MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 14

The ice storm in the state of Jammu strained many wires to the breaking point. In a particular situation, the transmission towers are separated by

$500\ m$ of wire. The top grounding wire

$15^{o}$ from horizontal at the towers, and has a diameter of

$1.5cm$ . The steel wire has a density of

$7860\ kg\ m^{-3}$ . When ice (density

$900\ kg\ m^{-3}$ ) built upon the wire to a diameter

$10.0\ cm$ , the wire snapped. What was the breaking stress (force/ unit area) in

$N\ m^{-2}$ in the wire at the breaking point? You may assume the ice has no strength.

0%
$7.4\ \times 10^{7}\ N\ m^{-2}$
0%
$4.5\ \times 10^{8}\ N\ m^{-2}$
0%
$2.6\ \times 10^{6}\ N\ m^{-2}$
0%
$1.15\ \times 10^{7}\ N\ m^{-2}$

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 dynes

$/cm^2$ is?

0%
$10\times 10^{12}$
0%
$100\times 10^{12}$
0%
$1\times 10^{12}$
0%
$2\times 10$

Explanation

Here,

$\dfrac{Delta V}{V}=\dfrac{-0.01}{100}$

$P=100\ atm=100\times 10^5\ atm$

$\beta=\dfrac{-100\times V}{\Delta V}$

$=100\times 10^5\times 10^4=10\times 10^10 N/m^2$

$dynes/cm^2$ multiply it by 10,

$=10^{12}\ dynes/cm^2$

The length of a wire increases by

$8mm$ when a weight of

$5kg$ is suspended from it. If other things remain the same but the radius of the wire is doubled, what will be the increase in its length?

0%
$5mm$
0%
$2mm$
0%
$7mm$
0%
$9mm$

Explanation

Stress =

$Y$ *Strain

$\frac{F}{A}$ =

$Y$

$(\frac{\Delta L}{L})$

Here,

$F$ =

$5*g$

$N$

$50$

$N$

$\frac{50}{A}$ =

$Y$

$(\frac{8 }{L})$ ....(1)

(

$(\Delta L)$ =

$8$ mm)

Now, when the radius of the wire is doubled ,

The area of cross section becomes

$4A$ .

Let the extension now be

${\Delta L'}$ .

Again , applying

$\frac{F}{A}$ =

$Y$

$(\frac{\Delta L}{L})$

$\frac{50}{4A}$ =

$Y$

$(\frac{\Delta L'}{L})$ ....(2)

Divding equation (1) by equation (2) :

$4$ =

$\frac{8}{\Delta L'}$

${\Delta L'}$ =

$2$ mm.

A student performs an experiment to determine the Young's modulus of a wire, exactly

$2m$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be

$0.8$ mm with an uncertainty of

$0.05 mm$ at a load of exactly

$1.0$ kg. The student also measures the diameter of the wire to be

$0.4mm$ with an uncertainty of

$0.01mm$ . Take

$g = 9.8 m s^{-2}$ (exact). The Young's modulus obtained from the reading is:

0%
$(2.0 \ \bar+ \ 0.3) \times 10^{11} N m^{-2}$
0%
$(2.0 \ \bar+ \ 0.2) \times 10^{11} N m^{-2}$
0%
$(2.0 \ \bar+ \ 0.1) \times 10^{11} N m^{-2}$
0%
$(2.0 \ \bar+ \ 0.05) \times 10^{11} N m^{-2}$

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

$\left(\dfrac{\Delta V}{V}\right)$ is proportional to.

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

Explanation

By a load, when the metal wire is stretched, in the transverse length, the fractional change is proportional to the longitudinal length's fractional change.

Let the cross functional area be denoted as “ $A$ ” and length is denoted as “l”. Wire's volume is shown as $Al$ .

Let's assume that there is no lateral strain when there is an occurrence of longitudinal strain.

Volume increase can be shown as:

$\triangle V = A \triangle l$

$\dfrac{ \triangle V}{V} = \dfrac{A\triangle l}{Al} = \dfrac{\triangle l}{l}$

So, $\dfrac{ \triangle V}{V}$ is directly proportional to $\dfrac{\triangle l}{l}$

Option (A) is correct.

A load is supported using three wires of same cross section area as shown. Then for :

	List - I		List - II
p	Equal tensile stress in all wires	1	$Y_{2}$ = 2 $Y_{1}$
q	Equal tension in all wires	2	$Y_{2}$ = 4 $Y_{1}$
r	Equal elastic potential energy in all wires	3	$Y_{2}$ = 8 $Y_{1}$
s	Equal elastic potential energy per unit volume in all wires	4	$Y_{2}$ = 16 $Y_{1}$

0%
P- 2, Q- 3, R- 1, S- 4
0%
P- 1, Q- 4, R- 2, S- 3
0%
P- 3, Q- 2, R- 4, S- 1
0%
P- 2, Q- 2, R- 3, S- 4

A uniform wire of length

$L$ and radius

$r$ is twisted by an angle

$\alpha$ . If modulus of rigidity of the wire is

$\eta$ , then the elastic potential energy stored in wire, is

0%
$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 2{ L }^{ 2 } }$
0%
$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 4{ L } }$
0%
$\dfrac { \pi \eta { r }^{ 4 }\alpha }{ 4{ L }^{ 2 } }$
0%
$\dfrac { \pi \eta { r }^{ 4 }{ \alpha }^{ 2 } }{ 2{ L } }$

Explanation

$\begin{array}{c}\text { Given, wire of length }=L \text { and } \\\text { radius }=r \\\text { medulus of rigidity }=\eta\end{array}$

Due to twisting there will shear stress and torque will be generated.

$\tau=\frac{\alpha \cdot \eta \cdot \gamma}{L}\quad\begin{array}{c}\text { where } \tau=\text { torque }\\n=\text { modulus of } \\\text { rigidity }\end{array}$

$\alpha= angle twrited$

$\begin{array}{l}\text { Now, elastic potential energy stored } \\\qquad E=\frac{\tau^{2}}{4 n} \times \text { volume }\end{array}$

$\begin{array}{l}=\frac{\alpha^{2} \cdot \eta^{2} \cdot r^{2}}{L^{2} \cdot 4\eta} \times \pi r^{2} L \\E=\frac{\pi \eta r^{4}\alpha^{2}}{4 L}\end{array}$

A mass m is hanging from a wire of cross sectional are A and length L. Y is young's modulus of wire. An external force F is applied on he wire which is then slowly further pulled down by

$\triangle x$ from its equilibrium position. Find the work done by the force F that the wire exerts on the mass:

0%
$mg\triangle x+\dfrac{YA}{2L}(\triangle x)^2$
0%
$mg\triangle x+\dfrac{YA(\triangle x)^2}{L}$
0%
$mg\dfrac{\triangle x }{2}+\dfrac{YA(\triangle x)^2}{L}$
0%
$mg\dfrac{\Delta x}{2}+\dfrac{YA(\Delta x)^2}{2L}$

Explanation

When external force is applied, work done is given by

Work done $W=mg\Delta x$

And

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

$\dfrac{1}{2}\times Y\times {{(strain)}^{2}}\times V$

$\dfrac{1}{2}\times Y\times {{(\dfrac{\Delta x}{L})}^{2}}\times AL$

$\dfrac{Y\Delta {{x}^{2}}A}{2L}$

So, total work done is

$W=mg\Delta x+\dfrac{Y\Delta {{x}^{2}}A}{2L}$

Two wires of the same radius and material and having length in the ratio

$8.9:7.6$ are stretched by the same force. The strains produced in the two cases will be in the ratio:

0%
$1:1$
0%
$8.9:1$
0%
$1:7.6$
0%
$1:3.2$

Explanation

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

As the materials are made up of same material Youngs modulus is same for both and as radius is same for both area is also same

Therefore strains are in the ratio of 1:1

Breaking stress for steel is

$F/A$ , where

$A$ is the area of cross-section of steel wire. A body of mass

$M$ is tied at the end of the steel wire of length

$L$ and whirled in a horizontal circle. The maximum number of revolution it can make per second is:

0%
$\sqrt { \dfrac { FA }{ { 4\pi }^{ 2 }ML } }$
0%
$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }MLA } }$
0%
$\sqrt { \dfrac { F }{ { 4\pi }^{ 2 }ML } }$
0%
$\sqrt { \dfrac { { 4\pi }^{ 2 } }{ MLA } }$

Explanation

Breaking stress for steel is

$\frac{F}{A}$ . On rotating the mass

$M$ ,

$F=m \omega^{2} L \quad \Rightarrow \quad w=\sqrt{\frac{F}{M L}}$

$\therefore$ Revolutions per second

$=f=\frac{w}{2 \pi}=\sqrt{\frac{F}{4 \pi^{2} M L}}$

A mild steel wire of length

$2l$ meter cross-sectional area

$A\,{m^2}$ is fixed horizontally between two pillars. A small mass

$m \ kg$ is suspended from the mid point of the wire. If extension in wire are within elastic limit. Then depression at the mid point of wire will be

0%
${\left( {\frac{{Mg}}{{YA}}} \right)^{1/3}}$
0%
${\left( {\frac{{Mg}}{{IA}}} \right)^{1/3}}$
0%
${\left( {\frac{{Mg{l^3}}}{{YA}}} \right)^{1/3}}$
0%
$\frac{{Mg}}{{2YA}}$

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

$Y_1$ and

$Y_2$ respectively. Their equivalent young's modulus of elasticity is :

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

The Young's modulus of brass and steel are respectively

$1.0\times {10}^{10}N/{m}^{2}$ and

$2\times {10}^{10}N/{m}^{2}$ . A brass wire and a steel wire of the same length are extended by

$1\ mm$ under the same force, the radii of brass and steel wires are

${R}_{B}$ and

${R}_{S}$ respectively. Then

0%
${R}_{S}=\sqrt{2}{R}_{B}$
0%
${R}_{S}=\cfrac{{R}_{B}}{\sqrt{2}}$
0%
${R}_{S}=4{R}_{B}$
0%
${R}_{S}=\cfrac{{R}_{B}}{4}$

Explanation

$\begin{array}{l}\text { Given, young modulus of brass }=10^{10}\mathrm{~N} / \mathrm{m}^{2} \\\text { Young modulus of steel }=2 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2} \\\text { Both wires have same length and } \\\text { extended by } 1 \mathrm{~mm} \text { under same force. }\end{array}$

$\begin{array}{l}F=Y_{B}\left(\pi R B^{2}\right)\left(\frac{\Delta L}{L}\right) \\F=Y_{S}\left(\pi R_{S}^{2}\right)\left(\frac{\Delta L}{L}\right)\\Y_{B}\left(R_{B}\right)^{2}=Y_{S}\left(R_{S}\right)^{2}\\R_{B}=\sqrt{2} R_{S} \quad \text { Hence, option } B \text { is}\\\text { the correct answer. }\end{array}$

A solid sphere of radius R made of a material of bulk modulus B surrounded by a liquid in a cylindrical container.A massless piston of area A floats on the surface of the liquid. Find the fractional decreases in the radius of the sphere

$\left( \frac { d R }{ R } \right)$ when a mass M is placed on the piston to compress the liquid:

0%
$\left( \dfrac { 3Mg }{ AB } \right)$
0%
$\left( \dfrac {2 Mg }{ AB } \right)$
0%
$\left( \dfrac { Mg }{ 3AB } \right)$
0%
$\left( \dfrac { Mg }{ 2AB } \right)$

Explanation

Given,

Radius of sphere, $R$

Mass placed on massless piston, $M$ .

Area of piston, $A$

Change in pressure $\Delta P=\dfrac{\Delta F}{A}=\dfrac{Mg-0}{A}=\dfrac{Mg}{A}$

Volume of sphere, $v=\dfrac{4}{3}\pi {{R}^{3}}$

Small decrease in volume, $-dv=d\left( \dfrac{4}{3}\pi {{R}^{^{3}}} \right)=4\pi {{R}^{2}}dR$

Bulk modulus, $B$

$B=\dfrac{dp}{-\dfrac{dv}{v}}=\dfrac{\dfrac{Mg}{A}}{-\dfrac{4\pi {{R}^{2}}dR}{\dfrac{4}{3}\pi {{R}^{3}}}}=\dfrac{Mg}{-3A\dfrac{dR}{R}}$

$-\dfrac{dR}{R}=\dfrac{Mg}{3AB}$

Hence, fractional decrease in radius of sphere is $\dfrac{Mg}{3AB}$

A uniform rod of length L, has a mass per unit length

$\lambda$ and area of cross-section A. The elongation in the rod is l due to its own weight, if it is suspended from the ceiling of a room.The Young's modulus of the rod is

0%
$\dfrac{ \lambda\,g L^2}{ 2 Al}$
0%
$\dfrac{3 \lambda\,g L^2}{Al}$
0%
$\dfrac{2 \lambda\,g L}{Al}$
0%
$\dfrac{ \lambda\,g L^2}{Al}$

Explanation

Given, a ned of length

$=L$

area of cross - sction

$=A$

Yourng's modulues

$=y$

Now , we can comert this

Sugstem into an equivalent spring - systerm of sprung equivalint

$K$ .

$\begin{array}{l} \text { Let } d x \text { be the elomgation in } \\ \text { sinall element } d z \text { . } \\ T_{1}=\left(\frac{M}{L}\right)(z)(g) \\ T_{1}=k(d x) \\ T_{1}=\left(\frac{y A}{d z}\right)(d x) \\ \frac{M g}{L} \int_{0}^{t} z d z=y A \int_{0}^{x} d x \end{array}$

$\frac{M g L}{2 A Y}=x$

$\text { } \begin{aligned} \text { Now , } & d m=\lambda d x \\ M &=\lambda \int_{0}^{0} d x \\ M &=\lambda L \end{aligned}$

$\begin{array}{l} \therefore \text { Total elongation } \\ \qquad \begin{aligned} l &=\frac{\lambda g L^{2}}{2 A Y} \\ \Rightarrow & y=\frac{\lambda g L^{2}}{2 A l} \end{aligned} \end{array}$

1996126_1087211_ans_c83a97b0ffbb466299a8c44e8e379bd0.PNG

$\delta$ is the depression produced in beam of length

$L$ , breath

$b$ and thickness

$d$ , when a load is placed at the mid point, then

0%
$\delta \propto L^3$
0%
$\delta \propto \dfrac{1}{b^3}$
0%
$\delta \propto \dfrac{1}{d}$
0%
All of these

A metallic ring of radius

$2cm$ and cross sectional area

$4cm^{2}$ is fitted into a wooden circular disc of radius

$4cm$ .If the Young's module of the material of the ring is

$2 \times 10^{11}N/m^2$ , the metal ring expand is:

0%
$2 \times 10^7N$
0%
$8 \times 10^7N$
0%
$4 \times 10^7N$
0%
$6 \times 10^7N$

Explanation

$\begin{array}{l} \text { Given, metal ring of radius }=2\mathrm{~cm} \\ \text { cross - sectional area }=4\mathrm{~cm}^{2} \\ \text { Young's modulus of ring }=2\times 10^{11} \frac{\mathrm{N}}{\mathrm{m}^{\prime}}\end{array}$

$\begin{array}{l} l_{i}=2 \pi r \\ l_{+}=2 \pi R\\\frac{\Delta l_{i}}{l_{i}}=\frac{R-r}{\gamma}=\frac{4-2}{2}=1 \\ \therefore \quad \text { Stress }=\quad Y\left(\frac{\Delta l}{l}\right)=2 \times 10^{\prime\prime}\\\therefore \quad \text { Force }\quad=2\times 10^{\prime \prime} \times 4 \times 10^{-4} \\ =8 \times 10^{7} \mathrm{~N} \end{array}$

The normal density of mercury is

$p$ and its bulk modulus is

$B$ . The increase in density of gold when a pressure

$p$ is applied uniformly on all sides is:

0%
$\dfrac{p^{P}}{B}$
0%
$\dfrac{2p^{p}}{B}$
0%
$\dfrac{p^{B}}{B}$
0%
$\dfrac{B}{p^{p}}$

The substances having very short plastic region are

0%
Ductile
0%
Brittle
0%
Malleable
0%
All of these

A neutron moving with a velocity

$v$ and kinetic energy

$E$ collides perfectly elastically head on with the nucleus of an atom of mass number

$A$ at rest. The energy received by the nucleus and the total energy of the system are related by

0%
$\cfrac { 4A }{ { \left( A+1 \right) }^{ 2 } }$
0%
${ \left( \cfrac { A-1 }{ A+1 } \right) }^{ 2 }$
0%
$\cfrac{(A+1)}{4{A}^{2}}$
0%
${ \left( \cfrac { A+1 }{ A-1 } \right) }^{ 2 }$

When a

$20g$ mass hangs attached to one end of a light spring of length

$10cm$ , the spring stretches by

$2cm$ . The mass is pulled down unitl the total length of the spring is

$14cm$ . The elastic energy, in Joule stored in the spring is -

0%
$4\times {10}^{-2}$
0%
$4\times {10}^{-3}$
0%
$8\times {10}^{-2}$
0%
$8\times {10}^{-3}$

One end of a uniform rod of mass

$M$ and cross-sectional area

$A$ is suspended from the other end. The stress at the mid-point of the rod will be :

0%
$\dfrac {2Mg}{A}$
0%
$\dfrac {3Mg}{2A}$
0%
$\dfrac {Mg}{A}$
0%
Zero

Explanation

The weight of suspended mass is given as,

${W_1} = mg$

The weight of the rod acting at the midpoint is given as,

${W_2} = \dfrac{{mg}}{2}$

The stress at the midpoint is given as,

$\sigma = \dfrac{{{W_1} + {W_2}}}{A}$

$\sigma = \dfrac{{mg + \dfrac{{mg}}{2}}}{A}$

$\sigma = \dfrac{{3mg}}{{2A}}$

A force

$F$ is applied along a rod of transverse sectional area

$A$ . The normal stress to a sectior

$PQ$ inclined

$\theta$ to transverse section will be maximum for

$\theta$ (in degree) is

0%
$0$
0%
$30$
0%
$45$
0%
$90$

A steel wire of length

$1.5\ m$ and area of cross section

$1.5\ mm^{2}$ is stretched by

$1.5\ cm$ . then the work done per unit volume.

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

0%
$1\times 10^{7}J/m^{3}$
0%
$2\times 10^{7}J/m^{3}$
0%
$3\times 10^{7}J/m^{3}$
0%
$4\times 10^{7}J/m^{3}$

A thick rope of rubber of density $1.5 \times {10^3}{\text{kg/}}{{\text{m}}^{\text{3}}}$ and Young's modulus $5 \times {10^6}{\text{N/}}{{\text{m}}^2}$ , $8 m$ length is hung from the ceiling of a room , the increases in its length due to its own weight is :

$\left( {g = 10{\text{m/}}{{\text{s}}^{\text{2}}}} \right)$

0%
$9.6 \times {10^{{\text{ - 2}}}}{\text{m}}$
0%
$19.2 \times {10^{{\text{ - 7}}}}{\text{m}}$
0%
$9.6 \times {10^{{\text{ - 7}}}}{\text{m}}$
0%
$9.6 m$

Explanation

Given that,

Density of rubber

$\rho=1.5\times10^{3}\ kg/m^3$

Young's modulus

$Y=5\times10^{6}\ Nm^2$

Length

$= 8 m$

We need to calculate the increase length

Mg acts at centre of gravity which is at

$\dfrac{L}{2}$

so we take length of rope

$=\dfrac{L}{2}$

we know

$Y=\dfrac{stress}{strain}=\dfrac{F/A}{\Delta l/l}=\dfrac{Fl}{\Delta l.A}$

$\Rightarrow \Delta l=\dfrac{Fl}{AY}=\dfrac{mg L}{2AY}=\dfrac{Al\rho g l}{2AY}$

$\Delta l=\dfrac{\rho gl^2}{2Y}$

$=\dfrac{1.5\times 10^3\times 10\times 8^2}{2\times 5\times 10^6}$

$\Rightarrow \Delta l=9.6\times 10^{-2}m$ .

So (A) is correct option.

1446121_1163043_ans_1ed419d3ef2d4b4da913f03385721fc3.png

The average density of Earth's crust

$10km$ benearth the surface is

$2.7gm/cm^3$ . The speed of longitudinal seismic waves at that depth is

$5.4km/s$ . The bulk modulus of Earth's crustconsidering its behaviour aas fluid at that depth is:

0%
$7.9\times 10^{10}Pa$
0%
$5.6\times 10^{10}Pa$
0%
$7.9\times 10^{7}Pa$
0%
$1.46\times 10^{7}Pa$

Explanation

$\begin{array}{l}\text { Wave speed, } v=\sqrt{\dfrac{B}{\mu}} \\\Rightarrow B=v^{2} \mu \\\Rightarrow B=\left(2.7\mathrm{gm} / \mathrm{cm}^{3}\right)(5.4\mathrm{~km} / \mathrm{s})^{2} \\\Rightarrow B=7.9\times10^{10} \text { pascal } \\\text { Answer }A\text{ . }\end{array}$

Two wires of the same material (young's modules

$Y$ ) and same length

$L$ but radii

$R$ and

$2R$ respectively are joined end to end and a weight

$W$ is suspended from the combination as shown in the figure. the elastic potential energy in the system in equilibrium is

0%
$\dfrac { 3{ W }^{ 2 }L }{ 4\pi { R }^{ 2 }Y }$
0%
$\dfrac { 3{ W }^{ 2 }L }{ 8\pi { R }^{ 2 }Y }$
0%
$\dfrac { 5{ W }^{ 2 }L }{ 8\pi { R }^{ 2 }Y }$
0%
$\dfrac { { W }^{ 2 }L }{ \pi { R }^{ 2 }Y }$

What is the approximate change in density of water in a lake at a depth of 400 m below the surface? The density of water at the surface is 1030 kg/

${m^3}$ and bulk modulus of water is

$2 \times {10^9}N/{m^3}$

0%
4 kg/ ${m^3}$
0%
2 kg/ ${m^3}$
0%
6 kg/ ${m^3}$
0%
8 kg/ ${m^3}$

Explanation

$\text { At a depth } x \text { m below the surface the pressure } P=\operatorname{\rho g} x$

$\begin{aligned}\text { Bulk strain} &=\frac{P}{B}=\frac{\operatorname{\rho g} x}{B}\\&\Rightarrow\frac{\Delta V}{V}=\frac{\operatorname{\rho g} x}{B} \\& \Rightarrow \Delta V=\frac{\operatorname{\rho g} x V}{B}\end{aligned}$

$\begin{aligned}&\text { Density at } x \text { m below the surface } \rho^{\prime}=\frac{\rho V}{v^{\prime}}\\&\left(v^{\prime}=v-\Delta v\right)\\&\Rightarrow \quad \dot{\rho}^{\prime}=\frac{\beta V}{V-\frac{\operatorname{\rho gx} V}{\beta}}\\&\Rightarrow\quad s^{\prime}=\frac{\rho B}{B\operatorname{\rho gx}}\end{aligned}$

$\begin{array}{l}\Rightarrow\quad\rho^{\prime}=\dfrac{1030\times2\times10^{9}}{2 \times 10^{9}-1030 \times 9.8 \times 400}\\\Rightarrow\quad \rho^{\prime}=\text { }1032\mathrm{~kg} / \mathrm{m}^{3} \\\Delta \rho =1032-1030=2\mathrm{~kg} / \mathrm{m}^{3} \\\text { Ans: } B\end{array}$

Assume that a block of very low shear modulus is fixed on an inclined place as shown. Due to elastic forces it will deform. What will be the shape of the block?

0%
0%
0%
none
0%
all of these

A material has normal density

$\rho$ and bulk modulus

$B$ . The increase in the density of the material, when it is subjected to an external pressure

$P$ from all sides is :

0%
$\cfrac { P }{ \rho B }$
0%
$\cfrac { BP }{ \rho }$
0%
$\cfrac { P \rho }{ B }$
0%
$\cfrac { B \rho }{ P }$

Explanation

Solution

We know that,

B (Bulk modulus) =

$\dfrac{-P}{\dfrac{\Delta V}{V}}$ ___ (1)

Also,

$\rho=\dfrac{m}{V}$

$\Rightarrow \rho= mV^{-1}$

$\Rightarrow \dfrac{\Delta \rho}{\rho}=\dfrac{\Delta m}{m}-\dfrac{\Delta V}{V}$

As mass is constant,

$\Delta m=0$

So,

$\dfrac{\Delta \rho}{\rho}=\dfrac{-\Delta V}{V}$ ___ (ii)

From (i) and (ii), we get

$B= \dfrac{-P}{\dfrac{-\Delta \rho}{\rho}}$

$\Rightarrow \dfrac{\Delta \rho}{\rho}=\dfrac{P}{B}$

$\Rightarrow \Delta \rho= \dfrac{P}{B} \rho$

Option - C is correct

A horizontal rod is supported at both ends and loaded at the middle. It L and Y are length and Young's modulus respectively, then depression at the middle is directly proportional to

0%
L
0%
${ L }^{ 2 }$
0%
Y
0%
$\dfrac { 1 }{ Y }$

Explanation

$\begin{array}{l}\text { Depression } \delta=\frac{W L^{3}}{4 b d^{3} y} \\\therefore\ \delta \alpha {L}{}\end{array}$

Hence option A is correct

1993293_1258287_ans_e81bbd9c6cd04883b706d84216bdaef4.PNG

A-3 Two wire of equal length and cross-sectional area suspended as shown in figure. Their Young's modulus are

$Y_1 and Y_2$ respectively. The equivelent Young's modulus will be

0%
$Y_1 + Y_2$
0%
$\frac { Y_ 1+Y_ 2 }{ 2 }$
0%
$\frac { Y_ 1Y_ 2 }{ Y_1 + Y_2 }$
0%
$\sqrt { Y_ 1Y_ 2 }$

Explanation

Consider whole combination a single rod

$\begin{array}{l} Y=\frac { { FL } }{ { 2A\ell } } =\frac { { \left( { { F_{ 1 } }+{ F_{ 2 } } } \right) L } }{ { 2A\ell } } =\frac { { { F_{ 1 } }L } }{ { 1A\ell } } \\ =\frac { { { Y_{ 1 } } } }{ 2 } +\frac { { { Y_{ 2 } } } }{ 2 } =\frac { { { Y_{ 1 } }+{ Y_{ 2 } } } }{ 2 } \end{array}$

If a rubber ball is taken down to a 100 m deep lake, its volume decreases by 0.1%. If

$g=10\quad m/{ s }^{ 2 }$ then the bulk modulus of elasticity for rubber, in N/

${ m }^{ 2 }$ , is

0%
${ 10 }^{ 8 }$
0%
${ 10 }^{ 9 }$
0%
${ 10 }^{ 11 }$
0%
${ 10 }^{ 10 }$

Explanation

A rubber ball is taken 1 km depth inside
water
Pressure on rubber ball ,P =

$P_0+\rho{gh}$
Here, Po is atmospheric pressure, p is
a density of water, g is acceleration due
to gravity and his depth inside water.
Given, h 1 km-1000 m
g 10 m/s2,

$\rho$ = 1000 kg/m3 density of
water

Now , pressure act on rubber ball , P = 1.03

$\times{10^5} + 1000 \times 10 \times{1000}$

$1.03\times10^5 + 100\times10^5 = 101.03\times10^5 N/m^2$

Given

$\dfrac{\Delta{V}}{V}=\dfrac{0.05}{100}$

and

$\Delta{Ρ}$ =

$P - P_o$ =

$10^7$

$N/m^2$

$\frac{\Delta{V}}{V} = 5 \times 10^{-4}$

Now, bulk modulus =

$\Delta{Ρ}/\Delta{V}/V$

$\dfrac{10^7}{5\times 10^{-4}}$

$2 \times 10^{10}N/m^2$

$2\ m$ long light metal rod

$AB$ is suspended from the ceiling horizontally by means of two vertical wires of equal length, tied to its ends. One wire is of brass and has cross-section of

$0.2 \times 10 ^ { - 4 }\ m ^ { 2 }$ and the other is of steel with

$0.1 \times 10 ^ { 4 }\ m^2$ cross-section. In order to have equal stresses in the two wires, a weight should be hung from the rod from end

$A$ a distance of

0%
$66.6\ cm$
0%
$133\ cm$
0%
$44.4\ cm$
0%
$155.6\ cm$

Explanation

$\begin{aligned}&\text { Since stress is equal }\rightarrow\\&\begin{aligned}\dfrac{T_{1}}{0.2\times10^{-4}}&=\dfrac{T_{2}}{0.1\times10^{-4}}\\\Rightarrow \quad T_{1}&=2T_{2}\longrightarrow(1)\end{aligned}\end{aligned}$

$\begin{array}{l}\text { Torque about } c \text { should be } 0 . \sum \tau=0 \\\qquad T_{1} x=T_{2}(2-x)\\\Rightarrow\quad T_{1} x=2 T_{2}-x T_{2} \\\Rightarrow x\left(T_{1}+T_{2}\right)=2 T_{2} \\\Rightarrow x=\dfrac{2 T_{2}}{3 T_{2}}=\dfrac{2}{3} m=66.6 \mathrm{~cm}\end{array}$

A particle of mass

$m$ is moving with constant speed

$v$ in a circular path on a smooth horizontal plane by a spring as shown. If the natural length of the spring is

$l _ 0$ and stiffness of the spring is

$k$ , the elongation of the spring is :

0%
$\dfrac { { k }^{ 2 }{ l }_{ 0 }^{ 2 }+4{ mv }^{ 2 }{ k }-{ kl }_{ 0 } }{ 2k }$
0%
$\dfrac { \sqrt { { k }^{ 2 }{ l }_{ 0 }^{ 2 }+{ 4mv }^{ 2 }k } -{ kl }_{ 0 } }{ 2k }$
0%
$\dfrac { \sqrt { { k }^{ 2 }{ l }_{ 0 }^{ 2 }+{ 4mv }^{ 2 }k-{ kl }_{ 0 } } }{ 2k }$
0%
$\dfrac { \sqrt { { k }^{ 2 }{ l }_{ 0 }^{ 2 }+{ 4mv }^{ 2 }k } +{ kl }_{ 0 } }{ 2k }$

Two wires of the same material (young's modules Y) and same length L but radii

$R$ and

$2R$ respectively are joined end to end and a weight

$W$ is suspended from the combination as shown in the figure. the elastic potential energy in the system in equilibrium is

0%
$\dfrac { 3{ W }^{ 2 }L }{ 4\pi { R }^{ 2 }Y }$
0%
$\dfrac { 3{ W }^{ 2 }L }{ 8\pi { R }^{ 2 }Y }$
0%
$\dfrac { 5{ W }^{ 2 }L }{ 8\pi { R }^{ 2 }Y }$
0%
$\dfrac { { W }^{ 2 }L }{ \pi { R }^{ 2 }Y }$

Explanation

$W = {\dfrac{1}{2}}\times{F}\times{e} = {\dfrac{w^2L}{2AY}}$

$W = {\dfrac{w^2L}{2(\pi{R^2})Y}} + {\dfrac{w^2L}{2{\pi}(2R)^2Y}}$

${\dfrac{5w^2L}{8{\pi}R^2Y}}$

The intensity at the maximum in a Young's double slit experiment is

${ I }_{ 0 }$ . Distance between two slits is

$d=5\lambda$ , where

$\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen at a distance

$D = 10 d$ ?

0%
${ I }_{ 0 }$
0%
$\cfrac { { I }_{ 0 } }{ 4 }$
0%
$\cfrac { 3 }{ 4 } { I }_{ 0 }$
0%
$\cfrac { { I }_{ 0 } }{ 2 }$

Explanation

$\begin{array}{l}\text { Here } D=5 \lambda, D=10 d,y=\frac{d}{2} \\\text { We have to find resultant intersily at } y=\frac{d}{2}\end{array}$

$\begin{aligned}\text { Path difference, } \Delta x=d\tan\theta=\frac{d y}{D}=\dfrac{d^{2}}{2 D} &=\frac{d^{2}}{2 \times 10 d} \\&=\frac{d}{20} \\&=\dfrac{\lambda}{4}\end{aligned}$

$\begin{array}{l}\phi=\dfrac{2 \pi}{\lambda},\Delta x=\dfrac{\pi}{2} \\I_{y}=I_{0} \cos ^{2}(\phi / 2)=I_{0} \cos ^{2}\left(\dfrac{\pi}{4}\right)=\dfrac{I_{0}}{2}\end{array}$

Find the stress in

$CD.$
Area of

$CD = 2\ m^2$

0%
$15\ N/m^2$
0%
$5\ N/m^2$
0%
$20\ N/m^2$
0%
$None \ of \ these$

Explanation

$\begin{array}{c}\text { Stress in } C D=\dfrac{30}{2}=15\mathrm{~N} / \mathrm{m}^{2} \\\text { Ans : } A \text { . }\end{array}$
2001025_1217670_ans_141fdf5b4b284040976aaa6fcf47ef6c.PNG

2001025_1217670_ans_141fdf5b4b284040976aaa6fcf47ef6c.PNG

The increase in length of a wire is

$\Delta \ell$ when force 'F' is applied on one end of the wire with its other end fixed with a rigid support. Ratio of energy stored in the wire to the work done in stretching it is

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

A uniform steel wire hangs from the ceiling and elongates due to its own weight. The ratio of elongation of the lower half of wire is

0%
4:1
0%
3:1
0%
3:2
0%
1:1

A wire is stretched by 0.01 m by a certain force 'F' another wire of same material whose diameter and lengths are double to original wire is stretched b the same force then its elongation will be-

0%
0.005 m
0%
0.01m
0%
0.02 m
0%
0.04 m

Explanation

$\begin{array}{l}\text { Given, Force applied = F }\\\text { Let, length }=L \\\text { Area of cross section }=A=\pi r^{2} \\\text { Young's modulus }=Y\end{array}$

$\begin{array}{l}\text { Case1) Given elongation= }0.01 \mathrm{~m} \\\text { elongation formula, }e=\frac{F L}{A y} \\\Rightarrow e=\frac{F L}{\pi r^{2} y}=0.01\end{array}$

$\begin{array}{l}\text { Case 2)} \text { Length and diameter are doubled. } \\\text {That means radius } P \text { s also doubled. } \\\text { Now elongation } e^{\prime}=\frac{F(2 L)}{\pi(2 r)^{2} y}\end{array}$

$\begin{aligned}e^{\prime} &=\frac{2 F L}{4 \pi r^{2} Y}=\frac{e}{2} \\e^{\prime} &=\frac{0.01}{2}\\\Rightarrow&e^{\prime}=0.005\mathrm{~m}\end{aligned}$

When a temperature of a gas is

$20^0C$ and pressure is changed from

$P_1 = 1.01 \times 10^5$ Pa to

$P_2 = 1.165 \times 10^5$ Pa then the volume changed by 10%. The bulk modulus is

0%
$1.55 \times 10^5$ Pa
0%
$0.115 \times 10^5$ Pa
0%
$1.4 \times 10^5$ Pa
0%
$1.01 \times 10^5$ Pa

A wire of length

$1\, m$ and its area of cross-section is

$1\, cm^3$ . The Young's modulus of the wire is

$10^{11}\, Nm^{-2}$ . Two forces each equal to

$F$ are applied on its two ends in the opposite directions. If the change in length be

$1\, mm$ , what is the value of

$F$ ?

0%
$0.5 \times 10^4 \, N$
0%
$10^4 \, N$
0%
$2 \times 10^4 \, N$
0%
None of these

A massless and thin string is wrapped several times around a disc kept on a rough horizontal surface. A boy standing at a distance 'd' for the cylinder holds free end of the string pulls the cylinder towards him. If there is no slipping, length of the string passed through the hand of the boy while the cylinder reaches his hands is

0%
$d$
0%
$2d$
0%
$3d$
0%
$4d$

One end of a uniform wire of length Land of weight Wis attached rigidly to a point in the roof and a weight

$W_1$ , is suspended from its lower end. If

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

$\cfrac{6L}{8}$ from its lower end is

0%
$\cfrac{W_1}{A}$
0%
$\cfrac{W_1+(W/4)}{A}$
0%
$\cfrac{W_1+(3W/4)}{A}$
0%
$\cfrac{W_1+W}{A}$

Y,K,n represent the Young's modulus, bulk modulus and rigidity modulus of a body respectively. If rigidity modulus is twice the bulk modulus, then

0%
$Y=5K/18$
0%
$Y=5 n/9$
0%
$Y=9K/5$
0%
$Y=18K/5$

An elastic string carrying a body of mass 'm' extends by 'e'. The body rotates in a vertical circle with critical velocity. The extension in the string at the lowest position is

0%
$2$ e
0%
$4$ e
0%
$6$ e
0%
$8$ e

In a Young's double slit experiment with sodium light, slits are 0.589 m apart. The angular separation of the maximum from the central maximum will be (given

$\lambda =589$ nm,):

0%
${ sin }^{ -1 }\left( { 0.33\times 10 }^{ 8 } \right)$
0%
${ sin }^{ -1 }\left( { 0.33\times 10 }^{ 6 } \right)$
0%
${ sin }^{ -1 }\left( { 3\times 10 }^{ 8 } \right)$
0%
${ sin }^{ -1 }\left( { 3\times 10 }^{ -6 } \right)$

Explanation

Using relation,

$d \sin \theta =n \lambda \Rightarrow \sin \theta =\dfrac {n \lambda }{d}$
For

$n=3. \sin \theta =\dfrac {3\lambda }{d}=\dfrac {3\times 589 \times 10^{-9}}{0.589}$

$=3\times 10^{-6}$ or

$\theta =\sin^{-1}(3\times 10^{-6})$

A rod of length

$L$ with square cross-section

$(a \times a)$ is bend to form a circular ring.
Then the value of stress developed at points

$E$ and

$F$ respectively, (

$Y$ is Young's modulus of metal rod)

0%
Zero in both
0%
$\dfrac{2\pi a Y}{L}$ compression, $\dfrac{2\pi a Y}{L}$ tension
0%
$\dfrac{\pi a Y}{L}$ compression, $\dfrac{\pi a Y}{L}$ tension
0%
$2\dfrac{\pi a Y}{L}$ compression in both.

A uniform slender rod of length L, cross-sectional area A and Young's modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is

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

Explanation

$\begin{array}{l} \text { Cross section area }(A)=A \\ \text { if can be break into two pices at } 0 \end{array}$

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

$\begin{aligned} \Delta L_{1} &=\frac{F_{A C} L_{A_{C}}}{A Y} \\ \Delta L_{1} &=\frac{B F \times \frac{2 L}{X}}{A Y} \\ \Delta L_{1} &=\frac{2 F L}{A Y} \end{aligned}$

$\begin{aligned} \text { Similarly } & \\ \qquad \Delta L_{2}=& \frac{F_{C B} L_{C B}}{A Y} \\ \Delta L_{2} &=\frac{2 F \times \frac{L}{3}}{A Y} \\ &=\frac{2 F L}{3 A Y} \end{aligned}$

$\begin{aligned} \Delta L_{\text {net }} &=\Delta L_{1}+\Delta L_{2} \\ &=\frac{2 F L}{A Y}+\frac{2 F L}{3 A Y} \\ \Delta L_{\text {net }} &\left.=\frac{F L}{A Y} \times \frac{8}{3}\right] \end{aligned}$

1990430_1327387_ans_dae494e57478466c9cef0f4548169a58.png

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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