MCQ 11 Mark
Following are the graphs of elastic materials. Which one corresponds to that of brittle material?
- ✓
- B
- C
- DNone of these
Answer
View full question & answer→Correct option: A.
Questions
M.C.Q (1 Marks)
Take a timed test
65 questions · auto-graded multiple-choice test.
MCQ 21 Mark
Two wires $A$ and $B$ of the same material have radii in the ratio $2 : 1$ and lengths in the ratio $4 : 1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is:
- ✓$1 : 1$
- B$2 : 1$
- C$1 : 2$
- D$1 : 4$
Answer
View full question & answer→Correct option: A.
$1 : 1$
MCQ 31 Mark
Modulus of rigidity of ideal liquids is
- AInfinity.
- ✓Zero.
- CUnity.
- DSome finite small non $-$ zero constant value.
Answer
View full question & answer→Correct option: B.
Zero.
As the liquid is ideal, hence it does not have frictional force among it's layers, thus the tangential forces are zero as there is no stress developed. This verifies.
MCQ 41 Mark
The load versus elongation graph for four wires of the same material is shown in the fig. The thinnest wire is represented by the line.
- A$OC$
- B$OD$
- ✓$OA$
- D$OB$
Answer
View full question & answer→Correct option: C.
$OA$
For a thinnest wire, the elongation in the wire will be maximum for a given load, which is so corresponding to line $OA.$
MCQ 51 Mark
A mild steel wire of length $2L$ and cross $-$ sectional area $A$ isstretched, well within elastic limit, horizontally between two pillars. A mass $m$ is suspended from the mid point of the wire.Strain in the wire is 
- ✓$\frac{\text{x}^2}{2\text{L}^2}$
- B$\frac{\text{x}}{\text{L}}$
- C$\frac{\text{x}^2}{\text{L}}$
- D$\frac{\text{x}^2}{\text{2L}}$
Answer
View full question & answer→Correct option: A.
$\frac{\text{x}^2}{2\text{L}^2}$
$\Delta\text{l}=(\text{AO+BO})-\text{AB}$
$\Delta\text{L}=2\text{AO}-\text{2l}=2[\text{AO}-\text{l}]$
$=2\Big[\big(\text{l}^2+\text{x}^2\big)^{\frac{1}{2}}-1\Big]$
$=2\text{l}\Big[\Big(1+\frac{\text{x}^2}{\text{l}^2}\Big)^{\frac{1}{2}}-1\Big]$
$\Delta\text{l}=2\text{l}\Big[1+\frac{\text{x}^2}{\text{2l}^2}-1\Big]$
$=2\text{l}.\frac{\text{x}^2}{2\text{l}^2}=\frac{\text{x}^2}{\text{l}}$
Strain $=\frac{\Delta\text{l}}{2\text{l}}=\frac{\frac{\text{x}^2}{\text{l}}}{\text{2l}}=\frac{\text{x}2}{2\text{l}^2}$
verifies the Option $(a).$
$\Delta\text{L}=2\text{AO}-\text{2l}=2[\text{AO}-\text{l}]$
$=2\Big[\big(\text{l}^2+\text{x}^2\big)^{\frac{1}{2}}-1\Big]$
$=2\text{l}\Big[\Big(1+\frac{\text{x}^2}{\text{l}^2}\Big)^{\frac{1}{2}}-1\Big]$
$\Delta\text{l}=2\text{l}\Big[1+\frac{\text{x}^2}{\text{2l}^2}-1\Big]$
$=2\text{l}.\frac{\text{x}^2}{2\text{l}^2}=\frac{\text{x}^2}{\text{l}}$
Strain $=\frac{\Delta\text{l}}{2\text{l}}=\frac{\frac{\text{x}^2}{\text{l}}}{\text{2l}}=\frac{\text{x}2}{2\text{l}^2}$
verifies the Option $(a).$
MCQ 61 Mark
Two wires of the same material and length but diameter in the ratio $1 : 2$ are stretched by the same load. The ratio of elastic potential energy per unit volume for the two wires is:
- A$1 : 1$
- B$2 : 1$
- C$4 : 1$
- ✓$16 : 1$
Answer
View full question & answer→Correct option: D.
$16 : 1$
MCQ 71 Mark
A spring is stretched by applying a load to its free end. The strain produced in the spring is
- AVolumetric.
- BShear.
- ✓Longitudinal and shear.
- DLongitudinal.
Answer
View full question & answer→Correct option: C.
Longitudinal and shear.
When a spring is stretched by a load its shape $($shear$)$ and length changes. So strain produced is shearing and longitudinal strain.
MCQ 81 Mark
The stress $-$ strain graphs for two materials are shown in $($assume same scale$).$
- AMaterial $(ii)$ is more elastic than material $(i)$ and hence material $(ii)$ is more brittle.
- BMaterial $(ii)$ is more brittle than material $(i).$
- CMaterial $(ii)$ is elastic over a larger region of strain as compared to $(i).$
- ✓$B$ and $C$
Answer
View full question & answer→Correct option: D.
$B$ and $C$
On comparing ultimate tensile strength of the materials, $(ii)$ is greater than $(i).$
Hence, material $(ii)$ is elastic over larger region as compare to $(i)$ so the material $(ii)$ is elastic over a larger region of strain as compared to $(i) ($verifies option $b).$
As the fracture point of material $(ii)$ is nearer than $(i),$ hence the material $(ii)$ is more brittle than material $(i).$
Hence, material $(ii)$ is elastic over larger region as compare to $(i)$ so the material $(ii)$ is elastic over a larger region of strain as compared to $(i) ($verifies option $b).$
As the fracture point of material $(ii)$ is nearer than $(i),$ hence the material $(ii)$ is more brittle than material $(i).$
MCQ 91 Mark
Elasticity of a material can be altered by:
- AAnnealing.
- BHammering.
- CAdding impurities.
- ✓All of the above.
Answer
View full question & answer→Correct option: D.
All of the above.
MCQ 101 Mark
A wire suspended vertically from one end, is stretched by attaching a weight $200N$ to the lower end. The weight stretches the wire by $1mm.$ The energy gained by the wire is:
- ✓$0.1J$
- B$0.2J$
- C$0.4J$
- D$10J$
Answer
View full question & answer→Correct option: A.
$0.1J$
MCQ 111 Mark
A wire of length $L$ and radius $r$ is rigidly fixed at one end. On stretching the other end of the wire with a force $F,$ the increase in its length is $l.$ If another wire of same material but of length $2L$ and radius $2r$ is stretched with a force of $2F,$ the increase in its length will be:
- ✓$\text{l}$
- B$2\text{l}$
- C$\frac{\text{l}}{2}$
- D$\frac{\text{l}}{4}$
Answer
View full question & answer→Correct option: A.
$\text{l}$
MCQ 121 Mark
$A$ and $B$ are two wires. The radius of $A$ is twice that of $B.$ They are stretched by the same load. Then, the stress on $B$ is:
- AEqual to that on $A.$
- ✓Four times that on $A.$
- CTwo times that on $A.$
- DHalf that on $A.$
Answer
View full question & answer→Correct option: B.
Four times that on $A.$
MCQ 131 Mark
A rod elongates by l when a body of mass $M$ is suspended from it. The work done is:
- A$\text{Mgl}$
- ✓$\frac{1}{2}\text{mgl}$
- C$\text{2Mgl}$
- DZero.
Answer
View full question & answer→Correct option: B.
$\frac{1}{2}\text{mgl}$
Work done $=\frac{1}{2}\text{F}\times\Delta\text{l}=\frac{1}{2}\text{Mgl}.$
MCQ 141 Mark
The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will
- ABe double.
- BBe half.
- CBe four times.
- ✓Remain same.
Answer
View full question & answer→Correct option: D.
Remain same.
$\text{Breaking stress}=\frac{\text{Breaking force}}{\text{Area of cross - section}}$
Since breaking force doesn't depend on length, hence changing the cross section has no effect.
So the breaking force remain same.
Since breaking force doesn't depend on length, hence changing the cross section has no effect.
So the breaking force remain same.
MCQ 151 Mark
A long spring is stretched by $2\ cm$ and its potential energy is $V.$ If the spring is stretched by $10\ cm,$ its potential energy will be:
- A$\frac{\text{V}}{5}$
- B$\frac{\text{V}}{25}$
- C$5\text{V}$
- ✓$25\text{V}$
Answer
View full question & answer→Correct option: D.
$25\text{V}$
$P.E.$ of a stretched spring, $\text{V}=\frac{1}{2}\text{kx}^2,$
where $k$ is the spring constant,
$\therefore\text{V}=\frac{1}{2}\text{k}\times2^2$ or $\text{k}=\frac{\text{V}}{2}$
And Now, $P.E., V' =\frac{1}{2}\text{k}\times10^2$
$=\frac{1}{2}\Big(\frac{\text{V}}{2}\Big)\times100$
$=25\text{V}$
where $k$ is the spring constant,
$\therefore\text{V}=\frac{1}{2}\text{k}\times2^2$ or $\text{k}=\frac{\text{V}}{2}$
And Now, $P.E., V' =\frac{1}{2}\text{k}\times10^2$
$=\frac{1}{2}\Big(\frac{\text{V}}{2}\Big)\times100$
$=25\text{V}$
MCQ 161 Mark
One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the roof and weight $W_1$ is suspended from the lower end. If $S$ is the area of cross$-$section of the wire, the stress in the wire at a height $\frac{3}{4}$ from its lower end is;
- A$\frac{\text{W}_1}{\text{S}}$
- B$\frac{\Big(\text{W}_1+\frac{\text{W}}{4}\Big)\text{S}}{\text{S}}$
- ✓$\frac{\Big(\text{W}_1+\frac{3\text{W}}{4}\Big)}{\text{S}}$
- D$\frac{(\text{W}_1+\text{W})}{\text{S}}$
Answer
View full question & answer→Correct option: C.
$\frac{\Big(\text{W}_1+\frac{3\text{W}}{4}\Big)}{\text{S}}$
Total force acting at a point $\frac{3\text{L}}{4}$ from its lowest end $=$ Weight suspended $+$ Weight of the wire of length $\frac{3\text{L}}{4}=\Big(\text{W}_1+\frac{3\text{W}}{4}\Big).$
Hence stress,
$=\frac{\text{Force}}{\text{Area}}=\frac{\Big(\text{W}_1+\frac{3\text{W}}{4}\Big)}{\text{S}}$
Hence stress,
$=\frac{\text{Force}}{\text{Area}}=\frac{\Big(\text{W}_1+\frac{3\text{W}}{4}\Big)}{\text{S}}$
MCQ 171 Mark
A wire fixed at the upper end stretches by length $I$ by applying a force $F.$ The work done in stretching is:
- A$\frac{\text{F}}{2\text{l}}$
- B$\text{F}\text{l}$
- C$2\text{Fl}$
- ✓$\frac{\text{Fl}}{2}$
Answer
View full question & answer→Correct option: D.
$\frac{\text{Fl}}{2}$
Work done $=$ Average force $\times$ Extension, $=\Big(\frac{0+\text{F}}{2}\Big)\times\text{l}=\frac{1}{2}\text{Fl}$
MCQ 181 Mark
The upper end of a wire of radius $4\ mm$ and length $100\ cm$ is clamped and its other end is twisted through an angle of $30^\circ .$ Then, angle of shear is:
- A$12^\circ$
- ✓$0.12^\circ$
- C$1.2^\circ$
- D$0.012^\circ$
Answer
View full question & answer→Correct option: B.
$0.12^\circ$
MCQ 191 Mark
A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of $1\ cm.$ The two wires will have
- AThe same stress and strain.
- ✓The same stress but different strain.
- CThe same strain but different stress.
- DDifferent strains and stress.
Answer
View full question & answer→Correct option: B.
The same stress but different strain.
MCQ 201 Mark
The length of a wire increases by $1\%$ by a load of $2\ \text{kg-wt.}$ The linear strain produced in the wire will be:
- A$0.02$
- B$0.001$
- ✓$0.01$
- D$0.002$
Answer
View full question & answer→Correct option: C.
$0.01$
MCQ 211 Mark
The potential energy $U$ between two molecules as a function of the distance $r$ between them has been shown in the adjoining figure. The two molecules are:
- AAttracted when $r$ lies between $A$ and $B$ and repelled when $r$ lies between $B$ and $C.$
- ✓Attracted when $r$ lies between $B$ and $C$ and repelled when $r$ lies between $A$ and $B.$
- CAttracted when they reach $B.$
- DRepelled when they reach $B.$
Answer
View full question & answer→Correct option: B.
Attracted when $r$ lies between $B$ and $C$ and repelled when $r$ lies between $A$ and $B.$
MCQ 221 Mark
A steel ring of radius $r$ and cross$-$section area $A$ is shifted on to a wooden disc of radius $R(R > r)$. If Young's modulus be $E$, then the force with which the steel ring is expanded is:
- A$\frac{\text{AER}}{\text{T}}$
- ✓$\frac{\text{AE}(\text{R}-\text{r})}{\text{r}}$
- C$\frac{\text{E}(\text{R}-\text{r})}{ \text{Ar}}$
- D$\frac{\text{Er}}{\text{AR}}$
Answer
View full question & answer→Correct option: B.
$\frac{\text{AE}(\text{R}-\text{r})}{\text{r}}$
MCQ 231 Mark
On suspending a weight $Mg,$ the length $l$ of elastic wire having area of cross$-$section $A,$ becomes double the initial length. The instantaneous stress action on the wire is:
- A$\frac{\text{Mg}}{\text{A}}$
- B$\frac{\text{Mg}}{2\text{A}}$
- ✓$\frac{2\text{Mg}}{\text{A}}$
- D$\frac{4\text{Mg}}{\text{A}}$
Answer
View full question & answer→Correct option: C.
$\frac{2\text{Mg}}{\text{A}}$
MCQ 241 Mark
The property of a body by virtue of which it tends to regain its original size and shape of a body when applied force is removed, is known as:
- AFluidity.
- ✓Elasticity.
- CPlasticity.
- DRigidity.
Answer
View full question & answer→Correct option: B.
Elasticity.
The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and the deformation caused is known as elastic deformation.
MCQ 251 Mark
A uniform bar of square cross$-$section is lying along a frictionless horizontal surface. A horizontal force is applied to pull it from one of its ends, then:
- AThe bar is under same stress throughout its length.
- BThe bar is not under any stress because force has been applied only at one end.
- CThe bar simply moves without any stress in it.
- ✓The stress developed gradually reduces to zero at the end of the bar where no force is applied.
Answer
View full question & answer→Correct option: D.
The stress developed gradually reduces to zero at the end of the bar where no force is applied.
MCQ 261 Mark
Young's modulus of a material has the same unit as:
- AStress.
- BEnergy.
- CCompressibility.
- ✓Pressure.
Answer
View full question & answer→Correct option: D.
Pressure.
Pressure.
MCQ 271 Mark
On applying a stress of $20 \times 10^8 Nm ^{-2}$, the length of a perfectly elastic wire is doubled. Its Young's modulus will be:
- A$40 \times 10^8Nm^{-2}$
- ✓$20 \times 10^8Nm^{-2}$
- C$10 \times 10^8Nm^{-2}$
- D$5 \times 10^8Nm^{-2}$
Answer
View full question & answer→Correct option: B.
$20 \times 10^8Nm^{-2}$
$20 \times 10^8Nm^{-2}$
MCQ 281 Mark
Amaterial has Poisson's ratio $0.5.$ If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is:
- A$0.6$
- B$0.4$
- C$0.2$
- ✓Zero.
Answer
View full question & answer→Correct option: D.
Zero.
As, the Poisson's ratio of material is $0.5$, so there is no change in volume.
MCQ 291 Mark
A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
- ATensile stress at any cross section $A$ of the wire is $F/ A.$
- BTension at any cross section $A$ of the wire is $F.$
- CTensile stress at any cross section $A$ of the wire is $2F/ A.$
- ✓$A$ and $B$
Answer
View full question & answer→Correct option: D.
$A$ and $B$
Stress $=\frac{\text{F}}{\text{A}}$ verifies option $(a).$
Here, Tension is balanced by force $F.$
Hence, $T = F$ verifies option $(d).$
Here, Tension is balanced by force $F.$
Hence, $T = F$ verifies option $(d).$
MCQ 301 Mark
Elasticity is due to:
- ADecrease of $PE$ with separation between atoms/ molecules.
- BIncrease of $PE$ with separation between atoms/ molecules.
- ✓Asymmetric nature of $PE$ curve.
- DNone of the above.
Answer
View full question & answer→Correct option: C.
Asymmetric nature of $PE$ curve.
MCQ 311 Mark
A uniform cube is subjected to volume compression. If each side is decreased by $1\%$, then bulk strain is:
- A$0.01$
- B$0.06$
- C$0.02$
- ✓$0.03$
Answer
View full question & answer→Correct option: D.
$0.03$
MCQ 321 Mark
A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of $1\ cm.$ The two wires will have
- AThe same stress.
- BDifferent strain.
- CThe same strain.
- ✓$A$ and $B$
Answer
View full question & answer→Correct option: D.
$A$ and $B$
$\because\text{stress}=\frac{\text{F}}{\text{A}}$
$\because$ area of cross section for both wire same and stretched by same force.
So their stress are equal verifies option $(a).$
$\text{strain}=\frac{\text{stress}}{\text{Y}}$
As stress for both wires are same,
$\text{strain}_\text{steel}\propto=\frac{\text{1}}{\text{Y}_\text{steel}}$ and $\text{(strain)}_\text{Al}\propto\frac{1}{\text{Y}_\text{Al}}$
$\frac{\text{strain}_\text{steel}}{\text{(strain)}_\text{Al}}=\frac{\text{Y}_\text{Al}}{\text{Y}_\text{steel}}$
$\text{Y}_\text{Al}<\text{Ys}$ So $\frac{\text{Al}}{\text{steel}}$
or $(\text{strain})_\text{steel}<(\text{strain})_\text{Al}$
Verifies option $(d).$
$\because$ area of cross section for both wire same and stretched by same force.
So their stress are equal verifies option $(a).$
$\text{strain}=\frac{\text{stress}}{\text{Y}}$
As stress for both wires are same,
$\text{strain}_\text{steel}\propto=\frac{\text{1}}{\text{Y}_\text{steel}}$ and $\text{(strain)}_\text{Al}\propto\frac{1}{\text{Y}_\text{Al}}$
$\frac{\text{strain}_\text{steel}}{\text{(strain)}_\text{Al}}=\frac{\text{Y}_\text{Al}}{\text{Y}_\text{steel}}$
$\text{Y}_\text{Al}<\text{Ys}$ So $\frac{\text{Al}}{\text{steel}}$
or $(\text{strain})_\text{steel}<(\text{strain})_\text{Al}$
Verifies option $(d).$
MCQ 331 Mark
Wire $A$ and $B$ are made from the same material $A$ has twice the diameter and three times the length of $B$. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in $A$ to that in $B$ is:
- A$2 : 3$
- ✓$12 : 1$
- C$3 : 2$
- D$6 : 1$
Answer
View full question & answer→Correct option: B.
$12 : 1$
Given $D_A=2 D ; I_A=3 I, D_B=D, I_B=I$
$F_A=F=F_B, Y_A=Y_B=Y$
Energy stored $(E) =\frac{1}{2}\times\frac{(\text{Stress})^2}{\text{Y}}\times\text{Volume}$
$\therefore\text{E}_\text{A}=\frac{\frac{\text{F}}{\pi}(2\text{D})^2}{\text{Y}}\times\frac{\pi(2\text{D})^2}{4}\times3\text{l}$
$\text{E}_\text{B}=\frac{1}{2}\frac{\frac{\text{F}}{\pi}\text{D}^2}{\text{Y}}\times\frac{\pi\text{D}^2}{4}\times\text{l}$
$\frac{\text{E}_\text{A}}{\text{E}_\text{B}}=\frac{12}{1}$
$F_A=F=F_B, Y_A=Y_B=Y$
Energy stored $(E) =\frac{1}{2}\times\frac{(\text{Stress})^2}{\text{Y}}\times\text{Volume}$
$\therefore\text{E}_\text{A}=\frac{\frac{\text{F}}{\pi}(2\text{D})^2}{\text{Y}}\times\frac{\pi(2\text{D})^2}{4}\times3\text{l}$
$\text{E}_\text{B}=\frac{1}{2}\frac{\frac{\text{F}}{\pi}\text{D}^2}{\text{Y}}\times\frac{\pi\text{D}^2}{4}\times\text{l}$
$\frac{\text{E}_\text{A}}{\text{E}_\text{B}}=\frac{12}{1}$
MCQ 341 Mark
A cube of aluminum of side $0.1m$ is subjected to a shearing force of $100N.$ The top face of the cube is displaced through $0.02\ cm$ with respect to the bottom face. The shearing strain would be:
- A$0.02$
- B$0.1$
- C$0.005$
- ✓$0.002$
Answer
View full question & answer→Correct option: D.
$0.002$
MCQ 351 Mark
A wire of diameter $1\ mm$ breaks under a tension of $1000N.$ Another wire of same material as that of the first one, but of diameter $2\ mm$ breaks under a tension of:
- A$500N$
- B$1000N$
- C$10000N$
- ✓$4000N$
Answer
View full question & answer→Correct option: D.
$4000N$
MCQ 361 Mark
A wire of length $L$ and cross section $A$ is make of material of Young's modulus $Y.$ It is stretched by an amount $x.$ the work done is:
- A$\frac{\text{YxA}}{2\text{L}}$
- B$\frac{\text{Yx}^2\text{A}}{\text{L}}$
- ✓$\frac{\text{Yx}^2\text{A}}{2\text{L}}$
- D$\frac{2\text{Yx}^2\text{A}}{\text{L}}$
Answer
View full question & answer→Correct option: C.
$\frac{\text{Yx}^2\text{A}}{2\text{L}}$
Here, $\Delta\text{l}=\text{x},\text{Y}=\frac{\frac{\text{F}}{\text{A}}}{\frac{\Delta\text{l}}{\text{L}}}\ \text{or}\ \frac{\text{YA}\Delta\text{l}}{\text{L}}$
Te work in done form $0$ to $x($change in length$)$,
so the average distance, $=\frac{0+\Delta\text{l}}{2}=\frac{\Delta\text{l}}{2}$
Work done $=$ Force $\times$ Distance,
$=\frac{\text{YA}\Delta\text{l}}{\text{L}}\times\frac{\Delta\text{l}}{2}$
$=\frac{\text{YA}(\Delta\text{l})^2}{2\text{L}}$
$=\frac{\text{YAx}^2}{2\text{L}}$
Te work in done form $0$ to $x($change in length$)$,
so the average distance, $=\frac{0+\Delta\text{l}}{2}=\frac{\Delta\text{l}}{2}$
Work done $=$ Force $\times$ Distance,
$=\frac{\text{YA}\Delta\text{l}}{\text{L}}\times\frac{\Delta\text{l}}{2}$
$=\frac{\text{YA}(\Delta\text{l})^2}{2\text{L}}$
$=\frac{\text{YAx}^2}{2\text{L}}$
MCQ 371 Mark
Dimensional formula of stress is same as that of:
- AImpulse.
- BStrain.
- CForce.
- ✓Pressure.
Answer
View full question & answer→Correct option: D.
Pressure.
MCQ 381 Mark
Two rods of different materials having coefficient of thermal expansion $\alpha_1,\alpha_2$ and Young's modulus $Y_1, Y_2$, respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If $\alpha_1:\alpha_2=2:3,$ the thermal stresses developed in the two rods are equal provided $Y_1 : Y_2$ is equal to:
- A$2 : 3$
- B$1 : 1$
- ✓$3 : 2$
- D$4 : 9$
Answer
View full question & answer→Correct option: C.
$3 : 2$
Expansion in rod due to rise in temperature $=$ compression in rod,
$\therefore\frac{\alpha_1}{\alpha_2}=\frac{\text{Y}_1}{\text{Y}_2}$ or $\frac{\alpha_2}{\alpha_1}=\frac{3}{2}$
$\therefore\frac{\alpha_1}{\alpha_2}=\frac{\text{Y}_1}{\text{Y}_2}$ or $\frac{\alpha_2}{\alpha_1}=\frac{3}{2}$
MCQ 391 Mark
Which of the following is not a unit of Young's modulus?
- A$\text{Nm}^{-2}$
- BMega Pascal $\text{(MPa)}$
- C$\text{dyne cm}^{-2}$
- ✓$\text{Nm}^{-1}$
Answer
View full question & answer→Correct option: D.
$\text{Nm}^{-1}$
MCQ 401 Mark
A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports. It can be done in one of the following three ways;
The tension in the strings will be
The tension in the strings will be
- AThe same in all cases.
- BLeast in $(a).$
- ✓Least in $(b)$.
- DLeast in $(c).$
Answer
View full question & answer→Correct option: C.
Least in $(b)$.
According to the $\text{FBD}$ diagram of the rectangular frame.
Let $M$ be the mass of rectangular frame and $0$ be the angle which the tension $T$ in the string makes with the horizontal
Balancing vertical forces,
$2\text{T}\sin\theta-\text{mg}=0 [T$ is tension in the string$]$
$\Rightarrow2\text{T}\sin\theta=\text{mg}$
Total horizontal force will be zero because of equal and opposite forces balance each other $(\text{T}\cos\theta).$
Now from Eq. $(i), \text{T}=\frac{\text{mg}}{2\sin\theta}$
Here, we know $mg$ is constant.
$\Rightarrow\text{T}\propto\frac{1}{\sin\theta}$
$T$ is least if $\sin\theta$ has maximum value.
$\text{T}_\text{min}=\frac{\text{mg}}{2\sin\theta_\text{max}}\text{ (since,}\sin\theta_{\text{max}}=1)$
$\sin\theta_\text{max}=1\Rightarrow\theta=90^\circ$
So the correct option is option $(c).$
Hence, tension is least for the case $(b).$
Let $M$ be the mass of rectangular frame and $0$ be the angle which the tension $T$ in the string makes with the horizontal
Balancing vertical forces,
$2\text{T}\sin\theta-\text{mg}=0 [T$ is tension in the string$]$
$\Rightarrow2\text{T}\sin\theta=\text{mg}$
Total horizontal force will be zero because of equal and opposite forces balance each other $(\text{T}\cos\theta).$
Now from Eq. $(i), \text{T}=\frac{\text{mg}}{2\sin\theta}$
Here, we know $mg$ is constant.
$\Rightarrow\text{T}\propto\frac{1}{\sin\theta}$
$T$ is least if $\sin\theta$ has maximum value.
$\text{T}_\text{min}=\frac{\text{mg}}{2\sin\theta_\text{max}}\text{ (since,}\sin\theta_{\text{max}}=1)$
$\sin\theta_\text{max}=1\Rightarrow\theta=90^\circ$
So the correct option is option $(c).$
Hence, tension is least for the case $(b).$
MCQ 411 Mark
For an ideal liquid
- AThe bulk modulus is infinite.
- BThe shear modulus is zero.
- CThe shear modulus is infinite.
- ✓Both $A$ and $B$
Answer
View full question & answer→Correct option: D.
Both $A$ and $B$
An ideal liquid is not compressible.
Bulk modulus $(\text{K})=\frac{-\text{p(V)}}{\Delta\text{V}} (\because\Delta\text{V=0})$
$\because\Delta\text{V=0}$ for ideal liquid.
$\therefore\text{K=}\infty$ for ideal liquid.
As there is no net tangential force on liquid $(S.T.$ is all around on a particle$)$ so shearing strain $\Delta\theta=0$ and $F=0.$
$\eta=\frac{\text{F/A}}{\Delta\theta}=\frac{0}{0}$ indeterminant value.
Hence, verifies option $(a)$ and $(b)$ and rejects option $(c).$
Bulk modulus $(\text{K})=\frac{-\text{p(V)}}{\Delta\text{V}} (\because\Delta\text{V=0})$
$\because\Delta\text{V=0}$ for ideal liquid.
$\therefore\text{K=}\infty$ for ideal liquid.
As there is no net tangential force on liquid $(S.T.$ is all around on a particle$)$ so shearing strain $\Delta\theta=0$ and $F=0.$
$\eta=\frac{\text{F/A}}{\Delta\theta}=\frac{0}{0}$ indeterminant value.
Hence, verifies option $(a)$ and $(b)$ and rejects option $(c).$
MCQ 421 Mark
Elasticity is shown by materials because inter$-$atomic or inter$-$molecular forces:
- AIncreases when a body is deformed.
- BDecreases when a body is deformed.
- CRemains same when a body is deformed.
- ✓Becomes non$-$zero when a body is deformed.
Answer
View full question & answer→Correct option: D.
Becomes non$-$zero when a body is deformed.
When a body is deformed, atoms/molecules are displaced from their equilibrium positions $(F = 0)$.
As a result, there is a force $(\text{F}\not=0)$ acts between them to restore their position.
As a result, there is a force $(\text{F}\not=0)$ acts between them to restore their position.
MCQ 431 Mark
Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. $A$ mass $M$ is attached to each of the free ends at the centre of the rods.
- ABoth the rods will elongate but there shall be no perceptible change in shape.
- BThe steel rod will elongate and change shape but the rubber rod will only elongate.
- CThe steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.
- ✓The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
Answer
According to the diagram shown below in which a mass $AT$ is attached at the centre of each rod, then both rods will be elongated. But due to different elastic properties of material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
View full question & answer→Correct option: D.
The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
According to the diagram shown below in which a mass $AT$ is attached at the centre of each rod, then both rods will be elongated. But due to different elastic properties of material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
MCQ 441 Mark
Young's modulus of a wire depends on:
- ✓Its material.
- BIts length.
- CIts area of cross-section.
- DBoth $(b)$ and $(c).$
Answer
View full question & answer→Correct option: A.
Its material.
MCQ 451 Mark
A wire suspended vertically from one of its ends is stretched by attaching a weight of $200N$ to the lower end. The weight stretches the wire by $1\ mm.$ Then the elastic energy stored in the wire is:
- A$0.2J$
- B$10J$
- C$20J$
- ✓$0.1J$
Answer
View full question & answer→Correct option: D.
$0.1J$
$\text{U}-=\frac{1}{2}\text{F}\times\Delta\text{l}$
$=\frac{1}{2}\times200\times10^{-3}$
$=0.1\text{J}.$
$=\frac{1}{2}\times200\times10^{-3}$
$=0.1\text{J}.$
MCQ 461 Mark
Bulk modulus of water is $2 × 10^9N/ m^2$. The pressure required to increase the density of water by $0.1% in N/ m^2$ is:
- A$2 \times 10^9$
- B$2 \times 10^8$
- ✓$2 \times 10^6$
- D$2 \times 10^4$
Answer
View full question & answer→Correct option: C.
$2 \times 10^6$
$\Delta\text{p}=\text{k}\frac{\Delta\rho}{\rho}=2\times10^9\times\Big(\frac{0.1}{100}\Big)$
$=2\times10^6\text{N}/\ \text{m}^2.$
$=2\times10^6\text{N}/\ \text{m}^2.$
MCQ 471 Mark
In the given figure, if the dimension of the wire are the same and materials are different, Young's modulus is more for:
- ✓$A$
- B$B$
- CBoth.
- DNone of these.
Answer
View full question & answer→Correct option: A.
$A$
MCQ 481 Mark
Stress-strain curves for the material $A$ and $B$ are shown below: Then,
- A$A$ is brittle material.
- B$B$ is ductile material.
- ✓$B$ is brittle material.
- DBoth $(a)$ and $(b).$
Answer
View full question & answer→Correct option: C.
$B$ is brittle material.
$B$ is brittle as there is no plastic region. However, $A$ is ductile as it has large plastic range of extension.
MCQ 491 Mark
Elastic limit is equal to:
- AYoung's modulus.
- BModulus of rigidity.
- ✓Stress.
- DStrain.
Answer
View full question & answer→Correct option: C.
Stress.
MCQ 501 Mark
Which of the following statements is incorrect?
- AYoung's modulus and shear modulus are relevant only for solids.
- BBulk modulus is relevant for liquids and gases.
- CMetals have larger values of Young's modulus than elastomers.
- ✓Alloys have larger values of Young's modulus than metals.
Answer
View full question & answer→Correct option: D.
Alloys have larger values of Young's modulus than metals.
Metals have larger values of Young's modulus than alloy and elastomers.
MCQ 511 Mark
A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l.$ Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
- A$\text{Y}_\text{copper}/\text{Y}_\text{iron}.$
- ✓$\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}.$
- C$\frac{\text{Y}^2_\text{iron}}{\text{Y}^2_\text{copper}}.$
- D$\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}.$
Answer
View full question & answer→Correct option: B.
$\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}.$
As the bar is supported symmetrically by the three wires, therefore extension in each wire is same. Let $T$ be the tension in each wire and diameter of the wire is $D,$ then Young’s modulus is
$(\text{Y})=\frac{\text{Stress}}{\text{Strain}}=\frac{\text{F/A}}{\Delta\text{L}/\text{L}}=\frac{\text{F}}{\text{A}}\times\frac{\text{L}}{\Delta\text{L}}$
$=\frac{\text{F}}{\pi(\text{D/2})^2}\times\frac{\text{L}}{\Delta\text{L}}=\frac{4\text{FL}}{\pi\text{D}^2\Delta\text{L}}$
$\Rightarrow\ \text{D}^2=\frac{4\text{FL}}{\pi\Delta\text{LY}}\Rightarrow\ \text{D}=\sqrt{\frac{4\text{FL}}{\pi\Delta\text{LY}}}$
As $F$ and $\frac{\text{L}}{\Delta\text{L}}$ are constants.
Hence, $\text{D}\propto\sqrt{\frac{1}{\text{Y}}}$
or $\text{D}=\frac{\text{K}}{\sqrt{\text{Y}}} (K$ is the proportionality constant$)$
Now we can find ratio as $\frac{\text{D}_\text{copper}}{\text{D}_\text{iron}}=\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}$
$(\text{Y})=\frac{\text{Stress}}{\text{Strain}}=\frac{\text{F/A}}{\Delta\text{L}/\text{L}}=\frac{\text{F}}{\text{A}}\times\frac{\text{L}}{\Delta\text{L}}$
$=\frac{\text{F}}{\pi(\text{D/2})^2}\times\frac{\text{L}}{\Delta\text{L}}=\frac{4\text{FL}}{\pi\text{D}^2\Delta\text{L}}$
$\Rightarrow\ \text{D}^2=\frac{4\text{FL}}{\pi\Delta\text{LY}}\Rightarrow\ \text{D}=\sqrt{\frac{4\text{FL}}{\pi\Delta\text{LY}}}$
As $F$ and $\frac{\text{L}}{\Delta\text{L}}$ are constants.
Hence, $\text{D}\propto\sqrt{\frac{1}{\text{Y}}}$
or $\text{D}=\frac{\text{K}}{\sqrt{\text{Y}}} (K$ is the proportionality constant$)$
Now we can find ratio as $\frac{\text{D}_\text{copper}}{\text{D}_\text{iron}}=\sqrt{\frac{\text{Y}_\text{iron}}{\text{Y}_\text{copper}}}$
MCQ 521 Mark
$K$ is a force constant of a spring. The work done in increasing its extension from $l_1$ to $l_2$ will be:
- A$\text{K}(\text{l}_1-\text{l}_2)$
- B$\frac{\text{K}(\text{l}_1+\text{l}_2)}{2}$
- C$\text{K}\big(\text{l}^2_1+\text{l}^2_1\big)$
- ✓$\frac{\text{K}\big(\text{l}^2_2+\text{l}^2_1\big)}{2}$
Answer
View full question & answer→Correct option: D.
$\frac{\text{K}\big(\text{l}^2_2+\text{l}^2_1\big)}{2}$
The energy stored in the spring extension $\text{l}_1=\frac{1}{2}\text{Kl}^2_1.$
The energy stored in the spring for extension $\text{l}_2=\frac{1}{2}\text{Kl}^2_2.$
Therefore work done in increasing its extension for $\text{l}_1\ \text{l}_2=\frac{1}{2}\text{Kl}^2_2+\frac{1}{2}\text{Kl}^2_1$
The energy stored in the spring for extension $\text{l}_2=\frac{1}{2}\text{Kl}^2_2.$
Therefore work done in increasing its extension for $\text{l}_1\ \text{l}_2=\frac{1}{2}\text{Kl}^2_2+\frac{1}{2}\text{Kl}^2_1$
MCQ 531 Mark
A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel $($wire $A)$ and aluminium $($wire $B)$ of equal lengths. The cross $-$ sectional areas of wires $A$ and $B$ are $1.0\ mm^2$ and $2.0\ mm^2,$ respectively. $(Y_{al} = 70 \times 10Nm^{-2}$ and $Y_{steel} = 200 \times 10^9Nm^{-2})$
- AMass $m$ should be suspended close to wire $A$ to have equal stresses in both the wires.
- BMass $m$ should be suspended close to $B$ to have equal stresses in both the wires.
- CMass $m$ should be suspended close to wire $A$ to have equal strain in both wires.
- ✓Both $B$ and $C$
Answer
View full question & answer→Correct option: D.
Both $B$ and $C$
According to the diagram a massless rod is suspended at its two ends by two wires of steel $($wire $A)$ and aluminum $($wire $B)$ of equal lengths.
Let the mass is suspended at $x$ from the end $B,$ which develop equal stress in wires.
Let $T_a$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.
stress in steel wire $A, \text{S}_\text{A}=\frac{\text{T}_\text{A}}{\text{A}_\text{A}}=\frac{\text{T}_\text{A}}{10^{-6}}$
stress in $Al$ wire $\text{S}_\text{B}=\frac{\text{T}_\text{B}}{\text{A}_\text{B}}=\frac{\text{T}_\text{B}}{2\times10^{-6}}$
where $A_A$ and $A_B$ are cross $-$ sectional areas of wire $A$ and $B$ respectively.
Also, from rotational equilibrium, net torque is zero,
i.e. $\text{T}_\text{B}\text{x}-\text{T}_\text{A}(\text{l}-\text{x})=0$
$\Rightarrow\frac{\text{T}_\text{B}}{\text{T}_\text{A}}=\frac{\text{l}-\text{x}}{\text{x}} ...(\text{i})$
For equal stress, $S_A$ = $S_B$
$\Rightarrow\text{S}_\text{A}=\text{S}_\text{B}$
$\Rightarrow\frac{\text{T}_\text{A}}{10^{-6}}=\frac{\text{T}_\text{B}}{2\times10^{-6}}$
$\Rightarrow\frac{\text{l}-\text{x}}{\text{x}}=2$
$\Rightarrow\frac{\text{l}}{\text{x}}-1=2$
$\Rightarrow\text{x}=\frac{\text{l}}{\text{3}}$
$\Rightarrow\text{l}-\text{x}=\text{l}-\frac{\text{l}}{\text{3}}=\frac{2\text{l}}{3}$
Hence, mass $m$ should be suspended close to wire $B (Al$ wire$).$
We know, Strain $=\frac{\text{Stress}}{\text{Y}}$
So, for equal strain in the wires,
$\Rightarrow\ \frac{\text{S}_\text{A}}{\text{Y}_\text{Steel}}=\frac{\text{S}_\text{B}}{\text{Y}_\text{Al}}$
$\Rightarrow\ \frac{\text{Y}_\text{Steel}}{\text{T}_\text{A}/\text{a}_\text{A}}=\frac{\text{Y}_\text{Al}}{\text{T}_\text{B}/\text{a}_\text{B}}$
$\Rightarrow\ \frac{\text{Y}_\text{Steel}}{\text{Y}_\text{Al}}=\frac{\text{T}_\text{A}}{\text{T}_\text{B}}\times\frac{\text{a}_\text{B}}{\text{a}_\text{A}}=\Big(\frac{\text{x}}{\text{l}-\text{x}}\Big)\Big(\frac{2\text{a}_\text{A}}{\text{a}_\text{A}}\Big)$
$\Rightarrow\ \frac{200\times10^9}{70\times10^9}=\frac{\text{2x}}{\text{l}-\text{x}}$
$\Rightarrow\frac{20}{7}=\frac{\text{2x}}{\text{l}-\text{x}}$
$\Rightarrow17\text{x}=10\text{l}$
$\Rightarrow\text{x}=\frac{10\text{l}}{17}$
$\Rightarrow\text{l}-\text{x}=\text{l}-\frac{10\text{l}}{17}=\frac{7\text{l}}{17}$
Hence, mass $m$ should be suspended close to wire $A ($steel wire$).$
Let the mass is suspended at $x$ from the end $B,$ which develop equal stress in wires.
Let $T_a$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.
stress in steel wire $A, \text{S}_\text{A}=\frac{\text{T}_\text{A}}{\text{A}_\text{A}}=\frac{\text{T}_\text{A}}{10^{-6}}$
stress in $Al$ wire $\text{S}_\text{B}=\frac{\text{T}_\text{B}}{\text{A}_\text{B}}=\frac{\text{T}_\text{B}}{2\times10^{-6}}$
where $A_A$ and $A_B$ are cross $-$ sectional areas of wire $A$ and $B$ respectively.
Also, from rotational equilibrium, net torque is zero,
i.e. $\text{T}_\text{B}\text{x}-\text{T}_\text{A}(\text{l}-\text{x})=0$
$\Rightarrow\frac{\text{T}_\text{B}}{\text{T}_\text{A}}=\frac{\text{l}-\text{x}}{\text{x}} ...(\text{i})$
For equal stress, $S_A$ = $S_B$
$\Rightarrow\text{S}_\text{A}=\text{S}_\text{B}$
$\Rightarrow\frac{\text{T}_\text{A}}{10^{-6}}=\frac{\text{T}_\text{B}}{2\times10^{-6}}$
$\Rightarrow\frac{\text{l}-\text{x}}{\text{x}}=2$
$\Rightarrow\frac{\text{l}}{\text{x}}-1=2$
$\Rightarrow\text{x}=\frac{\text{l}}{\text{3}}$
$\Rightarrow\text{l}-\text{x}=\text{l}-\frac{\text{l}}{\text{3}}=\frac{2\text{l}}{3}$
Hence, mass $m$ should be suspended close to wire $B (Al$ wire$).$
We know, Strain $=\frac{\text{Stress}}{\text{Y}}$
So, for equal strain in the wires,
$\Rightarrow\ \frac{\text{S}_\text{A}}{\text{Y}_\text{Steel}}=\frac{\text{S}_\text{B}}{\text{Y}_\text{Al}}$
$\Rightarrow\ \frac{\text{Y}_\text{Steel}}{\text{T}_\text{A}/\text{a}_\text{A}}=\frac{\text{Y}_\text{Al}}{\text{T}_\text{B}/\text{a}_\text{B}}$
$\Rightarrow\ \frac{\text{Y}_\text{Steel}}{\text{Y}_\text{Al}}=\frac{\text{T}_\text{A}}{\text{T}_\text{B}}\times\frac{\text{a}_\text{B}}{\text{a}_\text{A}}=\Big(\frac{\text{x}}{\text{l}-\text{x}}\Big)\Big(\frac{2\text{a}_\text{A}}{\text{a}_\text{A}}\Big)$
$\Rightarrow\ \frac{200\times10^9}{70\times10^9}=\frac{\text{2x}}{\text{l}-\text{x}}$
$\Rightarrow\frac{20}{7}=\frac{\text{2x}}{\text{l}-\text{x}}$
$\Rightarrow17\text{x}=10\text{l}$
$\Rightarrow\text{x}=\frac{10\text{l}}{17}$
$\Rightarrow\text{l}-\text{x}=\text{l}-\frac{10\text{l}}{17}=\frac{7\text{l}}{17}$
Hence, mass $m$ should be suspended close to wire $A ($steel wire$).$
MCQ 541 Mark
The temperature of a wire is doubled. The Young’s modulus of elasticity
- AWill also double.
- BWill become four times.
- CWill remain same.
- ✓Will decrease.
Answer
View full question & answer→Correct option: D.
Will decrease.
Key concept: Youngs modulus $(Y).$
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.
$\text{Y}=\frac{\text{Normal stress}}{\text{Longitudinal strain}}=\frac{\text{F/A}}{\Delta\text{L/L}}=\frac{\text{FL}}{\text{A}\Delta\text{L}}$
The fractional change in length of any material is defined as
$\frac{\Delta\text{L}}{\text{L}_0}=\alpha\Delta\text{T}$
where $\Delta T$ is change in the temperature, $L_o$ is original length, $\alpha$ is the coefficient of linear expansion of the given material and $L_o$ is the original length of material.
So, simply change in length is due to change in temperature.
$\Delta\text{A}=\text{L}_0\alpha\Delta\text{T}$
And Young's modulus
$(\text{Y})=\frac{\text{Stress}}{\text{Strain}}=\frac{\text{FL}_0}{\text{A}\times\Delta\text{L}}=\frac{\text{FL}_0}{\text{AL}_0\alpha\Delta\text{T}}\propto\frac{1}{\Delta\text{T}}$
As $\text{Y}\propto\frac{1}{\Delta\text{T}}$
It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.
$\text{Y}=\frac{\text{Normal stress}}{\text{Longitudinal strain}}=\frac{\text{F/A}}{\Delta\text{L/L}}=\frac{\text{FL}}{\text{A}\Delta\text{L}}$
The fractional change in length of any material is defined as
$\frac{\Delta\text{L}}{\text{L}_0}=\alpha\Delta\text{T}$
where $\Delta T$ is change in the temperature, $L_o$ is original length, $\alpha$ is the coefficient of linear expansion of the given material and $L_o$ is the original length of material.
So, simply change in length is due to change in temperature.
$\Delta\text{A}=\text{L}_0\alpha\Delta\text{T}$
And Young's modulus
$(\text{Y})=\frac{\text{Stress}}{\text{Strain}}=\frac{\text{FL}_0}{\text{A}\times\Delta\text{L}}=\frac{\text{FL}_0}{\text{AL}_0\alpha\Delta\text{T}}\propto\frac{1}{\Delta\text{T}}$
As $\text{Y}\propto\frac{1}{\Delta\text{T}}$
MCQ 551 Mark
When an elastic material with Young's modulus $Y$ is subjected to stretching stress $S,$ the elastic energy stored per unit volume of the material is:
- A$\frac{\text{YS}}{2}$
- B$\frac{\text{YS}^2}{2}$
- ✓$\frac{\text{S}^2}{2\text{Y}}$
- D$\frac{2}{2\text{Y}}$
Answer
View full question & answer→Correct option: C.
$\frac{\text{S}^2}{2\text{Y}}$
Elastic energy per unit volume,
$\text{u}=\frac{1}{2}\times\text{Stress}\times\text{Strain}$
$=\frac{1}{2}\times\text{Stress}\times\frac{\text{Stress}}{\text{Young modulus}}$
$=\frac{1}{2}\text{S}\times\frac{\text{S}}{\text{Y}}$
$=\frac{\text{S}^2}{2\text{y}}$
$\text{u}=\frac{1}{2}\times\text{Stress}\times\text{Strain}$
$=\frac{1}{2}\times\text{Stress}\times\frac{\text{Stress}}{\text{Young modulus}}$
$=\frac{1}{2}\text{S}\times\frac{\text{S}}{\text{Y}}$
$=\frac{\text{S}^2}{2\text{y}}$
MCQ 561 Mark
In solids, inter$-$atomic forces are:
- ATotally repulsive.
- BTotally attractive.
- ✓Combination of $(a)$ and $(b).$
- DNone of these.
Answer
View full question & answer→Correct option: C.
Combination of $(a)$ and $(b).$
MCQ 571 Mark
A wire of length $2m$ is made from $10\ cm^3$ of copper. A force $F$ is applied so that its length increases by $2\ mm.$ Another wire of length $8m$ is made from the same volume of copper. If the force $F$ is applied to it, its length will increase by:
- A$0.8\ cm$
- B$1.6\ cm$
- C$2.4\ cm$
- ✓$3.2\ cm$
Answer
View full question & answer→Correct option: D.
$3.2\ cm$
MCQ 581 Mark
Rigidity modulus of steel is $n$ and its Young's modulus is $Y.$ A piece of steel of cross$-$sectional area a is stretched into a wire of length $L$ are $\frac{\text{a}}{10}.$ Then:
- A$Y$ increases and $\eta$ decreases.
- ✓$Y$ and $\eta$ remain the same.
- C$Y$ decreases and $\eta$ increases.
- DBoth $Y$ and $\eta$ increases.
Answer
View full question & answer→Correct option: B.
$Y$ and $\eta$ remain the same.
Modulus of rigidity and Young's Modulus of elasticity are constants for a given material.
MCQ 591 Mark
The graph shows the behavior of a length of wire in the region for which the substance obeys Hooke's law. $P$ and $Q$ represent.
- A$P =$ applied force, $Q =$ extension.
- B$P =$ extension, $Q =$ applied force.
- ✓$P =$ extension, $Q =$ stored elastic energy.
- D$P =$ stored elastic energy, $Q =$ extension.
Answer
View full question & answer→Correct option: C.
$P =$ extension, $Q =$ stored elastic energy.
The graph between applied force and extension will be straight line because in elastic range,
Applied force $\propto$ extension,
But the graph between extension and stored elastic energy will be parabolic in nature.
As, $\text{U}=\frac{1}{2}\text{kx}^2$ or $\text{U}\propto\text{x}^2.$
Applied force $\propto$ extension,
But the graph between extension and stored elastic energy will be parabolic in nature.
As, $\text{U}=\frac{1}{2}\text{kx}^2$ or $\text{U}\propto\text{x}^2.$
MCQ 601 Mark
The upper end of a wire of radius $4\ mm$ and length $100\ cm$ is clamped and its other end is twisted through an angle of $30^\circ .$ The angle of shear is:
- A$12^\circ$
- B$1.2^\circ$
- ✓$0.12^\circ$
- D$0.012^\circ$
Answer
View full question & answer→Correct option: C.
$0.12^\circ$
Angle of twist at free end,
$=30^\circ=\frac{30}{180}\times\pi\text{ rad}=\frac{\pi}{6}\text{rad}$
Displacement of the free surface,
$\Delta\text{L}=\frac{2\not=\text{r}}{2\not=}\times\frac{\not=}{6}=\frac{\pi\text{r}}{6}=\frac{\pi\times0.4}{6}\text{cm}$
Angle of shear or shearing strain $=\frac{\Delta\text{L}}{\text{L}}$
$=\frac{\pi\times\frac{0.4}{6}}{100}\text{rad}$
$=\frac{\not=\times0.4}{6\times100}\times\frac{180}{\not=}\text{ degree}=0.12^\circ$
$=30^\circ=\frac{30}{180}\times\pi\text{ rad}=\frac{\pi}{6}\text{rad}$
Displacement of the free surface,
$\Delta\text{L}=\frac{2\not=\text{r}}{2\not=}\times\frac{\not=}{6}=\frac{\pi\text{r}}{6}=\frac{\pi\times0.4}{6}\text{cm}$
Angle of shear or shearing strain $=\frac{\Delta\text{L}}{\text{L}}$
$=\frac{\pi\times\frac{0.4}{6}}{100}\text{rad}$
$=\frac{\not=\times0.4}{6\times100}\times\frac{180}{\not=}\text{ degree}=0.12^\circ$
MCQ 611 Mark
A spring of force constant $k$ is cut into two equal parts. The force constant of each part is:
- A$\frac{\text{k}}{2}$
- B$k$
- ✓$2k$
- D$4k$
Answer
View full question & answer→Correct option: C.
$2k$
MCQ 621 Mark
When a pressure of $100$ atmosphere is applied on a spherical ball of rubber, then its volume reduces to $0.01\%.$ The bulk modulus of the material of the rubber in dyne $cm^{-2}$ is:
- A$10 \times 10^{12}$
- B$100 \times 10^{12}$
- ✓$1 \times 10^{12}$
- D$20 \times 10^{12}$
Answer
View full question & answer→Correct option: C.
$1 \times 10^{12}$
$1\text{atm}=10^5\text{Nm}^{-2}$
$\therefore100\text{atm}=10^7\text{Nm}^{-2}$ and $\Delta\text{V}=0.01\%\text{V}$
$\therefore\frac{\Delta\text{V}}{\text{V}}=0.0001$
$\text{B}=\frac{\text{P}}{\frac{\Delta\text{V}}{\text{V}}}=1\times10^{11}\text{Nm}^{-2}$
$=1\times10^{12}\frac{\text{dune}}{\text{cm}^2}$
$\therefore100\text{atm}=10^7\text{Nm}^{-2}$ and $\Delta\text{V}=0.01\%\text{V}$
$\therefore\frac{\Delta\text{V}}{\text{V}}=0.0001$
$\text{B}=\frac{\text{P}}{\frac{\Delta\text{V}}{\text{V}}}=1\times10^{11}\text{Nm}^{-2}$
$=1\times10^{12}\frac{\text{dune}}{\text{cm}^2}$
MCQ 631 Mark
In steel, the Young's modulus and the strain at the breaking point are $2 \times 10^{11}\ Nm^{-2}$ and $0.15,$ respectively. The stress at the breaking point for steel is therefore:
- A$1.33 \times 10^{11}Nm^{-2}$
- B$1.33 \times 10^{12}Nm^{-2}$
- C$7.5 \times 10^{-13}Nm^{-2}$
- ✓$3 \times 10^{10}Nm^{-2}$
Answer
View full question & answer→Correct option: D.
$3 \times 10^{10}Nm^{-2}$
MCQ 641 Mark
The nature of molecular forces resembles with the nature of the:
- AGravitational force.
- BNuclear force.
- ✓Electromagnetic force.
- DWeak force.
Answer
View full question & answer→Correct option: C.
Electromagnetic force.
MCQ 651 Mark
A steel rod of length $1m$ and radius $10mm$ is stretched by a force $100kN$ along its length. The stress produced in the rod is $Y_{steel} = 2 \times 10^{11}Nm^{-2}$.
- A$3.18 \times 10^6 \mathrm{Nm}^{-2}$
- B$3.18 \times 10^7 \mathrm{Nm}^{-2}$
- ✓$3.18 \times 10^8 \mathrm{Nm}^{-2}$
- D$3.18 \times 10^9 \mathrm{Nm}^{-2}$
Answer
View full question & answer→Correct option: C.
$3.18 \times 10^8 \mathrm{Nm}^{-2}$