more oxygen than would be expected from the oxidation number of the element (M). These have have been further classified into peroxides, and superoxides.

a : Peroxides : Metallic axides which on treatment with dilute acids produce hydrogen peroxide are called peroxides. For example, Na₂CO₂ and BaO₂. In these peroxides, the two oxygen atoms are linked by a single hand and each oxygen atom has an oxidation state -1. In other words, all peroxides contain a peroxide ion \(\left(0_{2}^{2-}\right)\) having the structure. In this structure,  all the electrons are paired and hence all peroxides are diamagnetic.

There are certain other oxides like PbCO₂ and MnO₂ which may be mistaken as peroxides. These compounds, however, do not give H₂O₂ on treatment with dilute acids. As such these compounds do not contain a peroxide ion \(\left(0_{2}^{2-}\right)\) and hence they cannot be called as peroxides. Actually in these compounds the two oxygen atoms are linked to the metal atom by a double bond and hence called dioxide i.e, 0 = Pb = 0 (Lead dioxide) and O = Mn = O (mangenses dioxide). In dioxides, the oxidation state of each oxygen atom is -2.

b. Superoxides: Besides peroxides, alkali metals also form higher oxides called superoxides. For example, potassium superoxide (KO₂), rubidium superoxide (RbO₂) cesium superoxide (CSO₂) etc. All these superoxides contain a superoxide ion, i.e., \(\mathrm{O}_{2}^{-}\) having the structure,  Thus all superoxides contain an odd number of electrons (i.e. 13) and hence are paramagnetic.

1st PUC Chemistry The S-Block Elements Give Reasons

Question 1.
The ionic compounds of alkali metals are colourless, why?
Answer:
Alkali metals form unipositive ions which have stable configuration of the nearest inert gas. Alkali metal salts are diamagnetic and colorless because they do not have impaired electrons.

Question 2.
Alkali metals are good conductors of electricity why ?
Answer:
Alkali metals have low ionization energy, Hence they show metallic character. They are good conductors of electricity due to the presence of mobile valance electrons