MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 9

Select the correct alternative(s).

0%
Elastic forces are always conservative
0%
Elastic forces are not always conservative
0%
Elastic forces are conservative only when Hooke's law is obeyed
0%
Elastic forces may be conservative even when Hooke's law is not obeyed

When the load on a wire is increasing slowly from

$2$ kg to

$4$ kg, the elongation increases from

$0.6$ mm to

$1$ mm. The work done during this extension of the wire is

$(g=10 m/s^2)$ .

0%
$14\times 10^{-3}$ J
0%
$0.4\times 10^{-3}$ J
0%
$8\times 10^{-2}$ J
0%
$10^{-3}$ J

Explanation

$W=(\dfrac{1}{2}F_2\times \Delta l_2)-(\dfrac{1}{2}F_1\times \Delta l_1)=20\times 10^{-3}=6\times 10^{-3}=14\times 10^{-3}$ J
Hence (A) is correct.

What is the percentage increase in length of a wire of diamater

$2.5\ mm$ , stretched by a force of

$100\ Kg\ wt$ ? Youngs modulus of elasticity of wire

$=1.25 \times {10}^{11}\ dyne/{cm}^{2}$

0%
$0.16\%$
0%
$0.32\%$
0%
$0.08\%$
0%
$0.12\%$

A steel wire has length 2.0 m, radius 1 mm and

$Y = 2 \times 10^{11} N/m^2$ A sphere of mass 1 kg is attached at one end of the wire which is then whirled in a vertical circle with speed 2 rev/sec. Elongation of the wire when mass is at the lowest position.

0%
1.0 mm
0%
2.0 mm
0%
0.1 mm
0%
0.01 mm

Explanation

$\begin{array}{l} \text { Given, lure of length }=2 \mathrm{~m} \\ \text { radius }=1 \mathrm{~mm} \\ y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \\ \text { sphere of mass }=1 \mathrm{~kg} \\ \text { frequency }=2 \end{array}$

$\begin{aligned} T-m g &=m \omega^{2} R \\ T=& m g+m \omega^{2} R \\ =& 10+16 \pi^{2}(2) \\ &=325 N \\ T=& Y A\left(\frac{\Delta L}{L}\right) \\ \frac{325 \times(2)}{2 \times 10^{11} \times \pi\left(10^{-6}\right)} &=\Delta L \\ \therefore & \Delta L=1 \mathrm{~mm} \end{aligned}$

Two copper wires having the length in ratio

$4:1$ and their radii ratio as

$1:4$ are stretched by the same force. Then the ratio of the longitudinal strain in the two will be

0%
$1:16$
0%
$16:1$
0%
$1:64$
0%
$64:1$

Explanation

Stress

$\propto$ Strain

$\propto \frac{F}{A}$

Ratio of strain

$=\frac{A_2}{A_1}=\left(\frac{r_2}{r_1}\right)^2=\left(\frac{4}{1}\right)^2=\left(\frac{16}{1}\right)$

A block of mass of

$2kg$ is attached to one end of a wire of cross sectional area

$1mm \,\,\,2$ and is. Find the elongation of the wire when the block is at top of the circle

$(Y=2\times 10^{11}Nm^{-2})$

0%
$0.2892 mm$
0%
$0.054 mm$
0%
$01446 mm$
0%
$0.0017 mm$

Scalar are quantity that are describe by

0%
Direction
0%
Magnitude and direction
0%
Magnitude
0%
Motion

Explanation

Scalar are quantity that are describe by Magnitude

Hence,

option

$C$ is correct answer.

A wire of density

$9\times{10^3}kg/{m^3}$ is stretched between two clamps 1 m apart and is stretched to an extension of

$4.9\times{10^{ - 4}}metre.$ Young's modulus of material is

$9\times{10^{10}}N/{m^2}$ . Then

0%
The lowest frequency of standing wave is 35 Hz
0%
The frequency of 1st overtone is 70 Hz
0%
The frequency of 1st overtone is 105 Hz
0%
The stress in the wire is $4.41\times{10^7}N/{m^2}$

Explanation

For wire if M = mass, r = density, A = Area of cross section V = volume, l = length, Dl = change

in length Then mass per unit length

$m = \dfrac{M}{l} = \dfrac{Al\rho }{l} = A\rho$ And Young? s modules of elasticity

$y = \dfrac{T/A}{\Delta l/l}Þ T = \dfrac{Y\Delta lA}{l}.$ Hence lowest frequency of vibration

$n = \dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$

$\dfrac{1}{2l} \sqrt{\dfrac{y\left(\dfrac{\Delta l}{l}\right)A}{A\rho }}= \dfrac{1}{2l}\sqrt{\dfrac{y\Delta l}{l\rho}}Þ\, n = \dfrac{1}{2\times 1}\sqrt{\dfrac{9\times 10^{10}\times 4.9\times 10^{-4}}{1\times 9\times 10^{3}}} = 35Hz$

A two meter long rod is suspended with the help of two wires of equal length. One wire is of steel and its cross-section is 0.1

$cm^{2}$ and another wire is of brass and its cross-section area is 0.2

$cm^{2}$ . If a load W is suspended from the rod and introduced in both the wires is same then the ratio of tensions in them will be

0%
(a) will depend on the position of W
0%
(b) $T_{1}/T_{2}=2$
0%
(c) $T_{1}/T_{2}=1$
0%
(d) $T_{1}/T_{2}=0.5$

Explanation

Given,

$A_S=0.1 cm^2$

$A_B=0.2 cm^2$

$T_1x=T_2.(l-x)$

Here 'x' is not given

So it will depend on the position of W.

A spherical ball contracts in volume by

$0.01$ % when subjected to a normal uniform pressure of

$100$ atmospheres. The bulk modulus of its material in dyne

${cm}^{-2}$ is:

0%
$10\times {10}^{11}$
0%
$100\times {10}^{2}$
0%
$1\times {10}^{11}$
0%
$2\times {10}^{11}$

Explanation

$\dfrac{\Delta V}{V}=\dfrac{0.01}{100}=10^{-4}$

$p=100atm$

$B=\dfrac{p}{\Delta V/V}=\dfrac{100 \times 10^5}{10^{-4}}=10^{11} N/m^2$

$=10^{12} dyne/cm^2$

The length of of a metal wire

$l$ when the tension in is

$'F'$ and

$'xl'$ when the tension is

$'yF'$ . Then the natural length of the wire is

0%
$\dfrac { \left( x-y \right) l }{ x-1 }$
0%
$\dfrac { \left( y-x \right) l }{ y-1 }$
0%
$\dfrac { \left( x-y \right) l }{ x+1 }$
0%
$\dfrac { \left( y-x \right) l }{ y+1 }$

Explanation

Let, Natural length ${{L}_{o}}$

Stress = Young modulus x strain

$\dfrac{F}{A}=Y\times \left( \dfrac{L-{{L}_{o}}}{{{L}_{o}}} \right)\ ......\ (1)$

$\dfrac{yF}{A}=Y\times \left( \dfrac{xL-{{L}_{o}}}{{{L}_{o}}} \right)\ ......\ (2)$

Divide equation (2) by (1)

$y=\dfrac{xL-{{L}_{o}}}{L-{{L}_{o}}}$

$\Rightarrow y\left( L-{{L}_{o}} \right)=\left( xL-{{L}_{o}} \right)$

$\Rightarrow {{L}_{o}}=\dfrac{\left( y-x \right)L}{y-1}$

Natural length of wire is $\dfrac{\left( y-x \right)L}{y-1}$

A tungsten wire of length 20 cm is stretched by 0.1 cm. Find the strain on the wire.

0%
0.002
0%
0.005
0%
0.001
0%
0.004

Explanation

Given,

Length of wire

$=20\,cm$

Elongation

$=0.1\,cm$

We have the equation,

$Strain=\dfrac{Elongation}{Length}=\dfrac{0.1}{20}=0.005$

When temperature of a gas is

$20^oC$ and pressure is changed from

${ P }_{ 1 }=1.0\times { 10 }^{ 5 } Pa$ to

${ P }_{ 2 }=1.65\times { 10 }^{ 5 } Pa$ and the volume is changed by 10%. The bulk modulus is :

0%
$1.55\times { 10 }^{ 5 }Pa$
0%
$1.15\times { 10 }^{ 5 }Pa$
0%
$1.4\times { 10 }^{ 5 }Pa$
0%
$1.01\times { 10 }^{ 5 }Pa$

Explanation

Given,

$T=20^0C$

$P_1=1.01\times 10^5 Pa$

$P_2=1.165\times 10^5 Pa$

$\Delta P=P_2-P_1=1.165\times 10^5-1.01\times 10^5 =0.155\times 10^5 Pa$

$\dfrac{\Delta V}{V}=10$ %

$=0.1$

Bulk modulus,

$B=\dfrac{\Delta P}{\Delta V/V}$

$B=\dfrac{0.65\times 10^5}{0.1}=1.55\times 10^5 Pa$

The correct option is A.

A solid cube is subjected to a pressure of

$\left( 5\times { 10 }^{ 5 }\quad N/{ m }^{ 2 } \right)$ . Each side of the cube is shortened by then volumetric strain and Bulk modulus of the cube are

0%
$0.03,5\times { 10 }^{ 5 }\quad N/{ m }^{ 2 }$
0%
$0.03,1.67\times { 10 }^{ 7 }\quad N/{ m }^{ 2 }$
0%
$3,1.67\times { 10 }^{ -7 }\quad N/{ m }^{ 2 }$
0%
$0.01,1.67\times { 10 }^{ 7 }\quad N/{ m }^{ 2 }$

Explanation

$P=5\times 10^5N/m^2$ , let initial side=

$a$

$V=a^3$

Shortened side=

$0.99a$

Shortened volume=

$(0.99)^3a^3$

$B=\frac { -V\triangle P }{ \triangle v } \Rightarrow \dfrac { -V\times 5{ \times 10 }^{ 5 } }{ -{ a }^{ 3 }\left( 1-{ (0.99) }^{ 3 } \right) } \\ \Rightarrow \dfrac { 5{ \times 10 }^{ 5 } }{ 0.030 } \Rightarrow 166.66{ \times 10 }^{ 5 }\\ \Rightarrow 1.67{ \times 10 }^{ 7 }N/{ m }^{ 3 }$

$\dfrac { \triangle v }{ V } =0.03\rightarrow$ Volumetric strain.

Three wires P,Q and R the same materials and length have radii

$0.12cm,0.2cm$ and

$0.3cm$ respectively. Which wire has the highest value of Young's modules of elasticity ?

0%
P
0%
Q
0%
R
0%
All have the same value

Explanation

We know Young's Modulus of elasticity

$E=\dfrac{stress}{strain}$ ------(A)

Also we know stress

$\sigma=\dfrac{F}{A}$ where F and A are the force and area of cross section.

And strain

$\epsilon=\dfrac{{l}_{0}}{\Delta l}$ where

${l}_{0}$ and

$\Delta L$ are the original length and elongation length.

Reducing the above found values of stress and strain in A we get,

$E=\dfrac{F{l}_{0}}{A\Delta l}$

Considering force applied in each wires to be same and given length of wires P,Q and R are same.

${E}_{P}=\dfrac{F{l}_{0}}{0.12\Delta {l}_{P}}$ -------(B)

${E}_{Q}=\dfrac{F{l}_{0}}{0.2\Delta {l}_{Q}}$ -------(C)

${E}_{R}=\dfrac{F{l}_{0}}{0.3\Delta {l}_{R}}$ -------(D)

Now comparing equation B,C and D we can say

$E\alpha \dfrac{1}{A\Delta l}$

Thus we can say that wire P will have the highest Young's modulus of elasticity.

The stress required to double the length of wire (or) to produce

$100\%$ longitudinal strain is:

0%
$Y$
0%
$\cfrac{Y}{2}$
0%
$2Y$
0%
$3Y$

Explanation

$\cfrac{Stress}{Strain}=Young's$

$modulus$

$Longitudinal$

$strain=\cfrac{Change\quad in\quad length}{Initial\quad length}$

Here,

$\cfrac{2L-L}{L}=\cfrac{L}{L}=1$

$\therefore Y=\cfrac{Stress}{1}$

A uniform rod of length

$60 cm$ and mass

$6kg$ is acted upon by two forces as shown in the diagram. The force exerted by

$45 cm$ part of the rod on

$15 cm$ part of the rod is

0%
$9 N$
0%
$18 N$
0%
$27 N$
0%
$30 N$

Explanation

The force exerted by $45\;{\rm{cm}}$ part of the rod on $15\;{\rm{cm}}$ art of the rod is given as,

$F = {F_1} \times \frac{{15}}{{60}} + {F_2} \times \frac{{45}}{{60}}$

$F = 21 \times \frac{{15}}{{60}} + 45 \times \frac{{45}}{{60}}$

$F = 45 \times \frac{{15}}{{60}} + 21 \times \frac{{45}}{{60}}$

$F = 27\;{\rm{N}}$

Two wires A and B are identical in shape and size are stretched by same magnitude of force. Then the extensions are found to be 0.2% and 0.3% respectively. Find the rate of their Young's modulii

0%
2 : 3
0%
3 : 2
0%
4 : 9
0%
9 : 4

Explanation

$\gamma{A}=\dfrac{stress}{strain}$

$stress_{A}=stress_{B}$

$\dfrac { { \gamma }_{ A } }{ { \gamma }_{ B } } =\dfrac { { strain }_{ B } }{ { stress }_{ A } } =\dfrac { 0.3\times { 10 }^{ -2 } }{ 0.2\times { 10 }^{ -2 } }$

$=3:2$

For a material

$Y={ 6.6\times 10 }^{ 10 }\ { N/m }^{ 2 }$ and bulk modulus

$K{ 11\times 10 }^{ 10 }\ { N/m }^{ 2 }$ , then its Poisson's ratio is:

0%
$0.8$
0%
$0.35$
0%
$0.7$
0%
$0.4$

Explanation

Given that,

Young’s modulus $Y=6.6\times {{10}^{10}}\,N/{{m}^{2}}$

Bulk modulus $B=11\times {{10}^{10}}\,N/{{m}^{2}}$

We know that,

$Y=3K\left( 1-2\mu \right)$

$6.6\times {{10}^{10}}=3\times 11\times {{10}^{10}}-66\times {{10}^{10}}\mu$

$-\mu =\dfrac{\left( 6.6-33 \right)\times {{10}^{10}}}{66\times {{10}^{10}}}$

$\mu =0.4$

Hence, the poisson’s ratio is $0.4$

A copper wire and an aluminium wire have lengths in the ratio 3:2, diameters in the ration 2:3 and forces applied in the ration 4:Find the ratio of the increase in the length of the two wires.

$\left( { Y }_{ cu }=1.1\times { 10 }^{ 11 }N{ m }^{ -2 },\quad { Y }_{ Al }=0.70\times { 10 }^{ 11 }N{ m }^{ -2 }\quad \right)$

0%
110:189
0%
180:110
0%
189:110
0%
80:11

Explanation

Youngs Modulus

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

$\Rightarrow \Delta L=\dfrac{FL}{AY}$

Given :-

$L_{cu}:L_m=3:2$

$d_{cu}:d_m=2:3$

$A_{cu}:A_{Al}=\dfrac{\pi}{4}(d_{cu})^2/\dfrac{\pi}{4}(d_{Al})^2$

$=\left(\dfrac{2}{3}\right)^2$

$F_{cu}:F_{Al}=4:5$

Ratio of increase in length of wires

$=\left(\dfrac{F_{cu}}{F_{Al}}\right)\left(\dfrac{L_{cu}}{L_{Al}}\right)\left(\dfrac{A_{Al}} {A_{Cu}}\right)\left(\dfrac{Y_{Al}}{Y_{Cu}}\right)$

$=\left(\dfrac{4}{5}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{9} {4}\right)\left(\dfrac{7}{11}\right)$

$=\dfrac{189}{110}$

1457466_1079326_ans_34796ee92ee84053b76cc569bb82d189.PNG

In the Searle's method to determine the Young's modulus of a wire, a steel wire of length

$156\ cm$ and diameter

$0.054\ cm$ is taken as experimental wire. The average increase in length for

$1.5\ kg\ wt$ is found to be

$0.050\ cm$ . Then the Young's modulus of the wire is

0%
$3.002\times 10^{11}N/m^{2}$
0%
$1.002\times 10^{11}N/m^{2}$
0%
$2.002\times 10^{11}N/m^{2}$
0%
$2.5\times 10^{11}N/m^{2}$

Explanation

Young’s modulus is given by

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

$Y=\dfrac{1.5\times 10\times 156}{3.14\times 2.7\times {{10}^{-4}}\times 5\times {{10}^{-4}}}$

$Y=1.002\times {{10}^{-11}}\,N/{{m}^{2}}$

Three bars having length

$l,2l$ and

$3l$ and area of cross-section

$A,2A$ and

$3A$ are joined rigidly and to end. Compound rod is subjected to a stretching force

$F$ . The increase in length of rod is (Young's modulles of material is

$Y$ and bars are massless)

0%
$\dfrac {13Fl}{2\ AY}$
0%
$\dfrac {3Fl}{AY}$
0%
$\dfrac {9Fl}{AY}$
0%
$\dfrac {13\ Fl}{AY}$

Explanation

Changed in length of compound bar

$=\Delta {l}_{1}+\Delta {l}_{2}+\Delta {l}_{3}$

$\Delta l=\cfrac{PL}{AE}=\cfrac{FL}{AE}=\cfrac{FL}{AY}$ (

$E=$ Young's modulus)

$=\cfrac{Fl}{AY}+\cfrac{F(2l)}{2AY}+\cfrac{F(3l)}{3AY}$

$=\cfrac{3Fl}{AY}$

A hydraullic press contains

$250 lit$ of oil. Find the decrease in volume of the oil when its pressure increases to

$10^7 Pa$ . The bulk modulus of the oil is

$K = 5 \times 10^5 Pa$

0%
$- 0.8 lit$
0%
$- 0.5 lit$
0%
$- 0.6 lit$
0%
$- 0.9 lit$

Explanation

Bulk modulus

$=-\dfrac{volume \times \Delta Pressure}{\Delta volume}$

$\Delta volume$

$=-\dfrac{volume \times \Delta Pressure}{Bulk \quad modulus}$

$\Delta volume=-\dfrac{250\times 10^7}{5\times 10^9}=-0.5$ liters.

The breaking stress of aluminium is

$7.5\times 10^{7}Nm^{-2}$ . The greatest length of aluminium wire that can hang vertically without breaking is(Density of auminium is

$2.7\times 10^{3}\ kg\ m^{-3}$ )

0%
$283\times 10^{3}\ m$
0%
$28.3\times 10^{3}\ m$
0%
$2.83\times 10^{3}\ m$
0%
$0.283\times 10^{3}\ m$

Explanation

$\begin{array}{l} \text { Given, breaking stress of aluminium }=7.5 \times 10^{7}\mathrm{~N}/\mathrm{m}^{2}\\\text { Density of aluminium }==2.7 \times 10^{3}\mathrm{~kg}/\mathrm{m}^{3}\\\text { Stress }=\frac{F}{A}=\frac{m g}{A} \\ d=\frac{m}{A \times L}\Rightarrow\frac{m}{A}=d x L \\ \therefore \quad 7.5\times 10^{73}=2 \cdot 7 \times 10^{3} \times L \times 100\\L=2.83\times 10^{3} \mathrm{~m} \text { . } \end{array}$

A uniform rod of mass m and length

$l$ is rotating with constant angular velocity

$\omega$ about an axis which passes through its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its young's modulus is Y. Neglect gravity. The strain at the mid point of the rod is :

0%
$\dfrac{m\omega^2l}{8AY}$
0%
$\dfrac{3m\omega^2l}{8AY}$
0%
$\dfrac{3m\omega^2l}{4AY}$
0%
$\dfrac{m\omega^2l}{4AY}$

Explanation

$\begin{array}{l}\text { Stress }=y(\text { strain }) \\\text { Strain }=\dfrac{T_{\text {mid-point }}}{A Y}\end{array}$

$\begin{array}{l}\text { Let us assume a small element of } \\\text { length } d x \text { and mass dm. } \\\text { Mass per unit length }=\frac{m}{L} \\\qquad \begin{array}{rl}d & m=\left(\frac{m}{L}\right)(d x) \\d T & =(d m)(l-x) \omega^{2} \\& =\left(\frac{m}{L}\right)(1-x)\omega^{2} d x\end{array}\end{array}$

$\begin{array}{l}\text { Now, centripetal force at mid point is }\\\qquad \begin{aligned}\int_{0}^{l / 2} d T &=\int_{0}^{l / 2}\frac{m}{L}(l-x) w^{2} d x \\&=\frac{m w^{2}}{L}\left[l x\frac{x^{2}}{2}\right]_{0}^{1 / 2}\end{aligned}\end{array}$

$\begin{array}{l}T_{\text {mid-point }}=\dfrac{M \omega^{2}}{L}\left(\dfrac{L^{2}}{2}-\dfrac{L^{2}}{8}\right)=\dfrac{3 M \omega^{2} L}{8} \\\text { Strain }=\dfrac{3 m \omega^{2} L}{8 A Y}\end{array}$

The length of a rubber cord is

$l_1$ metres when the tension in it is

$4N$ and

$l_2$ metres when the tension is

$5N$ . then the length in meters when the tension is

$9N$ is

0%
$3l_2 +4l_1$
0%
$3l_2 +2l_1$
0%
$5l_2 - 4l_1$
0%
$3l_2 - 2l_1$

Explanation

Let,

${ x }_{ 1, }\quad { x }_{ 2 },\quad and\quad { x }_{ 3 }\quad be\quad the\quad extension\quad with\quad 4N,\quad 5N\quad and\quad 9N\quad tension\quad respectively.\\ 4=\quad k{ x }_{ 1, },\quad 5=k{ x }_{ 2 }\quad and\quad 9=k{ x }_{ 3 }\quad \\$

Let l be the natural length of the rubber chord.

Therefore, we can say,

${ x }_{ 1 }+l=\quad { l }_{ 1 }-------(a)\\ { x }_{ 2 }+l=\quad { l }_{ 2 }--------(b)\\ { x }_{ 3 }+l=\quad { l }_{ 3 }--------(c)\\ \dfrac { 4 }{ 5 } =\dfrac { { x }_{ 1 } }{ { x }_{ 2 } } =\dfrac { { l }_{ 1 }-l }{ { l }_{ 2 }-l } \\ \quad \quad \quad =>l=5{ l }_{ 1 }-{ 4 }l_{ 2 }-------(1)\\ \\ \dfrac { 9 }{ 5 } =\dfrac { { x }_{ 3 } }{ { x }_{ 2 } } =\dfrac { { l }_{ 3 }-l }{ { l }_{ 2 }-l } \\ \qquad =>9{ l }_{ 2 }-9l=\quad 5{ l }_{ 3 }-5l\\ \qquad =>4l=9{ l }_{ 2 }-5{ l }_{ 3 }\\ From\quad (1)\quad =>\quad 4(5{ l }_{ 1 }-4{ l }_{ 2 })=9{ l }_{ 2 }-5{ l }_{ 3 }\\ \qquad \qquad \qquad =>20{ l }_{ 1 }=25{ l }_{ 2 }-5{ l }_{ 3 }------(2)\\$

$From\quad euation\quad (2,)\quad { l }_{ 3 }=5{ l }_{ 2 }-4{ l }_{ 1 }$

Therefore at 9N tension, extension is (

$5{ l }_{ 2 }-4{ l }_{ 1 }$ )m

The extension of wire by application of load is

$0.3cm$ The extension in a w wire of same material but of double the length and half the radius of cross section by the same load will be in (cm)

0%
$0.3$
0%
$0.6$
0%
$0.2$
0%
$2.4$

Explanation

given, extension of wire by application of a load is

$0.3 \mathrm{~cm}$ $

Now, on doutling length and making the radics half of its origenal, we get

$\begin{aligned}A_{f} &=\frac{A_{i}}{4} \quad L_{f}=2 L \\F &=Y A_{i}\left(\frac{0 \cdot 3}{L}\right) \\F &=Y \frac{A_{i}}{4}\left(\frac{\Delta L}{2 L}\right) \Rightarrow \frac{F}{0.3} \times \frac{\Delta L}{8} \\& \therefore \quad \Delta L=2 \cdot 4 \mathrm{~cm}\end{aligned}$

An aluminium rod has a breaking strain

$0.2\%$ . The minimum cross-sectional area of the rod in

$m^{2}$ in order to support a load of

$10^{4}N$ is if (Young's modulus is

$7\times 10^{9}Nm^{-2}$ )

0%
$1.7\times 10^{-4}$
0%
$1.7\times 10^{-3}$
0%
$7.1\times 10^{-4}$
0%
$1.4\times 10^{-4}$

Explanation

We know

Young's modulus

$(\gamma)=\cfrac{stress}{strain}$

For breaking (maximum) strain

$0.2$ % and load of

${10}^{4}N$ , let the minimum supportive cross-sectional area be

$A$ then,

$\gamma =\cfrac{\cfrac{L}{A}}{breaking\quad strain}$

$\Rightarrow$

$A=\cfrac{L}{r\times B.S}=\cfrac{{10}^{4}}{7\times{10}^{9}\times (\cfrac{2}{1000})}$

$A=7.1\times {10}^{-4}{m}^{2}$

Two wires A and B have young's modulii in the ratio

$1:2$ and ratio of the lengths is

$1:1$ . under the application of same stress the ratio of elongation is

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:4$

Explanation

Given,

$Y_A:Y_B=1:2$

$l_A:l_B=1:1$

Stress,

$s=\dfrac{F}{A}$

Young's modulus,

$Y=\dfrac{F/A}{\Delta l/l}$

$Y\propto \dfrac{l}{\Delta l}$

$\Delta l \propto \dfrac{l}{Y}$

$\dfrac{\Delta l_A}{\Delta l_B}=\dfrac{l_A}{l_B}.\dfrac{Y_B}{Y_A}$

$\dfrac{\Delta l_A}{\Delta _B}=\dfrac{1}{1}.\dfrac{2}{1}=\dfrac{2}{1}$

$\Delta l_A:\Delta l_B=2:1$

The correct option is B.

A steel wire

$2\ m$ in length is stretched by

$1\ mm$ . The cross-sectional area of the wire is

$2\ mm^{2}$ . If Young's modulus of steel is

$2\times 10^{11}\ N/m^{2}$ , then the elastic potential energy stored in the wire is

0%
$0.1\ J$
0%
$0.2\ J$
0%
$0.3\ J$
0%
$0.4\ J$

Explanation

given, Steel wire length

$=2 \mathrm{~m}$

$Area=2 \mathrm{~mm}^{2}$

Young modulus of steel

$=2 \times 10^{11}$

Using the spring equivalent concept:

$\begin{aligned} U &=\frac{1}{2} K(\Delta L)^{2} \quad \text { where } \quad K=\frac{Y A}{L} \\ &=\frac{1}{2}\left(\frac{Y A}{L}\right)(\Delta L)^{2} \\ &=\frac{1}{2} \times\left(\frac{2 \times 10^{11} \times 10^{-6} \times 2}{2}\left(10^{-6}\right)\right.\\ U &=0.1\mathrm{~J}\end{aligned}$ Elastic potential energy stored is

$\underline{0.1 \mathrm{I}}$

An aluminium wire and a steel wire of the same length and cross-section joined end to end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in the length of

$t\ h\ c$ composite wire is

$2.7\ mm$ , then the increase in the length of each wire is (in

$mm$ ).

$(Y_{At}=2\times 10^{11}Nm^{-3}, Y_{steel}=7\times 10^{11}Nm^{-2})$

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$1, 7, 1$
0%
$1.3, 1, 4$
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$1.5, 1, 2$
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$2.1, 1, 0.6$

Consider the bar in Fig. Cross-sectional area

$A_{e} = 1.2 in.^{2}$ , and Young's modulus

$E = 30\times 10^{6} psi$ . If

$q_{1} = 0.02\ in.$ and

$q_{2} = 0.025\ in.$ , determine the following (by hand calculation).

0%
The displacement at point $P$
0%
The strain $\epsilon$ and stress $\sigma$
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The element stiffness matrix, and
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The strain energy in the element

The force constant of a wire is

$k$ and that of another wire of the same material is

$2k$ . When both the wires are stretched, then work done is

0%
$W_{2}=1.5W_{1}$
0%
$W_{2}=2W_{1}$
0%
$W_{2}=W_{1}$
0%
$W_{2}=0.5W_{1}$

Explanation

$\begin{array}{l}\text { In wire } 1 \text { , force constant }=k \\\text { in wree } 2 \text { , force constant }=2 \mathrm{~K} \\\text { Now, using Hooke's law }\\\qquad\begin{aligned}F &=K x_{1} \\W_{1} &=\frac{1}{2} K(\Delta x)^{2}\\&=\frac{1}{2}(K)\left(\frac{F}{K}\right)^{2} \\W_{1} &=\frac{F^{2}}{2 K} -(1)\end{aligned}\end{array}$

$\begin{array}{l}\text { For wire } 2, \quad F=(2 k)\left(x_{2}\right) \\\qquad \begin{aligned}w_{2}=& \frac{1}{2}(2 k)\left(x_{2}\right)^{2} \\&=\frac{1}{2}(2 k)\left(\frac{F}{2 k}\right)^{2}=\frac{F^{2}}{4 k}-(2)\\\text { Woing equation }(1) \text { and }(2) \text { we get } \\w_{2} &=0.5 w_{1}\end{aligned}\end{array}$

A uniform cubical block is subjected to volumetric compression, which decreases its each side of

$2$ %. The Bulk strain produced in it is:

0%
$0.03$
0%
$0.02$
0%
$0.06$
0%
$0.12$

Explanation

$\begin{array}{l}\text { Given, a cutical block whose sides } \\\text { are reduced by } 2 \% \\\text { New side }=l-\frac{2 l}{100}=\frac{981}{100} \\\text { New volume}=\left(\frac{98 l}{100}\right)^{3} \\\Rightarrow \quad V_{f}=\left(\frac{98}{100}\right)^{3} \mathrm{~V}_{i} \\\end{array}$

$\Rightarrow \quad \frac{\Delta V}{V}=\begin{array}{l}\text { BUlR } \\\text { Strain }\end{array}\begin{aligned}=& 1-\left(\frac{98}{100}\right)^{3} =& 0.06 \end{aligned}$

There are two wires of same material. their radii and lengths are both in the ratio 1:if the extensions produced are equal, what is the ratio of the loads?

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1:2
0%
2:1
0%
1:4
0%
4:1

Explanation

Hint: According to the definition of Young's modulus, $Y=\dfrac{F. l}{A.\Delta l}$

Correct Answer: Option A

Explanation of Correct Option:

Let $\Delta l$ and $l$ be the elongation and initial length of the wire respectively,

$A$ and

$r$ be the area and radius of the wire respectively,

$F$ will be the load on the wire.

Since the ratio of their radii and length is $1:2$ , thus $r_1:r_2=1:2$ and $l_1:l_2=1:2$
As, the extensions produced is equal, thus $\Delta l_1=\Delta l_2=\Delta l$
Since $r_1:r_2=1:2$ and Area is the square of the radius, therefore $A_1:A_2=1:4$ .
Material of both the wires is same therefore, Young's modulus constant of both wires is equal

$Y_1=Y_2$

$\dfrac{F_1.l_1}{A_1.\Delta l}=\dfrac{F_2.l_2}{A_2.\Delta l}$

$\therefore \dfrac{F_1}{F_2}=\dfrac{A_1}{A_2}\times \dfrac{l_2}{l_1}$

$\therefore \dfrac{F_1}{F_2}=\dfrac{1\times 2}{4\times 1}=\dfrac{2}{4}=\dfrac{1}{2}$

Thus, the ratio of the loads is $1:2$

A thick rope of density

$\rho$ and length

$L$ is hung from a rigid support. The Young's modulus of the material of rope is

$Y$ . The increase in length of the rope due to its own weight is

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$\rho g{L}^{2}/4Y$
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$\rho g{L}^{2}/2Y$
0%
$\rho g{L}^{2}/Y$
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$\rho gL/Y$

Explanation

Take a small section

$\delta y$ at distance

$Y$ from top as shown in the figure.Let area of rope be

$A$

Stress due to this section

$=\rho Ag\dfrac{\delta y}{A}=\rho d \delta y$

Strain on section

$Y=\dfrac{\delta l}{y}$

Using stress

$=Y\times$ strain

$\Rightarrow \rho g\delta y=Y\times\dfrac{\delta l}{y}$

$\Rightarrow \Delta l=\dfrac{\rho g}{y}\displaystyle\int_{0}^{L}{y\delta y}$

$\Rightarrow \Delta l=\dfrac{\rho g{L}^{2}}{2Y}$

1267689_1108593_ans_01cd4903801048f796fafb1f3e1849d8.PNG

When a wire is stretched, its length increases by 0.3% and the diameter decreases by 0.1%. Poisson's ratio of the material of the wire is about

0%
0.03
0%
0.333
0%
0.15
0%
0.015

Explanation

$\begin{array}{l}\text { Given, increase in length of wire }=0.3\%\\\text { increase in diameter }=0.1 \% \\\text { We know that, } \\\text { } \quad \text { } \quad d=2r \\\qquad \ln (d)=\ln (2)+\ln (r) \\\text{Differentiating,}\qquad \frac{\Delta d}{d}=\frac{\Delta r}{r}\end{array}$

$\begin{aligned}&\therefore \Delta r=\frac{-0.1}{100}(r)=\frac{-r}{1000}\left(\begin{array}{l}\text { radius is } \\\text { decreasing })\end{array}\right.\\&\begin{aligned}\Delta l=\frac{0.3}{100}(l) &=\frac{3 l}{1000} \\\text { Now, poisson's ratio }(\sigma)&=\frac{\Delta r / r}{\Delta l / l} \\&=\frac{1 / 1000}{3 / 1000}=\frac{1}{3} \\&=0.333\end{aligned}\end{aligned}$

A steel ring of radius

$r$ and cross sectional area

$A$ is fitted on to a wooden disc of radius

$R(R> r)$ . If Young;s modulus be

$Y$ , then the force with which the steel ring is expanded, is

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$AY\cfrac{R}{r}$
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$AY(\cfrac{R-r}{r})$
0%
$\cfrac{Y}{A}\cfrac{(R-r)}{r}$
0%
$\cfrac{Yr}{AR}$

Explanation

Young modulus

$\begin{array}{l} \left( Y \right) =\frac { { FL } }{ { Ax } } \\ F=\dfrac { { AxY } }{ L } \end{array}$

But length

$=2\pi r$ (circumference of ring)

Change in length

$\left( x \right) =2\pi R-2\pi r=2\pi r=2\pi \left( { R-r } \right)$

Hence force of expansions is

$F=\dfrac { { YA\times 2\pi \left( { R-r } \right) } }{ { 2\pi r } } =\dfrac { { YA\left( { R-r } \right) } }{ r }$

Volume of a liquid when compressed by additional pressure of

$10^5 N/m^2$ is

$196$ cc and when compressed by a pressure of

$1.5 \times 10 ^5 N/m^2$ , the volume is

$194cc$ . The bulk modulus of the liquid is:

0%
$10^5N/m^2$
0%
$1.5 \times 10^6$
0%
$5 \times10^5N/m^2$
0%
$5 \times 10^6N/m^2$

Explanation

Bulk modulus is the ratio of the infinitesimal increase of pressure to the relative decrease in volume.

When calculating the pressure changes and volume changes we get the ratio as 1.5 ×

$10^6$ .

So the correct answer is ' 1.5 ×

$10^6$ '.

The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

0%
be double
0%
be half
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be four times
0%
remain same

Explanation

Hint: The stress is the ratio of force and the area.

Step 1: Explanation,

The expression or the stress can be written as,

$\begin{array}{l}\text { Stress }=\frac{F}{A}\\\qquad\Rightarrow \quad F=(\text { Stress }) \times(A) \\\text { On reducing the length of the wire } \\\text { to half , neither the stress change } \\\text { nor the area of cross- section. } \\\text { Maximum load wire can hold } \\\text { remains same. }\end{array}$

One end of a uniform wire of length L and of weight W you is attached rigidly see through a point in the roof and weight W1 is suspended from its lower end. If S is true area of cross section of the wire the stress in the wire at a height (3L/4) from its lower end is :

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$[W_1 + (W/4)] / S$
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$W_1/S$
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$[W_1 + (3W/4)] / S$
0%
$(W_1+W) / S$

A uniform rod of length

$L$ has a mass per unit length

$\lambda$ and area of cross section

$A$ . If the Young's modulus of the rod is

$Y$ . The elongation in the rod due to its own weight is

0%
$\dfrac{2\lambda gL^2}{AY}$
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$\dfrac{\lambda gL^2}{2AY}$
0%
$\dfrac{\lambda gL^2}{4 AY}$
0%
$\dfrac{\lambda gL^2}{6AY}$

Explanation

$F=\frac { { Mg } }{ 2 }$ (mass of rod will at

$\frac{L}{2}$ only

$\begin{array}{l} \Delta l=\frac { { FL } }{ { AY } } \Rightarrow \Delta l=\frac { { MgL } }{ { 2AY } } =\frac { { Mg{ L^{ 2 } } } }{ { 2AYL } } \\ \Rightarrow \frac { { \lambda g{ L^{ 2 } } } }{ { 2Ay } } \left( { \lambda =\frac { M }{ G } } \right) \end{array}$

The relationship between Young's modulus

$Y$ , Bulk modulus

$K$ and modulus of rigidity

$\eta$ is:-

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$Y = \dfrac{9 \eta K}{3 + k}$
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$Y = \dfrac{9 \eta K}{n + 3}$
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$Y=\dfrac{9 \eta K}{\eta+3 k}$
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$Y = \dfrac{3 \eta K}{9 \eta + k}$

Two wires of same material and radius have their lengths in ratio

$1:2$ . If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

0%
$2:1$
0%
$1:1$
0%
$1:2$
0%
$1:4$

Explanation

Hint: Young's modulus constant is the ratio of stress to the strain.

Correct Answer: Option B

Explanation of correct option:

As the material is the same for both the wires, Young's modulus

$(Y)$ remains constant.

Here,

$Y=\dfrac{stress}{strain}$

Also, force as well as radius is also same, the stress will be same for both the wires.

Since stress and Young's modulus are constant, the strain will also remain constant.

Thus, the strain produced in the two wires will be in the ratio of $1:1$ .

A rod of mass

$'M'$ is subjected to force

$'t'$ and

$'2f'$ at both the ends as shown in the figure. If young modulus of its material is

$'y'$ and its length is

$L$ find total elongation of rod.

0%
$\dfrac{fl}{2Ay}$
0%
$\dfrac{fl}{Ay}$
0%
$\dfrac{3fl}{2Ay}$
0%
$\dfrac{4fl}{2Ay}$

Explanation

To solve problem of these types we consider average force acting

$\begin{array}{l} \Delta l=\dfrac { { FL } }{ { Ay } } =\left( { \dfrac { { { f_{ 1 } }+{ f_{ 2 } } } }{ 2 } } \right) \dfrac { L }{ { Ay } } \\ =\left( { \dfrac { { f+2f } }{ 2 } } \right) \dfrac { L }{ { Ay } } \\ =\dfrac { { 3fL } }{ { 2Ay } } \\ Total\, elongation\, =\Delta l=\dfrac { { 3fL } }{ { 2Ay } } \\ Hence,\, the\, option\, \; C\, is\, the\, correct\, answer. \end{array}$

The bulk modulus of water is

$2.1 \times 10^9 N/m^2$ . The pressure required to increase the density of water by

$0.1$ % is:-

0%
$2.1 \times 10^5 N/m^2$
0%
$2.1 \times 10^3 N/m^2$
0%
$2.1 \times 10^6 N/m^2$
0%
$2.1 \times 10^7 N/m^2$

Explanation

$\begin{array}{l} \frac { { \Delta V } }{ V } =\frac { { -\Delta P } }{ P } =-{ 10^{ -3 } } \\ \Rightarrow \Delta P=-B\frac { { \Delta V } }{ V } =2.1\times { 10^{ 9 } }\times { 10^{ -3 } }=2.1\times { 10^{ 6 } } \end{array}$

A wire is stretched by

$0.01\ m$ by a certain force

$F$ . Another wire of same material whose diameter and lengths are doubled to the original wire is stretched by the same force. Then its elongation will be:

0%
$0.005\ m$
0%
$0.01\ m$
0%
$0.02\ m$
0%
$0.08\ m$

Explanation

Given that,

$\Delta {{l}_{1}}=0.01\,m$

Length ${{l}_{1}}=l$

Length ${{l}_{2}}=2l$

Radius ${{r}_{1}}=r$

Radius ${{r}_{2}}=2r$

We know that,

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

$\Delta l=\dfrac{Fl}{\pi {{r}^{2}}Y}$

$\Delta l\propto \dfrac{l}{{{r}^{2}}}$

Now, the diameter and lengths are doubled to the original wire is stretched by the same force

So,

$\dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=\dfrac{2l}{l}\times \dfrac{{{r}^{2}}}{4{{r}^{2}}}$

$\dfrac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=2\times \dfrac{1}{4}$

$\Delta {{l}_{2}}=\dfrac{0.01}{2}$

$\Delta {{l}_{2}}=0.005\,m$

Hence, the elongation is $0.005\ m$

If the temperature of a wire of length

$2m$ area of cross-section

$1{cm}^2$ is increased from

${0^ \circ}C$ to

${80^ \circ}C$ and is not allowed to increase in length, then force required for it is {

$Y = 10^{10}N/m^2, \alpha = 10^{ - 6}/^oC$ }

0%
$80N$
0%
$160N$
0%
$400N$
0%
$120N$

Explanation

80N

Formula,

$F=YA\alpha \Delta T$

$=10^{10}\times 0.0001\times 10^{-6}\times (80-0)$

$\therefore F=80N$

The pressure of a medium is changed from

$1.01\times 10^{5}\ Pa$ to

$1.165\times 10^{5}\ Pa$ and change in volume is

$10\%$ keeping temperature constant. The bulk modulus of the medium is

0%
$204.8\times 10^{5}\ Pa$
0%
$102.4\times 10^{5}\ Pa$
0%
$51.2\times 10^{5}\ Pa$
0%
$1.55\times 10^{5}\ Pa$

Explanation

$\textbf{Step 1 - Bulk modulus K}$

Given,

$P_{i} = 1.01\times 10^{5} Pa, P_{f} = 1.165\times 10^{5} Pa$

$\dfrac {\triangle V}{V} = 10\% = \dfrac {10}{100}$

$K = \dfrac {\triangle P}{\left (\dfrac {\triangle V}{V}\right )} = \dfrac {(P_{f} - P_{i})}{\left (\dfrac {\triangle V}{V}\right )} = \dfrac {(1.165 \times 10^{5} Pa - 1.01\times 10^{5} Pa)}{\left (\dfrac {10}{100}\right )}$

$\Rightarrow K = 1.55\times 10^{5} Pa$

A copper wire is having length

$2m$ and area of cross -section

$2mm^2$ . Then amount of work done (in joule ) in increasing its length by

$0.1mm$ will be (Young's modulus o elasticity for copper

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

0%
$6\times 10^4J$
0%
$6\times 10^{-3}J$
0%
$6\times 10^{3}J$
0%
$6\times 10^{-4}J$

Explanation

$W=\dfrac{1}{2}\times$ stress

$\times$ volume

$stress=y\times strain$

$W=\dfrac{1}{2}(strain)^2\times Vol$
Strain

$=\dfrac{\Delta L}{L}=\left(\dfrac{10-4}{2}\right)$

$vol=A\times L=2\times 10^{-6}\times 2$

$=4\times 10^{-6}$

$W=\dfrac{1.2\times 10^{11}}{2}\times \dfrac{10^{-8}}{4}\times 4\times 10^{-6}$

$=0.6\times 10^{-3}$

$W=6\times 10^{-4}J$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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