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

If the speed of longitudinal wave equals

$10$ times the speed of the transverse waves in a stretched wire of material which has modulus of elasticity E, then the stress in the wire is

0%
$100E$
0%
$\dfrac {E}{ 100 }$
0%
$\dfrac {E}{ 10 }$
0%
$10E$

Explanation

${v}_{l}=\sqrt{\dfrac{Y}{\rho}}$ and

${v}_{t}=\sqrt{\dfrac{Y\dfrac{\Delta l}{l}}{\rho}}={v}_{l}\sqrt{\dfrac{\Delta l}{l}}$

$\Rightarrow \sqrt{\dfrac{l}{\Delta l}}=\dfrac{{v}_{l}}{{v}_{t}}=10$

$\therefore \dfrac{\Delta l}{l}=\dfrac{1}{100}$

Stress

$=Y\dfrac{\Delta l}{l}=E\dfrac{\Delta l}{l}=\dfrac{E}{100}$

Two wires of the same material have masses in the ratio 3:The ratio of their extensions under the same load if their lengths are in the ratio 9:10 is

0%
5 : 3
0%
27 : 40
0%
6 : 5
0%
27 : 25

Explanation

$Y=\dfrac { FL }{ \left( Ae \right) }$

For same material

$Y$ is constant and

$A$ is constant for two similar wires.

$\dfrac { FL }{ e } =$ constant

here,

$F=mg$

$\dfrac { { e }_{ 1 } }{ { e }_{ 2 } } =\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } \times \dfrac { { L }_{ 1 } }{ { L }_{ 2 } }$

$=\dfrac { 3 }{ 4 } \times \dfrac { 9 }{ 10 }$

$=\dfrac { 27 }{ 40 }$

$40\ cm$ wire having a mass

$3.2\ gm$ and area of cross-section

$1\ mm^{2}$ is stretched between two supports

$40.05\ cm$ apart. In its fundamental note, the wire vibrates with a frequency

$220\ Hz$ . The Young's molulus is :

0%
$1.98\times 10^{11}\ N/m^{2}$
0%
$2.2\times 10^{11}\ N/m^{2}$
0%
$3.96\times 10^{11}\ N/m^{2}$
0%
$3.2\times 10^{11}\ N/m^{2}$

Explanation

$l = 40\,cm$

$\Delta l = 0.05\,cm$

$f = \dfrac{v}{{2l}}$

$\Rightarrow v = 22 \times 8 = 176m/s$

Now

$,$

$v = \sqrt {\dfrac{T}{\mu }}$

$\Rightarrow T = 176 \times 176 \times \dfrac{{3.2 \times {{10}^{ - 3}}}}{{40 \times {{10}^{ - 2}}}}$

$= \dfrac{{176 \times 176 \times 32}}{{40 \times 10 \times 10}}$

$= \dfrac{{176 \times 176 \times 32}}{{4 \times {{10}^3}}} = 247.8\,N$

$\therefore \gamma = \dfrac{{247.8 \times 40}}{{1 \times {{10}^{ - 6}} \times 0.05}}$

$= 1.98 \times {10^{11}}\,N/{m^2}$

Hence,

option

$(A)$ is correct answer.

The extension produced in a wire by the application of a load is

$3.0$ mm. The extension produced in a wire of the same material and length but half the radius by the same load is:

0%
$5$ mm
0%
$9$ mm
0%
$12$ mm
0%
$13$ mm

Explanation

Formulae,

$\dfrac{\Delta l_2}{\Delta l_1}=\dfrac{a_1}{a_2}$

$=\dfrac{a_1}{\dfrac{a_1}{4}}$

$\Rightarrow \Delta l_2=4\Delta l_1$

$\therefore \Delta l_2=4\times 3=12mm$

The

$'\sigma '$ of a material is 0.If a longitudinal strain of

$4.0\times { 10 }^{ -3 }$ is caused, by what percentage will the volume change -

0%
0.48 %
0%
0.32 %
0%
0.24 %
0%
0.50 %

Explanation

$\begin{array}{l}\text { Longitudenal } \text { Strain }=4\times 10^{-3} \\\text {Lateral Strain }=\sigma \times 0.4\% \\=0.08 \%\end{array}$

$\begin{aligned}\text { Volumetric strain } &=\text { long}\text {itudinal Strain }-2 \times \text { lateral Strain } \\&=4 \times 10^{-3}-2 \times 8 \times 10^{-4}\\&=0.24\%\end{aligned}$

The force required to punch a square hole 2 cm side in steel sheet 2 mm thick is:-
(shearing stress of steel sheet

$=3.5 \times10^8 N/m^2)$

0%
$5.6\times10^4 N$
0%
$3.4\times10^4 N$
0%
$9.1\times10^4 N$
0%
$6.8\times10^4 N$

Explanation

$\begin{array}{l}\text { Breaking area }=2 \pi r\times L \\=2 \pi x \\\text { Breaking area = (Side length) }\times \text { (thickness) } \times 4\\=2\times 10^{-2} \times 2 \times 10^{-3} \times 4\\=4 \times 10^{-5} \times 4 \\=16 \times 10^{-5}\end{array}$

$\begin{aligned}F &=(\text { Breaking stress })\times \text { Area } \\&=16 \times 10^{-5} \times 3.5 \times 10^{8} \\F &=5.6 \times 10^{4}\mathrm{~N}\end{aligned}$

In case of bending of a beam, depression

$\delta$ depends on Young modulus of elasticity

$Y$ as

0%
$\propto Y$
0%
$\propto Y^{2}$
0%
$\propto Y^{-1}$
0%
$\propto Y^{-2}$

Explanation

In case of bending of a beam, depression

$\delta$ depends on Young modulus of elasticity

$Y$ as

$\delta = \dfrac{PL^3}{48YI}$

$\therefore \delta \propto Y^{-1}$

Depression δ=WL34bd3Y⇒δ∝1Y Depression δ=WL34bd3Y⇒δ∝1

When a wire is stretched, an amount of work is done. What is the amount of work done in stretching a wire through

$0.1mm$ , if its length is

$2m$ and area of cross - section,

$10^-6m^2 ( Y=2 \times10^11 N/m^2)$

0%
$5 \times 10^-1 J$
0%
$5 \times 10^-2$
0%
$5 \times 10^-3 J$
0%
$5 \times 10^-4 J$

Explanation

Young Modulus of Elasticity

$Y=\frac{FL}{Al}$

$F=\frac{YAl}{L}$

$F=\frac{2\times 10^{11}\times 1\times 10^{-6}\times 10^{-3}\times 10^{-1}}{2}$

$F= 100\,N$

Work Done =

$\frac{1}{2}\times 100\times 1\times 10^{-3}\times 10^{-1}$

$= 0.005 \, J$

Work done =

$5\times 10^{-3}\, joule$

1222123_1280286_ans_cdb96471037f452fb510dc6d0cb33e81.jpg

On suspending a weight

$Mg$ , the length

$l$ of elastic wire and area of cross-section

$A$ its length becomes double the initial length. the instantaneous stress action on the wire is

0%
$Mg/A$
0%
$Mg/2A$
0%
$2Mg/A$
0%
$4Mg/A$

Explanation

Final stress :

$\dfrac {F}{A}$
Volume of classic wire remain same

$A\times l = 2l\times A'$

$\therefore A' = \left (\dfrac {A}{2}\right )$

$\therefore Final\ stress = \dfrac {Mg}{\dfrac {A}{2}} = \left (\dfrac {2Mg}{A}\right )$ .

A bar is subjected to an axial forces as shown in figure. Find the total elongation in the bar. (E is the modulus of elasticity of the bar and A is its cross-section)

0%
$\dfrac{FI}{AE}$
0%
$\dfrac{2FI}{AE}$
0%
$\dfrac{3FI}{AE}$
0%
$\dfrac{4FI}{AE}$

Explanation

In given figure,

Considering the two halves separately.

Since, this heavy need to be in equilibrium.

An additional force facts towards light.

So,

$\Delta h = \frac{F}{A} \times \frac{l}{E} = \frac{{3Fl}}{{AE}} = \frac{{3Fl}}{{AE}}$

The other haly becomes.

$\Delta {h_2} = \frac{{Fl}}{{AE}}$

Net elongation

$= \frac{{4Fl}}{{AE}}$

1209867_1278350_ans_707ff333c5be4e0c8dc0253ef3baa350.png

When a weight of 10 kg is suspended from a copper wire of length 3 meters and diameter 0.4 mm, its length increases by 2.4 cm. If the diameter of the wire is doubled, then the extension in its length will be-

0%
9.6 cm
0%
4.8 cm
0%
1.2 cm
0%
0.6 cm

Explanation

$l\ \propto\ \dfrac{1}{r^2}$ (

$F,L$ and

$Y$ are constant)

$\dfrac{l_2}{l_1} =\bigg(\dfrac{r_1}{r_2}\bigg)^{2} = \bigg(\dfrac{1}{2}\bigg)^{2}$

$\implies l_2 = \dfrac{l_1}{4} = \dfrac{2.4}{4} \implies l_2 =0.6cm$

A wire is subjected to a tensile stress. If A represents area of cross-section, L represents original length, I represents extension and Y is Young's modulus of elasticity, then elastic potential energy of the stretched wire is

0%
$U=\frac{2L}{AY} I^2$
0%
$U=\frac{AL}{2Y} I^2$
0%
$U=\frac{AY}{2L} I^2$
0%
$U=\frac{1}{4} \frac{AY}{L} I^2$

Explanation

Work done

$=\dfrac{1}{2}\times stress \times strain \times volume$

$=\dfrac{1}{2}\times Y\times \left( strain \right)^2\times volume$

$=\dfrac{1}{2}\times Y\times \left( \dfrac{I}{L}\right)^2 \times AL$

$=\dfrac{1}{2}\times Y\times \dfrac{I^2}{L^2}\times AL$

$=\dfrac{YI^2A}{2L}$

Hence, the answer is

$\dfrac{YI^2A}{2L}.$

A square frame of ABCD consisting of five steel bars of cross section area

$400 \,mm^2$ and joined by pivot is subjected to action of two forces

$P =40 \, kN$ in the direction of the diagonal as shown.Find change in angle at

$A$ if Young's modulus

$Y=2\times 10^5 \,N/min.$

0%
$\dfrac{1}{2000}$ rad
0%
$\dfrac{1}{1000}$ rad
0%
$\dfrac{\sqrt{2}}{1000}$ rad
0%
$\dfrac{\sqrt{3}}{1000}$ rad

The strain produced in the wire of length 60 cm is 1%, the change in the length of wire is

0%
0.6 cm
0%
6 cm
0%
60 cm
0%
0.016 cm

Explanation

$\begin{array}{l}\text { Strain } =1 \% \\\text { dength }=60 \mathrm{~cm} \\\text { Strain }=\dfrac{\text { change in length }}{\text { length }} \\\Rightarrow \Delta l=\left(10^{-2}\right) \times 60 \mathrm{~cm}\end{array}$

$=0.6 \mathrm{~cm}$

For maximum extension with given load for a wire with same value of Young's modulus of elasticity, which is true?

0%
$l=300 cm, d= 10 cm$
0%
$l=200 cm, d=5 cm$
0%
$l=100 cm, d=2 cm$
0%
$l=50 cm, d=0.5 cm$

Explanation

$\begin{array}{l} \text { Given } \\ \text { - load is same for all } \\ \text { - young modulus of elasticity is also same } \end{array}$

We know.

$Y=\frac{\text { stress }}{\text { strain }}$

$Y=\frac{F_{\text {load }}}{A\left(\frac{\Delta l}{l}\right)}$

$\left[\begin{array}{l}\text { Stress }=\frac{\text { Fioad }}{\text { Area of cross section }} \\ \text { Strain }=\frac{\text { change in length }}{\text { total length }}\end{array}\right.$

$\Delta l=\frac{F_{\text {load }} l}{A Y}=\frac{4 F_{\text {load }} l}{\pi d^{2} Y}$

$\begin{array}{l} \text { - For } \Delta l \text { to be maximum }\left(\frac{l}{d^{2}}\right) \text { should be maximum } \\ \text { By checking option (d) will be correct option } \end{array}$

A steel bar of cross-section $500$ mm and length $1$ m acted upon by two forces as shown If Young modulus of elasticity of steel is $200 \times 10^9$ $N/m^2$ then elongation of the rod is

0%
$0.05$ mm
0%
$0.5$ mm
0%
$0.1$ mm
0%
$0.2$ mm

Explanation

$\begin{array}{l}\text { Tension in } b a r=50\times10^{3}N\\\text { length }=12 m \\\text { Area of cross section =500mm }^{2} \\\qquad Y=200\times10^{9}\mathrm{~N} / \mathrm{m}^{2}\end{array}$

$\begin{aligned}\Delta l &=\frac{F L}{A Y}\\&=\frac{50_{}\times 10^{3} \times 1}{\operatorname{500}\times10^{6}\times200\times10^{9}}\\&=\frac{1}{2} \times 10^{-3} \\\Delta l&=0.5\mathrm{~mm}\end{aligned}$

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

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

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

0%
The lowest frequency of standing waveis 35 Hz
0%
The frequency of 1 st overtone is 70 Hz
0%
The frequency of 1 st overtone iis 105 Hz
0%
A and B both

Explanation

$\begin{array}{l}\text { This is a fundamental standing wave } \\\qquad \begin{aligned}\Rightarrow L &=\frac{\lambda}{2}\\& \lambda=2 L \\\text { we know that, }&V=n\lambda\\\text { Also } & Y=\sqrt{\frac{T}{\mu}}\end{aligned}\end{array}$

$\begin{aligned}n_{0} &=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\\\text { Given, } & \rho=a \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} \\& l=1 \mathrm{~m} \\& \Delta l=4.9\times10^{-4} \mathrm{~m} \\&Y=9\times10^{10}\mathrm{~N} / \mathrm{m}^{2}\end{aligned}$

$\begin{aligned}n_{0} &=\frac{1}{2 \times 1} \sqrt{\frac{V{A }\Delta l}{l A \rho}} \\&=\frac{1}{2}\sqrt{\frac{9\times10^{10} \times 4.9 \times 10^{-4}}{1\times 9 \times 10^{3}}}{10}\\n_{0}&=35\mathrm{~Hz}\end{aligned}$

A metal wire having Poisson's ratio

$\frac { 1}{ 4 }$ Young's modulud

$8\times {10 }^{ 10} N/m^2$ is stretched force, which produces a lateral strain of 0.02 it. The elastic potential energy stored per volume in wire is

$[in J/m^3]$

0%
$2.56\times 10^4$
0%
$1.78\times 10^2$
0%
$3.72\times 10^2$
0%
$3.18\times 10^5$

Explanation

$\begin{array}{c}\text { Given poisson ratio }=1 / 4 \text { , young modulus }=8 \times 10^{10} \mathrm{N} /\mathrm{m}^{2}\\\Rightarrow 1 / 4=\dfrac{\text { lateral strain }}{\text { longi} \text {tudinal Strain }\mathrm{}}\end{array}$

$\begin{array}{l}\text { given lateral strain }=0.02\%=2\times 10^{-4} \\\Rightarrow \text { longi} \text {tudinal strain }=8 \times 10^{-4}\end{array}$

$\begin{aligned}\text {PE density } &=\frac{1}{2} \times Y\times(\operatorname{Strain})^{2} \\&=\frac{1}{2}\times4\times10^{10}\times\left(8\times10^{-4}\right)^{2}\\&=4\times10^{10}\times64\times10^{-8}\\&=256\times10^{2}\\&=2.56 \times 10^{4} \mathrm{~J} / \mathrm{m}^{3}\end{aligned}$

The bulk modulus of gas is

$6 \times 10^3N/m^2$ . The additional pressure needed to reduce the volume of liquid by 10% is

0%
$1200 N/m^2$
0%
$600 N/m^2$
0%
$2400 N/m^2$
0%
$1600 N/m^2$

Explanation

Given,

$B=6\times 10^3 N/m^2$

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

$=\dfrac{10}{100}=0.1$

Bulk modulus,

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

$\Delta P=B\dfrac{\Delta V}{V}=6\times 10^3\times 0.1$

$\Delta P=600N/m^2$

The correct option is B.

A wire is stretched under a force. If the wire suddenly snaps, the temperature of the wire

0%
Remains the same
0%
Decreases
0%
Increases
0%
First decreases then increases

The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

0%
$\frac{p}{B}$
0%
$\frac{B}{3p}$
0%
$\frac{3p}{B}$
0%
$\frac{p}{3B}$

When a rubber cord is stretched, the change in volume with respect to change in its linear dimensions is negligible. The Poisson's ratio for rubber is

0%
1
0%
0.25
0%
0.5
0%
0.75

Explanation

$V = \pi {r^2}I$

$\frac{{\Delta V}}{V} = \frac{{\Delta \left( {\pi {r^2}I} \right)}}{{\pi {r^2}I}}$

$\frac{{\Delta V}}{V} = \frac{{{r^2}\Delta I + 2rI\Delta r}}{{{r^2}I}}$

$\frac{{\Delta V}}{V} = \frac{{\Delta I}}{I} + \frac{{2\Delta r}}{r}$

But

$\frac{{\Delta V}}{V} = 0$

therefore

$,$

$\frac{{\Delta I}}{I} = \frac{{ - 2\Delta r}}{r}$

Now

$,$ Poisson's ratio

$,$

$\sigma = \frac{{\frac{{ - \Delta r}}{r}}}{{\frac{{\Delta I}}{I}}}$ -------------------

$(1)$

from equation

$(1),$

$\sigma = - \left( {\frac{{\frac{{\Delta r}}{r}}}{{\frac{{ - 2\Delta r}}{r}}}} \right) = \frac{1}{2} = 0.5$

Hence,

option

$(C)$ is correct answer.

The maximum intensity of fringes in Young's experiment is l. If one of the slits is closed , then intensity at that place becomes

${ I }_{ 0 }$ . then relation between I and

${ I }_{ 0 }$ is

0%
$I={ I }_{ 0 }$
0%
$I=2{ I }_{ 0 }$
0%
$I=4{ I }_{ 0 }$
0%
there is no relation between $I$ and $I_0$

Explanation

Suppose slit width's are equal so they produces waves of equal intensity say

$I'$ , Resultant intensity at any point

$I_R =4\ I' \cos^2 \phi$ where

$\phi$ is the phase difference between the waves at the point of observation.
For maximum intensity

$\phi =0^o \Rightarrow I_{max} =4\ I' =I......(I)$
If one of slit is closed. Resultant intensity at the same point will be

$I'$ only i.e.,

$I'=I_0..(ii)$
Comparing equation (i) and (ii) we get

$I =4\ I_0$

When the load on a wire is increased from

$3 \mathrm { kg }-wt$ to

$8\mathrm { kg } -wt$ , the elongation increases from

$0.61 \mathrm { mm }$ to

$1.02 \mathrm { mm }$ . The required work done during the extension of the wire, is

0%
$16 \times 10 ^ { - 3 } \mathrm { J }$
0%
$8 \times 10 ^ { - 2 } \mathrm { J }$
0%
$20 \times 10 ^ { - 2 } \mathrm { J }$
0%
$31 \times 10 ^ { - 3 } \mathrm { J }$

Explanation

$\begin{aligned} &\Delta l_{1}=0.61 \mathrm{~mm} \quad \Delta \mathrm{l}_{2}=1.02 \mathrm{~mm}\\ &\text { (work done) }=\bigcup_{1}-\bigcup_{2} \ldots . .(i)\\ \end{aligned}$

$\begin{aligned} U &=\frac{1}{2} \times \text { stress } x \text { strain } x \text { volume } \\ &=\frac{1}{2} \times \frac{F}{A-1} \times \frac{\Delta L}{-L} \times A \cdot K \\ U &=\frac{1}{2} \times F \times \Delta L \end{aligned}$

$\begin{aligned} U_{1} &=\frac{1}{2} \times 3 \times 10 \times 0.61 \times 10^{-3} \\ &=9.15 \times 10^{-3} \mathrm{~J} \\ &=\text { work done dussing } \\ & \text { case } 1 \end{aligned}$

$\begin{aligned} U_{2} &=\frac{1}{2} \times 8 \times 10 \times 1.02 \times 10^{-3} \\ U_{2} &=40.8 \times 10^{-3} \mathrm{~J} \\=& \text { work done during } \\ & \text { case II } \end{aligned}$

$\begin{aligned} (\text { work done) }& U_{2}-U_{1} \text { during case II to case } 1 &=(40.8-9.15) \times 10^{-3} \mathrm{~J} \\ &=31.65 \times 10^{-3} \mathrm{~J} \end{aligned}$

1990442_1344294_ans_e2d2a3f0b9e64d99b75ec5bbe8bdba4b.png

A copper wire and a steel wire of the same diameter and length 1 m and 2 m respectively are connected end and a force is applied which stretches their combined length by 1 cm. How much each wire is elongated respectively. Y of copper =

$1.2\times { 10 }^{ 10 }$

$M{ m }^{ -2 }$ and Y of steel +

$2.0\times { 10 }^{ 10 }\quad N{ m }^{ -2 }$

0%
0.45 cm, 0.55 cm
0%
0.55 cm, 0.45 cm
0%
0.045 cm , 0.55 cm
0%
0.45 cm, 0.055 cm

Explanation

$\begin{array}{l} \text { Y of copper }=1.2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ \text { Yof steel }=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\ \text { L of copper }=\mathrm{tm} \\ \text { L of steel }=2 \mathrm{~m} \\ \text { both have same diameder }=\mathrm{d} \\ \Delta \mathrm{L}_{\text {net }}=1 \mathrm{~cm} \end{array}$

$\text { Let } F \text { force is applied, }$

$\begin{array}{l} \Delta L_{\text {net }}=\Delta L_{1}+\Delta L_{2} \\ \text { we know } \\ Y=\frac{F L}{A \Delta L} \end{array}$

$\Delta L=\frac{F L}{A Y} \quad \Delta L \alpha \frac{L}{Y} \quad \begin{array}{c} (\text { since } F \text { and } A \text { are same } \\ \text { For both }) \end{array}$

$\begin{array}{l} \text { For steel , For copper } \\ \Delta L_{1}=\frac{x L_{1}}{Y_{1}} & \Delta L_{2}=x \frac{L_{2}}{Y_{2}} \end{array}$

$\begin{aligned} \operatorname{1x} 10^{-2} &=\frac{2 \times x}{2 \times 10^{10}}+\frac{x \times 1}{1.2 \times 10^{10}} \\ x &=\frac{6}{11} \times 10^{8} \end{aligned}$

$\begin{aligned} \Delta L_{n e t}=& 1 \mathrm{~cm} \\ &\left\{\begin{array}{c} \{x \text { is proportionality } \\ \text { constant }) \end{array}\right.\\ \therefore \quad \Delta L_{1}=\frac{6}{11} \times \frac{2}{2 \times 10^{10} \times 10^{8}}=0.055 \mathrm{~cm} \\ \Delta L_{2}=\frac{6}{11} \times 10^{8} \times \frac{1}{1.2 \times 10^{10}}=0.045 \mathrm{~cm} \end{aligned}$

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

$\mathrm { cm } ^ { - 2 }$ is

0%
$10 \times 10 ^ { 12 }$
0%
$100 \times 10 ^ { 12 }$
0%
$1 \times 10 ^ { 12 }$
0%
$20 \times 10 ^ { 12 }$

Explanation

$K =\dfrac{100}{0.01/100} =10^\ atm=10^{11} N/m^2 = 10^{12} dyne/cm^2$

If rigidity modulus is 2.6 times of youngs modulus then the value of poission's ratio is

0%
0.2
0%
0.3
0%
0.5
0%
0.1

Explanation

$\begin{array}{l}\text{Young's Modulus}= 2 \times\text { Rigidity Modulus } \times \text { (1+ poisson's Ratio)}\\\text { Given , } \text{ Young's Modulus }=2.6 \times \text {Rigidity Modulus}\\ 2.6 = 2 \times (1+ \sigma)\\ 1.3 = 1 + \sigma \\ \sigma =0.3 \end{array}$

A steel wire of mass 3.16 Kg is stretched to a tensile strain of 1 x

${ 10 }^{ -3 }$ . What is the elastic deformation energy if density p=7.9g/cc and Y=2x

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

0%
4 KJ
0%
0.4 KJ
0%
0.04 KJ
0%
40 J

Explanation

$\begin{array}{l}\text { Given } \\\text { mass }=3.16 \mathrm{~kg} \\\text { Strain }\left(\frac{\Delta L}{L}\right)=1\times10^{-3}\end{array}$

$\rho(\text { density })=7.8 \mathrm{~g} /\mathrm{cc}$

$Y(\text { young modulus })=2\times 10^{11}$

$f=\frac{7.9 \mathrm{gram}}{\mathrm{cm}^{3}}=\frac{7.9\times 10^{-3}\mathrm{~kg}}{10^{-6}\mathrm{~m}^{3}}$

$=7.9 \times 10^{3} \mathrm{~kg}\mathrm{m}^{3}$

$\begin{array}{l}\text { Now, } \\\text { Energy stored }=\frac{1}{2} \times \text {stress }\times \text { strain } \times \text { volume }\end{array}$

$\begin{array}{c}\therefore \quad Y=\frac{\text { stress }}{\text { strain }} \\\text { Stress }=Y\text { strain }\end{array}$

$E=\frac{1}{2} \times Y\times(\operatorname{strain})^{2}\times\frac{(\rho V)}{\rho}$

$\therefore \rho V=m$

$=\frac{1 \times 2 \times 10^{11} \times\left(1\times 10^{-3}\right)^{2}\times 3.16}{2 \times7.9\times 10^{+3}}$

$\begin{array}{l}=0.4 \times 10^{2} \\E=40 \mathrm{~J}\end{array}$

Find the intensity on the screen at

$O$ if only

$S_{3}$ is covered

0%
$\dfrac {\sqrt {3}I_{o}}{7}$
0%
$\dfrac {\sqrt {3}I_{o}}{\sqrt {7}}$
0%
$\dfrac {3I_{2}}{7}$
0%
$\dfrac {I_{o}}{2}$

A 2.0 m long steel cable has cross-sectional area of

$0.30 cm^2$ . A 550 kg load is now hung from the cable. Assume that the cable behaves like a solid rod with an uniform cross-sectional area. Then the stress of the cable is:

0%
$1.8 \times 10^8 Pa$
0%
$2.8 \times 10^8 Pa$
0%
$18 \times 10^8 Pa$
0%
$28 \times 10^8 Pa$

Two wires of length

$l$ , radius

$r$ and length

$2l$ , radius

$2r$ having same Young's modulus

$Y$ are hung with weight

$mg$ . Net elongation is

0%
$3\ mgl/2\pi r^{2}Y$
0%
$1\ mgl/2\pi r^{2}Y$
0%
$2\ mgl/3\pi r^{2}Y$
0%
$3\ mgl/3\pi r^{2}Y$

Explanation

$\begin{aligned}\text { Net elongation } &=\Delta L_{1}+\Delta L_{2} \\\Delta L_{1} &=\text { elongation is } 1^{\text {st }} \text { wire } \\\Delta L_{2} &=\text { elongation is } 2^{\text {nd }} \text { wire }\end{aligned}$

$\begin{array}{l}\text { For } 1^{\text {st }}\text { wire, }\\\qquad \begin{array}{l}Y=\frac{M g L}{\pi r^{2} \Delta L_{1}}\\\Delta L_{1}=\frac{M g L}{\pi r^{2} Y}\end{array}\end{array}$

$\begin{array}{l}\text { For } 2^{\text {nd }}\text { wire } \\Y=\frac{\operatorname{mg} 2 L}{\pi(2 r)^{2} \Delta L_{2}} \\\Delta L_{2}=\frac{2 m g L}{4 \pi r^{2} Y}\end{array}$

$\begin{array}{c}\Delta L_{\text {net }}=\frac{m g L}{\pi r^{2} Y}+\frac{m g L}{2 \pi r^{2} Y} \\\Delta L_{\text {net }}=\frac{3}{2}\frac{m g L}{\pi r^{2} Y}\end{array}$

Hence option A is correct

The time after which the block reaches the position where is laming maximum elongation is

0%
$2\pi \sqrt { \dfrac { m }{ k } }$
0%
$\pi \sqrt { \dfrac { m }{ k } }$
0%
$\frac { \pi }{ 2 } \sqrt { \dfrac { m }{ k } }$
0%
$\pi \sqrt { \dfrac { m }{ 2k } }$

All three sliits are now uncovered and a transparent plate of thickness

$1.4\ \mu$ and refreactive index

$1.25$ is placecd in front of

$S_{2}$ . Rresultant intensity at point

$O$ is

0%
$3I_{0}/7$
0%
$4I_{0}/7$
0%
$5I_{0}/7$
0%
$6I_{0}/7$

Find the intensity on the screen at

$O$ if

$S_{1}$ and

$S_{3}$ are covered.

0%
$\dfrac {I_{0}}{\sqrt {7}}$
0%
$\dfrac {I_{0}}{{7}}$
0%
$\dfrac {I_{0}}{\sqrt {6}}$
0%
$\dfrac {I_{0}}{{6}}$

A metallic beam having Young's modulus

$Y$ is supported at the two ends. It is loaded at the centre. The depression of the centre is proportional to

0%
$Y$
0%
$1/Y$
0%
$Y^{2}$
0%
$1/Y^{2}$

Explanation

Depression in the beam is given by

$\delta=\dfrac{W{L}^{3}}{4Yb{d}^{3}}$

where

$W$ is the weight,

$L$ is length,

$b$ is breadth,

$Y$ is young's modulus and

$d$ is the depth and

$\delta$ is the depression in the beam.

Here, it is clear that,

Depression in the beam is inversely proportional to Young's modulus

$\delta\propto \dfrac{1}{Y}$

What happens when the applied load increases and upto breaking stress in the experiment to determine the Young's modulus of elasticity?

0%
The area of wire goes in decreasing and wire extends and breaks.
0%
The area of wire goes in increasing and wire breaks.
0%
The wire extends and area remains constant.
0%
The area remains same and wire length is also same.

Explanation

$\begin{array}{l}\text { In searl's experiment, when applied load } \\\text { increases upto breaking stress. then, area } \\\text { of wire goes on decreasing and length } \\\text { will extend and finally break. }\end{array}$

Hence option A is correct

Calculate the speed of sound in oxygen at

$0^{\circ}C$ and 1 atm. (Bulk modulus of elasticity of

$O_{2}$ is

$1.41\times 10^{5}$ Pa and density is 1.43 kg/

$m^{3}$ )

0%
300 m/s
0%
340 m/s
0%
314 m/s
0%
none of these

Explanation

$\begin{aligned} v=\sqrt{\frac{\beta}{\rho}} &=\sqrt{\frac{1.41 \times 10^{5}}{1.43}}=\sqrt{986} \cdot \times 10 \\ &=31.4 \times 10 \\ &=314 \mathrm{~m} / \mathrm{s} \end{aligned}$

Which of the following substances has the highest value of the young's modulus?

0%
Steel
0%
Rubber
0%
wood
0%
Plastic

Explanation

$\begin{array}{l}\text { we know, } \\\text { young's modulus }=\frac{\text { Stress }}{\text { strain }}\end{array}$

$\begin{array}{l}Y=\frac{F L}{A \Delta L} \\Y\alpha \frac{1}{\Delta L}\end{array}$

$\begin{array}{l}\text { By observation, }\Delta \mathrm{L} \text { in case of steel is }\\\text { smallest. So } Y \text { (young's modulus) will be } \\\text { highest for steel. }\end{array}$

The isothermal bulk modulus of a gas at atmospheric pressure is

0%
1 mm of Hg
0%
13.6 mm of Hg
0%
$1.013\times { 10 }^{ 5 }N/{ m }^{ 2 }$
0%
none of these

Explanation

Isothermal elasticity

$K_i = P = 1atm = 1.013 \times 10^5 N/m^2$

A wire of mass

$M ,$ density

$\rho$ and radius

$R$ is stretched. If

$r$ is the change in the radius and

$l$ is the change in its length, then Poisson's ratio is given by :

0%
$\dfrac { \pi l } { \rho M r R ^ { 3 } }$
0%
$\dfrac { R M \pi } { l \rho r ^ { 3 } }$
0%
$\dfrac { r M } { \pi l \rho R ^ { 3 } }$
0%
$\dfrac { l M } { \pi l \rho R ^ { 3 } }$

Explanation

Given. mass - M

$\operatorname{den} s i t y-p$ radius -

$R$ Change in radius -

$\gamma$ Change in length -

$l$

$\text { Volume }=\frac{M}{\rho}=\pi R^{2} L$

Poissons Ratio =

$\frac{\frac{\Delta d}{d}}{\frac{\Delta l}{l}}$

$\begin{array}{l} =\frac{\frac{2 \gamma}{2 R}}{\frac{l}{L}}=\frac{\gamma L}{l R} \\ =\frac{\gamma}{l R} \cdot \frac{M}{P \pi R^{2}}=\frac{M \gamma}{\ell \pi \rho R{3}} \end{array}$

Two wires of equal length and cross-section area suspended as shown in figure. Their Young's modulus are

$Y_1$ and

$Y_2$ respectively. The equivalent Young's modulus will be

0%
$Y_1 + Y_2$
0%
$\dfrac{Y_1 + Y_2}{2}$
0%
$\dfrac{Y_1Y_2}{Y_1 + Y_2}$
0%
$\sqrt{Y_1Y_2}$

Explanation

We know that

$K = \dfrac{F}{ \triangle l}$ =

$\dfrac{YA}{l}$

and,

$K_{eq}=k_1+k_2$

$\Rightarrow$

$\dfrac{Y_{eq}2A}{l}$ =

$\dfrac{Y_1A}{l}$ +

$\dfrac{Y_2A}{l}$

$\Rightarrow$

$Y_{eq} = \dfrac{Y_1 + Y_2}{2}$

A thin ring of radius R is made up of a material of density

$\rho$ and young's modulus Y. If the ring is rotated about its centre in its own plane with an angular velocity

$\omega$ then the small increase in radius

$(\Delta R)$ of the ring is

0%
$\cfrac { \rho { \omega }^{ 2 }{ R }^{ 3 } }{ Y }$
0%
$\cfrac{\rho\omega^2}{YR^3}$
0%
$\cfrac{Y}{\rho\omega^2R^3}$
0%
None

Explanation

Consider an element PQ of length dl. Let T be the tension and A the area of cross-section of the wire .

Mass of element

$dm=volume \times density = A(dl) \times \rho$

The component of T, towards the centre provides the necessary centripetal force to portion PQ

$F= 2T sin \left( \dfrac{d \theta}{2} \right) = (dm) R \omega^2$

For small angles

$\dfrac{sin(d \theta)}{2} = \dfrac{d \theta}{2} = \dfrac{dl/R}{2}$

$d \theta = \dfrac{d}{R}$

Substituting in Eq. (i), we have

$T \ times \dfrac{dl}{R} = A(dl)( \rho R \omega^2$

$T= A \rho \omega^2 R^2$

Let

$\Delta R$ be the increase in radius.

Longitudinal strain=

$\dfrac{ \Delta l}{l} = \dfrac{ \Delta(2 \pi R)}{ 2 \pi R} = \dfrac{ \Delta R}{R}$

Now,

$Y= \dfrac{T/A}{ \Delta R/ R}$

$\Delta R = \dfrac{T R}{AY} = \dfrac{A \rho \omega^2 R^2 ) R}{ AY}$ or

$\Delta R= \dfrac{ \rho \omega^2 R^3}{Y}$

2066437_1428825_ans_f388c5b926fa450786e61808d821b517.png

The Young's modulus of a wire is numerically equal to the stress which will

0%
not charge the length of the wire
0%
double the length of the wire
0%
increase the length by 50%
0%
charge the radius of the wire to half

Explanation

$\frac{\Delta l}{l}=1$

$l_{f i n a l}=l+\Delta l=2 l$ Therefere. length of wire will be doubled.

The SI unit of stress is same as the SI unit of

0%
Strain
0%
Modulus of elasticity
0%
Pressure
0%
Both (2) and (3)

Explanation

The

$SI$ unit of stress is same as the

$SI$ unit of pressure

Stress

$=\dfrac { F }{ A } =\dfrac { Force }{ acceleration }$

Similarly, Pressure

$=\dfrac { Force }{ acceleration }$ .

A 1000 kg lift is tied with metallic wires of maximum safe stress of

$1.4 \times 10^8 N/m^2$ . If the maximum acceleration of the lift is

$1.2 cm^{-2}$ , then the minimum diameter of the wire is

0%
$0.01 m$
0%
$0.01 cm$
0%
$0.001 m$
0%
$0.02 cm$

Explanation

Tension in wire

$,$

$T = 1000\left( {9.8 + 1.2} \right) = 11000N$

$Maximum\,\,stress = \dfrac{T}{{\pi {{\left( {{s_{\min }}} \right)}^2}}}$

$= 11000 \times \dfrac{7}{{22 \times {{\left( {{s_{\min }}} \right)}^2}}}$

$0r,\,{\left( {{s_{\min }}} \right)^2} = \dfrac{{{{\left( {50} \right)}^2}}}{{{{\left( {{{10}^4}} \right)}^2}}}$

$\therefore$ minimum radius

$= 50 \times {10^{ - 4}}$

$\therefore$ minimum diameter

$= 2 \times {s_{\min }}$

$= 2 \times 50 \times {10^{ - 4}}$

$= 0.01\,m$

Hence,

option

$(A)$ is correct answer.

Two wires of different material and radius have their length in ratio of

$1:2.$ if these were stretched by the same force

$,$ the strain produced will be in the ratio

$.$

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

Explanation

We know that the young's modulus is given by

$,$

$Y = stress / strain = (F/A) / strain$

$=> strain = F/AY$

where

$,$

$=$ force applied

$=$ area of cross section

since the wires are made of same material so there young's modulus will be same

$.$

Also the force applied and the area of cross section is same

$($ because radius is same

$)$

Hence

$,$

$strai{n_1}/strai{n_2}= F₁A₂Y₂/F₂A₁Y₁ = 1$

$,$ the ratio of strain will be 1:1 despite of the ratio of length

$.$

Hence,

option

$(B)$ is correct answer.

The bulk modulus of elasticity for a monoatomic ideal gas during an isothermal process is (P = pressure of the gas)

0%
$P$
0%
$\dfrac{2P}3$
0%
$\dfrac{5P}{3}$
0%
$\dfrac{7P}{5}$

Explanation

$\text { Fer isothermal process of an ideal gas }$

$\begin{aligned} P V &=K \quad \text { (constant) } \\ P &+V \frac{d P}{d V}=0 \\ V \frac{d P}{d V}=&-P \end{aligned}$

$\text { Bulk modulus }(\beta)=-\frac{d P}{\frac{d V}{V}}=-\frac{V}{d V}=P$

Statement I:- Bulk modulus of an ideal fluid is infinite.
Statement II:- An ideal fluid is compressible.

0%
Statement I is true,statement II is true and statement II is a correct explanation for statement I.
0%
Statement I is true,statement IIis true and statement II is NOT the correct explanation for statement I.
0%
Statement I is true, statement II is false.
0%
Statement I is false,statement II is true.

Explanation

$\begin{array}{l} \text { Statement I - } \\ \text { For ideal liquid } \\ \qquad \Delta V=0 \\ \beta=\text { Bulk modulus }=-\frac{V P}{d V} y_{0}=\infty \end{array}$

$\begin{array}{l} \text { Statement II - } \\ \text { (compressibility }(k)=\frac{1}{\beta}=0 \end{array}$

$\begin{array}{l} \text { For ideal fluid, } k=0 \\ \text { i.e. idial fluid is incompressible. } \\ \text { statement } I \text { is true, statement } II \text { is false. } \end{array}$

A telephone wire between the two poles is 50 m long and 1 mm in radius.When it is stretched by a load of 65 kg,its length becomes 50.12 m.Calculate Young's modulus of elasticity.

0%
$4.22\times10^{10} Nm^{-2}$
0%
$8.44\times10^{10} Nm^{-2}$
0%
$4.22\times10^{9} Nm^{-2}$
0%
$8.44\times10^{9} Nm^{-2}$

Explanation

$\begin{array}{l} y=\frac{F l}{A \Delta l}=\frac{650 \times 50}{\pi \times 10^{-6} \times 0.12} \\ y=\frac{65 \times 5}{\pi \times 12} \times 10^{4}=8.44 \times 10^{4} \end{array}$

Pick up the correct statement:

0%
Cubic crystals exhibit isotropic nature with respect to therma expansion
0%
Anisotropic solids expand in one direction and contract in perpendicular direction
0%
Many of the crystals are anisotropic solids
0%
Rock salt crystal exhibits isotropic nature with respect to thermal expansion

Explanation

$\begin{array}{l} \text { Many of the crystals are anisotropic in nature. } \\ \text { because some properties like electrical resistance } \\ \text { or refractive index of crystalline sulids are. } \\ \text { different when measured along different direction. } \\ \text { This is called anisotropy. } \end{array}$

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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