MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 10

A plank of mass

$M$ is suspended horizontally by using two wires as shown in the figure.They have same length

$(L)$ and same cross sectional area

$(A)$ . Their Young's modulus are

$Y_1$ and

$Y_2$ respectively. The elastic potential energy of the system will be

0%
$\dfrac{2M^2g^2L}{A(Y_1+Y_2)}$
0%
$\dfrac{m^2g^2L}{A(Y_1+Y_2)}$
0%
$\dfrac{m^2g^2L}{2A(Y_1+Y_2)}$
0%
$\dfrac{M^2g^2L(Y_1+Y_2)}{2AY_1Y_2}$

Explanation

In both wires tension =

$\left(\dfrac{m_2}{2}\right)=F$

Total strain energy

$=S_1+S_2$

$=\dfrac{2F^2l}{Y_1}+\dfrac{2F^2l}{y_2}$

$=\dfrac{2M^2g^2l}{AY_1}+\dfrac{2M^2g^2l}{YA}$

$=\dfrac{M^2y^2l}{2A(Y_1+Y_2)}$

1090135_1137846_ans_535729f0ec7b466aaf5559e22094ddd3.png

To wires

$A$ and

$B$ have the same length and area of cross section. But Young's modulus of

$A$ is two times the Young's modulus of

$B$ . Then the ratio of force constant of

$A$ to that of

$B$ is

0%
$1$
0%
$2$
0%
$\dfrac{1}{2}$
0%
$12\ 2$

Explanation

$\begin{array}{l}\text { Given, } Y_{A}=2 Y_{B} \\\text { Same length and area of cross - section of wires } A \text { & } B \\\text { According to the spring equivalent concept ,}\end{array}$

$\begin{array}{l}F=k x \quad(x=\text { extension in spring })\\F=\left(\frac{Y A}{L}\right)(\Delta L) \\\text { From above equations, we conclude } \\\qquad k=\frac{Y A}{L}=\text { force constant } \\\therefore \quad \frac{Y_{A}}{Y_{B}}=\frac{k_a}{k_{b}}=2\quad(\text { area and length } \\\text { are same. })\end{array}$

One end of uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight

$W_1$ is suspended from its lower end. If s is the area of cross section of the wire, the stress in the wire at a height

$\dfrac{L}{4}$ from its lower end is -

0%
$\dfrac {W_1} {s}$
0%
$\dfrac { \left[ { W }_{ 1 }+\dfrac { W }{ 4 } \right] }{ s }$
0%
$\dfrac { \left[ { W }_{ 1 }+\dfrac { 3W }{ 4 } \right] }{ s }$
0%
$\dfrac { { W }_{ 1 }+W }{ 4 }$

Explanation

Consider the FBD as:

Let x be the weight of wire part of

$l=l/4$

As wire is uniform,

$\dfrac{x}{\dfrac{l}{4}}=\dfrac{w}{l}$

$\Rightarrow x=w/4$

For equilibrium,

$F=w_1+x=w_1+w/4$

So stress at point P

$=\dfrac{F}{s}=\dfrac{(w_1+w/4)}{s}$

option B is correct.

1130803_1157232_ans_3721d3ace0034790999c674bcc95f4e7.jpg

When a metal wire is stretched by a load, the fractional change in its volume

$\Delta V/V$ is proportional to?

0%
$-\dfrac{\Delta l}{l}$
0%
$\left(\dfrac{\Delta l}{l}\right)^2$
0%
$\sqrt{\Delta l/l}$
0%
None of these

Explanation

$v=\dfrac { \pi { d }^{ 2 }l }{ 4 }$

$⟹\dfrac { ΔV }{ V } =\dfrac { 2Δd }{ d } +\dfrac { Δl }{ l }$

$⟹d\frac { ΔV }{ V } =\dfrac { (1−2σ)Δl }{ l }$

$(\dfrac { Δd }{ d } =\dfrac { −σΔl }{ l } )$

where $σ$ is Poisson's ratio.

A metal rod of Young's modulus

$2\times { 10 }^{ 10 }N{ m }^{ -2 }$ undergoes an elastic strain of 0.02%. The energy per unit volume stored in the rod, in

$Joules$ is ?

0%
$400$
0%
$800$
0%
$1200$
0%
$1600$

Explanation

$\text{Energy per unit voume = Energy density}$

$\begin{array}{l}=\dfrac{1}{2} \times \text { stress } \times \text { strain } \\=\dfrac{1}{2}\times\dfrac{F}{A} \times \dfrac{\Delta l}{l}\\=\dfrac{1}{2} \times\left(\dfrac{\gamma}{l}\right)\left(\dfrac{\Delta l}{l}\right) \\=\dfrac{1}{2} Y\left(\dfrac{\Delta l}{l}\right)^{2} \\=\dfrac{1}{2} \times 2 \times 10^{10} \times\left(\dfrac{0.02}{100}\right)^{2}\end{array}$

$=400 \mathrm{Joule}$

$\text{Ans: (A)}$

A wire elongates by

$l$ mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (In mm) -

0%
$l$
0%
$2l$
0%
zero
0%
$\dfrac{l}{2}$

Explanation

$\begin{array}{l}\text { Hence at equilibriun }\\\qquad \begin{array}{l}T=W \\\frac{W}{A}=Y\frac{l_{1}}{L} \\\Rightarrow l_{1}=\frac{W L}{A Y}\end{array}\end{array}$

$\begin{array}{l}\text { At equilibnumu } T=W \text{ for each } \\\text { side of wire }\rightarrow\\\frac{W}{A}=Y \frac{l_{2}}{L}\\\Rightarrow l_{2}=\frac{W L}{A Y}\end{array}$

$\text { hence elongation will be same }\Rightarrow\text { ans } \rightarrow(A)$

2000630_1153312_ans_9dc7a384a9bc4b0cb5e9a7cf64feb169.PNG

Write Copper, Steel, Glass and Rubber in order of increasing coefficient of elasticity

0%
Steel, Rubber, Copper, Glass
0%
Rubber. Copper, Glass, Steel
0%
Rubber. Glass, Steel, Copper
0%
Rubber. Glass, Copper, Steel

A metal wire of length

$2.5 m$ and area of cross section

$1.5\times { 10 }^{ -6 }{ m }^{ -2 }$ , is stretched through

$2 mm$ . Calculate the work done during stretching.

$Y=1.2\times { 10 }^{ 11 }N{ m }^{ -2 }$

0%
$0.15 J$
0%
$0.51 J$
0%
$1.51 J$
0%
$5.1 J$

A mild steel wire of length

$2L$ and cross-sectional area

$A$ is stretched, well within elastic limit, horizontally between two pillars. A mass

$m$ is suspended from the mind point of the wire. Strain in the wire is

0%
$\dfrac{x^{2}}{2L^{2}}$
0%
$\dfrac{x}{L}$
0%
$\dfrac{x^{2}}{L}$
0%
$\dfrac{x^{2}}{2L}$

Explanation

$\begin{array}{l}\text { Increase in Length }=\Delta L=A B+B C - A C=2 A B-A C \\\qquad A B=\sqrt{x^{2}+L^{2}} \quad , A C=2 L\\\Rightarrow\quad \Delta L=2 \sqrt{x^{2}+L^{2}}-2 L\\\Rightarrow \Delta L=2 L\left(1+\left(\dfrac{x}{L}\right)^{2}\right)^{1 / 2}-2 L \\\Rightarrow\Delta L=2 L\left \{\sqrt{1+\dfrac{x^{2}}{L^{2}}}-1\right\}.\end{array}$

$\begin{array}{l}\text { Using Binomial theorem. }\\\left(1+\dfrac{x^{2}}{L^{2}}\right)^{\dfrac{1}{2}}=1+\dfrac{x^{2}}{2 L^{2}} \\\Delta L=2 L\left\{1+\dfrac{x^{2}}{2 L^{2}}-1\right\}=\dfrac{x^{2}}{L}\\\text { strain }=\dfrac{\Delta l}{2 L}=\dfrac{x^{2}}{2 L^{2}}\end{array}$

$\text { Ans } \rightarrow A$

A brass wire of diameter

$1 mm$ and of length

$2 m$ is stretched by applying a force of

$20 N$ . If the increase in length is

$0.51 mm$ . Then the Young's modulus of the wire is

0%
$8.848 \times 10^{10} N/m^2$
0%
$7.984 \times 10^{10} N/m^2$
0%
$6.984 \times 10^{10} N/m^2$
0%
$9.984 \times 10^{10} N/m^2$

Explanation

Given,

$F=20N,l=2m,d=1mm,\Delta l=0.51mm$

We know,

$Y=\dfrac{Fl}{A\Delta l}$

$Y=\dfrac{20\times 2}{\pi 5^2 \times 10^{-8}\times 0.51\times 10^{-3}}$

$Y=\dfrac{40}{25\pi\times 0.51}\times 10^11$

$Y=0.9984\times 10^11=9.984\times 10^{10}$

Option

$\textbf D$ is the correct answer

A force of

$15 N$ increases the length ofa by

$1 \mathrm { mm }$ . The additional force require increase the length by

$2.5 \mathrm { mm }$ in N is

0%
$2.25$
0%
$22.5$
0%
$37.5$
0%
$3.75$

Explanation

We know,

$Y=\dfrac{FL}{A\Delta L}$

For a wire, A,L,Y are constant.

$\dfrac{F}{\Delta L}=$ constant

Let extra Force needed be

$F$ ,

$\dfrac{15}{1}=\dfrac{F+15}{2.5}$

$F+15=2.5\times 15$

$F=1.5\times 15=22.5$

Option

$\textbf B$ is the correct answer

Ball A strikes with velocity u elastically with identical ball B at rest, inclined at an angle of with line joining the centers of two balls. What will be the speed of ball B after the collision:

0%
u
0%
$\dfrac{u\sqrt{3}}{2}$
0%
$\dfrac{u}{2}$
0%
$\dfrac{u}{\sqrt{2}}$

Explanation

It is not possible for ball

$A$ to collide with ball

$B$ at an angle to the line joining their centers because it will always consider with the line joining their centers

$\rightarrow$ It will always collide along the line joining their centers.

As the collision is elastic the velocity of ball

$B$ will be

$u$ .

1362010_1184933_ans_e5420fb8dcae410eaff443cb4f478e95.png

A mass of

$0.5\ kg$ is suspended from wire, then length of wire increase by

$3\ mm$ then find out work done.

0%
$4.5\times 10^{-3} J$
0%
$7.5\times 10^{-3} J$
0%
$9.3\times 10^{-2} J$
0%
$2.5\times 10^{-2} J$

Explanation

$\begin{aligned}\text { Work done in stretching of a wire }&=\frac{1}{2} F l \\&=\frac{1}{2} m g l \\&=\frac{1}{2}(0.5)(10)\times\left(3\times10^{-3}\right)\\&=7.5\times10^{-3}\\\text {Hence answer is } B &\end{aligned}$

A material has poisson's ratio

$0.3$ . If a uniform rod of it suffers a longitudinal strain of

$25\times 10^{-3}$ , then the percentage increase in its volume is

0%
$1\%$
0%
$2\%$
0%
$3\%$
0%
$4\%$

Explanation

$\begin{aligned}\mu &=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow& \frac{3}{10}=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow & \frac{Y}{6 B}=\frac{1}{2}-\frac{3}{10}=\frac{1}{5} \\\Rightarrow & \frac{Y}{6 B}=\frac{1}{5}\\\Rightarrow & \frac{Y}{B}=\frac{6}{5}\end{aligned}$

$\begin{aligned}\Rightarrow \frac{(\Delta V / V)}{(\Delta l / l)} &=\frac{6}{5} \\\Rightarrow \quad(\Delta v / v)&=\frac{25 \times 10^{-3} \times 6}{5}=30\times10^{-3}\\\Rightarrow \quad \frac{\Delta V}{V}\%&=30 \times 10^{-3} \times 100=3 \% \\A n s: &(C)\end{aligned}$

If the interatomic spacing in a steel wire is

$2.8\times 10^{-10}m$ and

$Y_{real}=2\times 10^{11}N/m^{2}$ , then force constant in N/m is -

0%
$5.6$
0%
$56$
0%
$0.56$
0%
$560$

Explanation

$\begin{aligned}\text { Force constant } K, &=Y \times\text{ normal distance between atoms}\\&=2.8\times10^{-10}\times2\times10^{11}\\&=5.6\times10 \\&=56 \mathrm{~N} / \mathrm{m} \\\text{Hence answer is B}\end{aligned}$

The Young's modulus a rubber string 8 cm long and density 1.5 kg/

$m^{3}$ is

$5\times 10^{8}N/m^{2}$ , is suspended on the ceiling in a room. The increases in length due to its own weigth will be

0%
$9.6\times 10^{-5}$ m
0%
$9.6\times 10^{-11}$ m
0%
$9.6\times 10^{-3}m$
0%
9.6 m

Explanation

Given that,

Length of rubber band, $l=8\,cm$

Density, $d=1.5\,Kg/{{m}^{3}}$

Young’s modulus, $Y=5\times {{10}^{8}}\,N/{{m}^{2}}$

The relation for increase in length and young’s modulus is as follows:

$l=\dfrac{{{L}^{2}}dg}{2Y}$

$l=\dfrac{{{(8\times {{10}^{-2}})}^{2}}1.5\times 10}{2\times 5\times {{10}^{8}}}$

$l=\dfrac{64\times {{10}^{-4}}\times 1.5\times 10}{{{10}^{9}}}$

$l=96\times {{10}^{-12}}$

$l=9.6\times {{10}^{-11}}\,m$

When a weight of

$10 \,kg$ is suspended from a copper wire of length

$3 \,m$ and diameter

$0.4 \,mm$ . Its length increases by

$2.4 \,cm$ . If the diameter of the wire is doubled, then the extension is its length will be :

0%
$7.6 \,cm$
0%
$4.8 \,cm$
0%
$1.5 \,cm$
0%
$0.6 \,cm$

Explanation

Mass

$= 10 \,kg$

$D = 0.4 \,mm$

$f = mxg = 100 N$

$A = \pi r^2 = \dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} = \dfrac{\pi D^2}{4}$

$L = 3$

$T = \dfrac{fL}{A\Delta L}$

$Y = \dfrac{100 \times 3}{\dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} \times \Delta L}$

$\Delta L = \dfrac{300 \times 4}{\pi (0.4 \times 10^{-3})^2 \times 4}$

Now,

$Y$ is same for the material.

$Y = \dfrac{F \times L'}{A' \times \Delta L'}$

$\Delta L' = \dfrac{f \times \dfrac{L}{4}}{4 \times A \times Y}$

$\Delta L' = \dfrac{\dfrac{100 \times 3}{4}}{4 \times 1.25 \times 10^{-7} \times 1 \times 10^{11}}$

$\Delta L' = \dfrac{75}{4 \times 1.25 \times 10^4}$

$\Delta L' = 1.5 \times 10^{-3}m$

$\Delta L' = 1.5 \,cm$

The Poisson's ratio of the material of a wire is

$0.25 .$ If it is stretched by a force F, the longitudinal strain produced in the wire is

$5 \times 10 ^ { - 4 } .$ What is the percentage increase in its volume?

0%
$0.2$
0%
$2.5 \times 10 ^ { - 2 }$
0%
Zero
0%
$1.25 \times 10 ^ { - 6 }$

Explanation

$\sigma =.25$

$\Rightarrow \ \dfrac {-\Delta R1R}{\Delta l 10}$

$\Rightarrow \ \dfrac {\Delta R}{R}=-.25\dfrac {\Delta l}{l}$

$=-.25\times 5\times 16^4$

$v=\pi R^2 l$

$\Rightarrow \ \dfrac {\Delta v}{v}\times 100 \left (2\dfrac {\Delta R}{R} + \dfrac {\Delta l}{l}\right)\times 100$

$=(2\times (-.25\times 5\times 10^{-4})+5\times 10^{-4})\times 150$

$=2.5\times 10^{-2}=.025\%$

The length of a steel wire is

$l_{1}$ when the stretching force is

$T_{1}$ and

$l_{2}$ when the stretching force is

$T_{2}$ The natural length of the wire is

0%
$\dfrac{l_{1}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$
0%
$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}+T_{2}}$
0%
$\dfrac{l_{2}T_{1}+l_{2}T_{2}}{T_{1}-T_{2}}$
0%
$\dfrac{l_{2}T_{1}-l_{1}T_{2}}{T_{1}-T_{2}}$

Explanation

assume natural length be

$l$ .

Hook's law :-

$T=K\triangle x$ (T= Tension,

$\triangle x$ = elongation)

in case

$1$ ,

$T=T_{1}$ and

$\triangle x= l_{1}-l$

so,

$T_{1}=K(l_{1}-l)\rightarrow (1)$

similarly for case 2,

$T_{2}=K(l_{2}-l)\rightarrow (2)$

Divide

$(1)\div (2)$ ,

$\frac{T_{1}=K(l_{1}-l)}{T_{2}\, K(l_{2}-l)}\Rightarrow T_{1}(l_{2}-l)=T_{2}(l_{1}-l)$

$\Rightarrow l=\frac{T_{1}l_{2}-T_{2}l_{1}}{T_{1}-T_{2}}$

1106826_1179709_ans_93b5c1490b8d42f0b5a22021cf3338cd.JPG

Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is

$2.2\times 10^{4}Nm^{-2}$

0%
$4.4\times 10^{7}Nm^{-2}$
0%
$2.2\times 10^{7}Nm^{-2}$
0%
$3.3\times 10^{7}Nm^{-2}$
0%
$1.1\times 10^{7}Nm^{-2}$

Explanation

Bulk modulus (B) is defined as :-

$B = \frac{-P}{\triangle v/v}$ when,

$\frac{\triangle v}{v} = 0.02 (2\%)$

$\Rightarrow P = B\times \frac{Av}{v} = 2.2\times 10^{4}\times 0.02 = 4.4\times 10^{2} Atm$

$= 4.4\times 10^{7}pa$

$[1atm = 10^{5}pa]$

1106837_1179969_ans_f09393375e614d689484280fa706b43c.JPG

A uniform cyclinder rod of length

$L$ , cross -sectional area

$A$ and youngs modules

$Y$ is acted upon by the forces shown in fig. The elongation of the rod is

0%
$\dfrac{3FL}{5AY}$
0%
$\dfrac{2FL}{5AY}$
0%
$\dfrac{3FL}{8AY}$
0%
$\dfrac{8FL}{3AY}$

Explanation

$\begin{array}{l}\text {Using } \dfrac{F}{A}=Y \dfrac{\Delta l}{\ell} \\\text { (1) } \Delta l_{1}=\dfrac{F_{1} l_{1}}{A{Y}} \\\Rightarrow \Delta l_{1}=\dfrac{3 F(2 L / 3)}{A Y}\\\Rightarrow \Delta l_{1}=\dfrac{2 F L}{A Y}\end{array}$

$\begin{array}{l}\text { (2) } \Delta l_{2}=\dfrac{F_{2} l_{2}}{A Y} \\\Rightarrow \Delta l_{2}=\dfrac{2 F(L/3)}{A Y}\\\Rightarrow \Delta l_{2}=\dfrac{2 F L}{3 A Y}\end{array}$

$\begin{aligned}\text { Total elongation } &=\dfrac{2 F L}{A Y}+\dfrac{2 F L}{3 A Y}=\dfrac{8 F L}{3 A Y}\end{aligned}$

To apply a maximum stress of

$1.2\times10^8N/m^2$ under the action of force

$40N$ the minimum radius of wire required is

0%
$0.32$ mm
0%
$0.66$ mm
0%
$0.44$ mm
0%
$0.22$ mm

Explanation

$\begin{array}{l}\text { Max stress }=1.2 \times 10^{8}\mathrm{~N} / \mathrm{m}^{2} \\\text { Force}=40\mathrm{~N} \\\Rightarrow\text { } A=\dfrac{F}{\text { Stress }}=\dfrac{40}{1.2 \times 10^{8}}=\dfrac{400}{12\times10^{8}}\end{array}$

$\begin{aligned}&\Rightarrow \quad \pi r^{2}=\dfrac{400}{12\times 10^{8}}\\&\Rightarrow \quad r=\sqrt{\dfrac{400}{12\times 10^{8} \times \pi}}\\&\Rightarrow\text { } \quad r=0.32 \mathrm{~mm}\\&\text { Ans :(A) }\end{aligned}$

When a mass of 8 kg is suspended from a string, its length is

$l_1$ . If a mass 10 kg is suspended, its length is

$l_2$ . Length, when a mass of 16 kg is suspended from it, is given by

0%
$2l_2 - l_1$
0%
$2l_2 + l_1$
0%
$4l_2 + 3l_1$
0%
$4l_2 - 3l_1$

Explanation

$\begin{array}{l}m_{1}=8 \mathrm{~kg},\ l_{1}\\m_{2}=10\mathrm{~kg},\ l_{2} \\m_{3}=16 \mathrm{~kg} ,\ \mathrm{y}_{1}\end{array}$

$\begin{array}{l}\text { Let original length of string} =2\\\text { To find:}\ y_{1} \text { in terms of } L_1\text { and } L_2\end{array}$

$\begin{array}{l}\text { using } \dfrac{F}{A}=Y \dfrac{\Delta l}{l} \\\text { (1) } \dfrac{8 g}{A}=Y \dfrac{\ell_{1}-x}{x}\\\text { (2) } \dfrac{10 g}{A}=Y \dfrac{l_{2}-x}{x} \\\text { (3) } \dfrac{16 g}{A}=Y \dfrac{y_{1}-x}{x}\end{array}$

$\begin{array}{l}\text { Using }(1) \text { and }(2) \text { find } x \text { , and eliminate } x \text { in }(3) \text { , we get } \\y_{1}=2 l_{2}-l_{1}\quad \\\text { Ans }: \text { (A) }\end{array}$

A wire 2 m in length suspended vertically stretches by 10 mm when the mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is? (take g = 10

$m/s^2$ )

0%
0.5 J
0%
5 J
0%
50 J
0%
500 J

Explanation

Formulae:

spring force by wire = weight of mass attached

$k\Delta x=mg$

$k(0.01)=10 \times 9.8$

$k=9800N/m$

elastic potential energy,

$U=(0.5)k\Delta x^2$

$=(0.5)\times 9800 \times (0.01)^2$

$\therefore U=0.49\approx 0.5J$

One end of uniform wire of length

$L$ and of weight

$W$ is attached rigidly to a point in the roof and a weight

$W_{1}$ is suspended from its lower end. If

$s$ is the area of cross section of the wire, the stress in the wire at a height

$(3L/4)$ from its lower end is

0%
$\dfrac {W_{1}}{s}$
0%
$\left [W_{1} + \dfrac {W}{4}\right ]s$
0%
$\left [W_{1} + \dfrac {3W}{4}\right ]s$
0%
$\dfrac {W_{1} + W}{s}$

Explanation

The total weight at

$(3L/4)$ from the lower end:

$Weight = W_{1}+\dfrac{3L \times W}{4 \times L}$

So, the

$stress=weight/area$

i.e,

$stress=\dfrac{W_{1}+ \dfrac{3W}{4}}{s}$

hence Option C is correct

The isothermal Bulk modulus of an ideal gas at pressure

$'P'$ is

0%
$P$
0%
$\gamma P$
0%
$P/2$
0%
$P/\gamma$

Explanation

$\begin{array}{l}\text { Isothermal Bulk modulus is ratio of volumetric}\\\text { stress and volumetric strain at constant temprature}\end{array}$

$B=-\frac{\Delta P}{\Delta V / V} \longrightarrow \text { (1) }$

$\text{for iso thecunal process $\Rightarrow p v=$ constant}$

$\begin{aligned}\\\text{ Differentiating}\\& P(\Delta V)+(\Delta P) V=0 \\\Rightarrow & P(\Delta V)=-\Delta P(V)\\\Rightarrow & \frac{\Delta V}{V}=-\frac{\Delta P}{P}\rightarrow(2)\end{aligned}$

$\begin{array}{l}B=-\frac{\Delta P}{(-\Delta P / P)} \text {........ using (1) and }(2)\\\Rightarrow B=P \\\text { Hence ans is (A) }\end{array}$

What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt? Young's modulus of elasticity of wire = 12.5 x

$10^{11} dyne/cm^2$

0%
0.16%
0%
032%
0%
0.18%
0%
0.12%

Explanation

$\begin{array}{l}\text { Diameter }d=2.5\mathrm{~mm}=2.5\times 10^{-3} \mathrm{~m} \\\text { Force},\mathrm{f}=100\mathrm{~kg}\mathrm{wt}=100\times9.8\mathrm{~N}=980 \mathrm{~N} \\Y \quad=12.5 \times 10^{11} \text { dyne } / \mathrm{cm}^{2}=12.5 \times 10^{10} \mathrm{~N} /\mathrm{m}^{2}\end{array}$

$\begin{aligned}&\dfrac{\Delta L}{L}=\dfrac{F}{A Y}=\dfrac{980}{12.5 \times 10^{10} \times \pi\left(\dfrac{2.5 \times10^{-3}}{2}\right)^{2}}=0.0016\\&\Rightarrow \dfrac{\Delta l}{l}\times 100=0.16 \% \rightarrow \text { percentage increase in length.}\\&\text {Ans : A }\end{aligned}$

A material has Poisson's ratio

$0.5$ . If a uniform rod of it suffers a longitudinal strain of

$3\times 10^{-3}$ , what will be percentage increase in volume?

0%
$2\%$
0%
$3\%$
0%
$5\%$
0%
$0\%$

Explanation

$\begin{array}{l}\mu \text { , poisson's ratio }=0.5\\\dfrac{\Delta l}{l}=3 \times 10^{-3} \\\text {To find: } \quad \dfrac{\Delta V}{V} \times 100=\text { percentage increase in volume. } \\\mu=\dfrac{1}{2}-\dfrac{Y}{6 B}\end{array}$

$\begin{array}{l}\Rightarrow \dfrac{1}{2}=\dfrac{1}{2}-\dfrac{Y}{6 B} \\\Rightarrow \quad \dfrac{Y}{6 B}=0\\B\rightarrow \infty \\\left|\dfrac{F}{A}\right|=\left|B- \dfrac{\Delta V}{V}\right|\\\Rightarrow\quad\left|\dfrac{\Delta V}{V}\right|=\left|\dfrac{F}{B A}\right| \\\quad a s \quad B\longrightarrow \infty \\\dfrac{\Delta V}{V} \longrightarrow 0 .\end{array}$

$\begin{array}{l}\text { Hence : } \dfrac{\Delta V}{V}\times100=0 \% \\\text { Ans : } D\end{array}$

A rod of length

$3$ in and uniform cross-section area

$1mm^{2}$ is subjected by four forces at different cross section as shown in the figure. Yungs modulus of the rod is

0%
$5.235\ mJ$
0%
$7.25\ mJ$
0%
$3.6250\ mJ$
0%
$1.8125\ mJ$

A rod of uniform cross sectional area

$A$ and length

$L$ has a weight

$W$ . It is suspended vertically from a fixed support vertically from a fixed support. If Young's modulus for rod is

$Y$ , then elongation produced in rod is

0%
$\dfrac{WL}{YA}$
0%
$\dfrac{WL}{2YA}$
0%
$\dfrac{WL}{4YA}$
0%
$\dfrac{3WL}{4YA}$

Explanation

$\text { Each cross-section of rod is experiencing different amoun of stress. Hence,}$

$\begin{array}{l}\text { Let us consider a small element of differential length } d x \\\text { at a distance } x \text { from free end of the rod.The stress }\end{array}$

$\begin{array}{l}\text { at the position of this element is produced by the weight } \\\text { of the rod of length }x\text { lying below it i.e }(W / L) x\end{array}$

$\sigma=\frac{(W / L) x}{A}=\dfrac{W x}{A L}$

$\text { elongaluon } d \delta \text { is produced } \text { in differential}$

$\text { element } d x=\dfrac{\sigma d x}{Y}=\dfrac{Ww x d x}{A Y L}$

$\begin{aligned}\delta &=\int d\delta\\&=\int_{0}^{L}\frac{W x d x}{A Y L}=\frac{W}{Y A L}\left[\frac{x^{2}}{2}\right]_{0}^{L}=\frac{(W ) L}{2A Y}\end{aligned}$

Length of wire becomes 16 cm when certain tension is applied. If the tension is now doubled then wire is further elongated by 1 cm then original length of wire is

0%
14 cm
0%
15 cm
0%
15.5 cm
0%
16.5 cm

Explanation

$\text { Let the original length of the wire be }x\mathrm{~cm} \text { . }$

$\dfrac{F}{A}=Y \dfrac{16-x}{x} \quad \longrightarrow(1)$

$\text { if Tension doubles i.e, } 2 \mathrm{~F} \text { , final length of wire becomes } 17 \mathrm{~cm} \text { . }$

$\begin{array}{l}\dfrac{2 F}{A}=Y \dfrac{17-x}{x}\rightarrow\text { (2) } \\\text { Using (2) }\&\mathbb{(1) } \\\dfrac{2(16-x)}{ x}=\dfrac{17-x}{x}\\\Rightarrow 32-2 x=17-x\\\Rightarrow\quad15=x\end{array}$

The bulk modulus of liquid is

$1.2\times10^9N/m^2$ what is change in volume produced on a volume produced on a volume of

$5$ liters by a pressure equal to

$10^6N/m^2$ ?

0%
$1.2$ cc
0%
$3.2$ cc
0%
$4.2$ cc
0%
$5$ cc

Explanation

$\begin{array}{l}B=1.2 \times 10^{9} \mathrm{~N} /\mathrm{m}^{2} \\P=10^{6} \mathrm{~N} /\mathrm{m}^{2}\\V=5\mathrm{}l\\\Delta V=? \\P=-B\dfrac{\Delta V}{V}\end{array}$

$\begin{array}{l}\Rightarrow \Delta V=\dfrac{P V}{B}\\\Rightarrow \Delta V=\dfrac{10^{6} \times 5}{1.2\times10^{9}} \\\Rightarrow \triangle V=\dfrac{5}{1200}\\\Rightarrow \quad \Delta V=0.0042 \ell=4.2\mathrm{cc} .\end{array}$

If the interatomic spacing in a steel wire is

${ 2.8\times 10 }^{ -10 }$ m. and

${ Y }_{ steal }=2\times 10^{ 11 }N/m^{ 2 }$ ,then force constant in N/m is-

0%
5.6
0%
56
0%
0.56
0%
560

Explanation

Interatomic force constant

$= Y \times inter\ atomic \ distance$

$= 2\times 10^{11} \times 2.8\times 10^{-10}$

$= 2\times2.8\times 10$

$= 56 N/m$

A brass bar, having cross sectional area 10

$cm^2$ is subjected to axial forces as shown in figure. Total elongation of the bar is (Take Y =

$8 \times 10^2 \ t/m^2$ ).

0%
$0.0775\ cm$
0%
$7.5\ cm$
0%
$0.75\ cm$
0%
$0.075\ cm$

Explanation

$\text { Since the system is in equilibrium } \sum F_{\text {right }}=\sum F_{\text {left }}$

$\begin{array}{l}\text { Note: at section } B \text { net force } 3 t \text { is right is present.} \\\text { Similarly for section } C \text { too. }\end{array}$

$\begin{aligned}\Delta l_{1} &=\dfrac{P_{1} L_{1}}{A Y}=\dfrac{(5 t)(0.6)}{A \times 8 \times 10^{2} \times t}=\dfrac{3}{800 A} \rightarrow \mathbb{(1)} \\\Delta l_{2}&=\dfrac{(2 t)(1)}{A \times 8 \times 100 t}=\dfrac{1}{4004} \rightarrow(2) \\\Delta l_{3} &=\dfrac{(1 t)(1.2)}{A\times 800 t}=\dfrac{1.2}{800 A}\quad\rightarrow(3)\end{aligned}$

$\begin{array}{l}\text { Since each cases is elongation, we add } \rightarrow \\\qquad(1+2+3)=\dfrac{3+2+1.2}{800A}=\dfrac{6 \cdot 2}{800 \times 10 \times 10^{-4}}=0.0775 cm\end{array}$

In a Young's double slit experiment, the intensity at the cetral maximum is

$l_{0-}$ . The intensity ata distance

$\beta/4$ from the central maximum is (

$\beta$ is frige width)

0%
$l_0$
0%
$\dfrac{l_0}{2}$
0%
$\dfrac{l_0}{\sqrt2}$
0%
$\dfrac{l_0}{4}$

Explanation

In a Young 's double slit experimental the intensity at the central maximum is

$l_0$ . The intensity at a distance

$=\beta/4$

From the central maximum is

(

$\beta$ is fringe width)

Solution

The intensity of central maxima is

$l_0$ , let

$l_1$ and

$I_2$ be the intensity emitted by the two slits

$S_1$ and

$S_2$ repectively. The expression for resultant intensity is

$l=l_1l_2+2\sqrt{l_1l_2\cos\phi}$

For central maxima

$l=l_0$ and

$\phi=0$

In Searle's apparatus we have two wires. During experiment we study the extension in one wire. The use of second wire is-

0%
to support the apparatus because it is heavy and may not break single wire
0%
to compensate the changes in length caused by changes in temperature of atmosphere during experimentation
0%
to keep the apparatus in level so that extension is measured accurately
0%
all the three above

The Young's modulus of a rubber string 8 cm long and density 1.5

$kg/m^{3}$ is

$5\times 10^{8} N/m^{2}$ , is suspended on the ceiling in a room. The increase in length due to its own weight will be:-

0%
$9.6\times 10^{-5}m$
0%
$9.6\times 10^{-11}m$
0%
$9.6\times 10^{-3}m$
0%
9.6 m

Explanation

$y=5\times { 10 }^{ 8 }N/{ m }^{ 2 };a=g;l=8cm=8\times { 10 }^{ -2 }m;d=1.5kg/{ m }^{ 3 }\quad$

$y=\cfrac { stress }{ strain } =\cfrac {\dfrac{F}{A} }{ \dfrac{\Delta l}{l} } =\cfrac { \cfrac { F\times d }{ A\times d } }{ \dfrac{\Delta l}{l} } =\cfrac { \cfrac { F\times d }{ v } }{ \dfrac{\Delta l}{l} } =\cfrac { F\times d\times l }{ v\times \Delta l } =dg\times \cfrac { A }{ dl } \quad$

$=5\times { 10 }^{ 8 }=\cfrac { 1.5\times 64\times { 10 }^{ -4 }\times 10 }{ \Delta l } \quad$

$\Delta l=\cfrac { 1.5\times 64\times { 10 }^{ -4 }\times 10 }{ 2\times 5\times { 10 }^{ 8 } } =\cfrac { 0.3\times 64\times { 10 }^{ -12 }\times 10 }{ 2 }$

$\therefore \Delta l=9.6{ 10 }^{ -11 }m$

In a young's double-slit experiment, the slits are separated by

$0.5 mm$ , and the screen is placed

$150 cm$ away. A beam of light consisting of two wavelengths,

$650 nm$ , and

$520 nm$ , is used to obtain interference fringes on the screen. The least distance from the common central maxima to the point where the bright fringes due to both the wavelengths coincide is:

0%
$9.75 mm$
0%
$15.6 mm$
0%
$1.56 mm$
0%
$7.8mm$

Explanation

Given,

$d=0.5mm=0.5\times 10^{-3}m$

$D=150cm=1.5m$

$\lambda_1=650nm$

$\lambda_2=520nm$

For common maxima,

$n_1\lambda_1=n_2\lambda_2$

$\dfrac{n_1}{n_2}=\dfrac{\lambda_2}{\lambda_1}$

$\dfrac{n_1}{n_2}=\dfrac{520}{650}=\dfrac{4}{5}$

Now,

$n_1=4, \lambda_1=650nm$

Condition of maxima,

$\dfrac{yd}{D}=n_1\lambda_1$

$y=\dfrac{n_1\lambda_1 D}{d}$

$y=\dfrac{4\times 650\times 10^{-9}\times 1.5}{0.5\times 10^{-3}}=7.8mm$

The correct option is D.

An iron wore and copper wire having same length and cross-section are suspended from same roof Young's modulus of copper is

$\dfrac 13$ that of iron. Then the ratio of the weights to be added at their ends so that their ends are at the same level is

0%
$1:3$
0%
$1:9$
0%
$3:1$
0%
$9:1$

Explanation

Correct Answer: Option C
Step 1: Relate the change in length
Refer to figure

$1$
The initial lengths of both the copper and iron wires are same and finally their ends are at same level.
Let the initial length of each be

$l$
Change in length of iron be

$\Delta l_1$ and that of copper be

$\Delta l_2$
so,

$l + \Delta l_1 = l + \Delta l_2$
Therefore,

$\Delta l_1 = \Delta l_2$ ..........

$(I)$

Step2: Use Young's Modulus
We know,

Young's Modulus

$Y = \dfrac{Stress}{Strain}$

Let the weight added to Iron be

$F_1$ and that to copper be

$F_2$
Also, their cross section area are same.

so,

$Y_1 = \dfrac{F_1l}{A \Delta l_1}$ and

$Y_2 = \dfrac{F_2l}{A \Delta l_2}$
It is given that

$Y_1 = 3Y_2$
therefore,

$\dfrac{F_1}{\Delta l_1} = 3 \times \dfrac{F_2}{\Delta l_2}$
as

$\Delta l_1 = \Delta l_2$

$F_1 = 3F_2$
so,

$\dfrac{F_1}{F_2} = \dfrac 31$
Hence, Option C is the correct answer

2191088_1224616_ans_897304b68d2a49f8b4678a349f2263fc.png

If the work done in stretching a wire by

$1\ mm$ is

$2\ J$ , then work necessary for stretching another wire of same material but with double radius of cross-section and half of the length by

$1\ mm$ is

0%
$8\ J$
0%
$16\ J$
0%
$4\ J$
0%
$32\ J$

A body of mass 1 Kg is fastended to one end of a steel wire of cross - sectional area 3

$\times 10^{-6}$ m

$^{2}$ and it rotated in horizontal circle of radius 20 cm with a constant speed of 2 m/s .The elongation in the wire is (Y = 2

$\times 10^{11}$ N / m

$^{2}$ )

0%
$0 . 33\times 10^{-5}$ m
0%
$0 . 67 \times 10^{-5}$ m
0%
$2 \times 10^{-5}$ m
0%
$4\times10^{-5}$ m

A steel wire (original length = 2m, diameter = 1mm) and a copper wire ( original length =1m, diameter =2mm) are loaded as shown in the figure . Find the ratio of extension of steel wire to that of copper wire. Given , Young's modulus of steel =

$2\times 10^{11} Nm^{-2}$ and that of copper is

$10^{11} Nm^{-2}$ .

0%
$\frac{10}{3}$
0%
$5$
0%
$\frac{20}{3}$
0%
$\frac{14}{3}$

An elongation of

$0.1^\circ$ in a wire of cross sectional area

$10^{-6}m^{2}$ causes a tension of 100n. The Young's modulus is-

0%
$10^{12}N/M^{2}$
0%
$10^{11}N/M^{2}$
0%
$10^{10}N/M^{2}$
0%
$10^{2}N/M^{2}$

Explanation

Given,

$\dfrac{\Delta l}{l}\times 100=0.1$ %

$\dfrac{\Delta l}{l}=\dfrac{0.1}{100}=1\times 10^{-3}$

Cross-sectional area,

$A=10^{-6}m^2$

$T=100N$

Young's modulus,

$Y=\dfrac{T/A}{\Delta l/l}=\dfrac{100/10^{-6}}{10^{-3}}=\dfrac{100}{10^{-9}}$

$Y=10^{11}N/m^2$

The correct option is B.

A material has poisson's ratio 0.If a uniform rod of it suffers a longitudinal strain of

$2\times { 10 }^{ -3 }$ then the percentage increase in its volume is

0%
0%
0%
10%
0%
20%
0%
5%

A man grows into a giant such that his linear dimensions increase by a factor of

$9$ . Assuming that his density remains same, the stress in the leg will change by a factor of?

0%
$81$
0%
$\cfrac{1}{81}$
0%
$9$
0%
$\cfrac{1}{9}$

Explanation

$\begin{aligned}\text { Stress } &=\frac{\text { Force }}{\text { Area }}=\frac{(\text { height })\times \text { Area }\times \text { density }\times g}{\text { Area }} \\&=\text { (height) }(g\times\text { density })\end{aligned}$

$\begin{array}{l}\text { Stress a height } \\\text { So stress in leg will also change by a factor of } 9 .\end{array}$

The Young's modulus of the material of a rod is

$20 \times 10^{10}$ pascal. When the longitudinal strain is 0.04%, The energy stored per unit volume is

0%
$4 \times 10^{3} \ J/m^3$
0%
$8 \times 10^{-3} \ J/m^3$
0%
$16 \times 10^{-3} \ J/m^3$
0%
$16 \times 10^{3} \ J/m^3$

Explanation

$\begin{array}{l}\text { Energy stored per unit volume }=\frac{1}{2} \times \text { stress}\times\text{strain. } \\\qquad \begin{array}{c}\text { Giuen } y=20 \times 10^{10}\mathrm{Pa}\text { and } \text { strain }=0.04\% \\y=\frac{\text { Stress }}{\text { strain. }} \Rightarrow \text { stress }=y(\text { strain })\end{array}\end{array}$

$\begin{aligned}\Rightarrow \text { Energy stored }&\text { Per unit volume }=\frac{1}{2}\times Y(\operatorname{strain} )^{2} \\&=\frac{1}{2} \times 20 \times 10^{10} \times\left(4 \times 10^{-4}\right)^{2} \\&=10 \times 10^{10} \times 16\times 10^{-8} \\&=16 \times 10^{3} \mathrm{~J} /\mathrm{m}^{3}\end{aligned}$

A woman with a mass of

$65\ kg$ puts all her weight on one heel of her high-heel shoe. The cross-sectional area of the heel is

$1\ cm^{2}$ . According to the table, if she is standing on a pane of glass that is flat against the ground, does the glass break :

0%
No, because the stress is less than the ratio for the ultimate strength to Young's modulus for glass
0%
No, because the stress is less than the ultimate strength of glass
0%
Yes, because the stress is greater than the ratio for the ultimate strength to Young's modulus for glass
0%
No, because the stress is greater than the ultimate strength of glass

Explanation

$(B)$ No, because the stress is less than the ultimate strength of glass.

$m=65kg$

$A=1{ cm }^{ 2 }$

$=1\times { \left( { 10 }^{ -2 } \right) }^{ 2 }$

$=1\times { 10 }^{ -4 }{ m }^{ 2 }$

A force F doubles the length of wire of cross-section a. The Young modulus of wire is

0%
$\dfrac{F}{a}$
0%
$\dfrac{F}{3a}$
0%
$\dfrac{F}{2a}$
0%
$\dfrac{F}{4a}$

Explanation

As youngsters modulus is given as

$,$

$Y=f/a.$

$L/\Delta L$

From above we can conclude that

$Y' = 2Y$

That is youngs modulus becomes

$2$ times to its original value

$.$

Hence,

option

$(C)$ is correct answer.

A metal wire of length

$1\ m$ and cross-section area

$2\ mm^{2}$ and Young's modulus of elasticity

$Y=4\times 10^{11}\ N/m^{2}$ is stretched by

$2\ mm$ . Then

0%
the restoring force developed in the wire is $1600\ N$
0%
the energy density in the wire is $4\times 10^{5}\ J/m^{3}$
0%
the restoring force developed in the wire is $400\ N$
0%
the total elastic energy stored in the wire is $1.6\ J$

Explanation

$L=1m$

$\alpha=2{ mm }^{ 2 }=\left( \dfrac { 2 }{ { \left( 1000 \right) }^{ 2 } } \right) =2\times { 10 }^{ -6 }m$

$Y=4\times { 10 }^{ 11 }N/{ m }^{ 2 }$

$\Delta l=2mm=\dfrac { 2 }{ 10.00 } =0.002m$

$F=\dfrac { Y\alpha l }{ L }$

$=\dfrac { 4\times { 10 }^{ 11 }\times 2\times { 10 }^{ -6 }\times 2\times { 10 }^{ -3 } }{ 1m }$

$=16\times { 10 }^{ 11-6-3 }$

$=16\times { 10 }^{ 11-9 }$

$=16\times { 10 }^{ 2 }$

$=1600N$

Where

$\alpha =$ cross sectional area

$=2\times { 10 }^{ -6 }m$

$l=2mm=2\times { 10 }^{ -3 }m$

$L=1m$

An elastic metal rod will change its length when it

0%
falls vertically under its weight
0%
is pulled along its lengthy by a force acting at one end
0%
both
0%
none

Explanation

$\begin{array}{l}\text { An elastic metal rod will change it's length when } \\\text { it is pulled along it's length by a force acting at one } \\\text { end. }\end{array}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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