MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 13

In solids, inter-atomic forces are

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Totally repulsive
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Totally attractive
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Combination of (a) and (b)
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None of these

Two wires of the same material have lengths in the ratio

$1:2$ and their radii are in the ratio

$1:\sqrt{2}$ . If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

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$2:\sqrt{2}$
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$\sqrt{2}:2$
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$1:1$
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$1:2$

Explanation

$l=\dfrac{FL}{\pi r^2 Y} \implies l\ \alpha\ \dfrac{L}{r^2}$ (

$F$ and

$Y$ are constant)

$\dfrac{l_1}{l_2} =\dfrac{L_1}{L_2}\bigg(\dfrac{r_2}{r_1}\bigg)^{2} = \dfrac{}{2}(\sqrt{2})^2$

$\therefore \dfrac{l_1}{l_2} = 1:1$

An area of cross-section of rubber string is

$2cm^2$ . Its length is doubled when stretched with a linear force of

$2 \times 10^5 dynes$ . The Young's modulus of the rubber in

$dyne/cm^{2}$ will be

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$4 \times 10^{5}$
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$1 \times 10^{5}$
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$2 \times 10^{5}$
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$1 \times 10^{4}$

Explanation

If length of the wire is doubled than strain =

$1$

$\therefore Y=Stress = \dfrac{Force}{Area} = \dfrac{2 \times 10^5}{2} = 10^{5}\dfrac{dyne}{cm^{2}}$

If a spring is extended to length

$l$ then according to Hook's law

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$F=kl$
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$F=\dfrac{k}{l}$
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$F=k^2l$
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$F=\dfrac{k^2}{l}$

Explanation

According to Hook's law within the elastic limit, stress is directly proportional to strain. So, we can say that,

$Stress \propto Strain$

or.

$\frac{Stress}{Strain}=K$

This constant of proportionality is called the modulus of elasticity.

Stress is given by force per area and strain is given by the change in length by total length. Let's assume the cross sectional area as

$A$ and the total length as

$L$ . Hence,

$\frac{\frac{F}{A}}{\frac{l}{L}}=K$

or,

$\frac{F L}{Al}=K$

or,

$F=\frac{AK}{L}l$

The area total length and the modulus of elasticity are all constants. Let's replace them with

$k=\frac{AK}{L}$

Hence, we have

$F=kl$ .

Two wires

$A$ and

$B$ have equal lengths and are made of the same material, but the diameter of wire

$A$ is twice that of wire

$B$ . Then, for a given load:

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the extension of $B$ will be four times that of $A$
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the extensions of $A$ and $B$ will be equal
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the strain in $B$ is four times that in $A$
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the strains in $A$ and $B$ will be equal

Explanation

$Y=\dfrac{F/\pi \left ( \dfrac{d}{2} \right )^2}{\Delta l/l}$

$Y=\dfrac{F/\pi \left ( \dfrac{d_A}{2} \right )^2}{\Delta l_A/l}$ -----(1)

$Y=\dfrac{F/\pi \left ( \dfrac{d_B}{2} \right )^2}{\Delta l_B/l}$ -----(2)

Dividing (1) by (2)

$1=\dfrac{\pi \left ( \dfrac{d_B}{2} \right )^2}{\pi \left ( \dfrac{d_A}{2} \right )^2}\times \dfrac{\Delta l_B}{\Delta l_A}$

$1=\left ( \dfrac{d_B}{d_A} \right )^2\times \dfrac{\Delta l_B}{\Delta l_A}$

$1=\left ( \dfrac{1}{2} \right )^2\times \dfrac{\Delta l_B}{\Delta l_A}$

$\dfrac{\Delta l_A}{\Delta l_B}=\dfrac{1}{4}$

strain

$=\dfrac{\Delta l}{l}$

So, extension in B will be four times of A and strain in B will be four times of A since

$\Delta_B=4\Delta_A$ and length of A and B are same.

A uniform rod of mass m, length L , area of cross-section A and is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity

$\omega$ in a horizontal plane. If Y the Youngs modulus of the material of rod, the increase in its length due torotation of rod is :

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$\frac{m\omega ^{2}L^{2}}{AY}$
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$\frac{m\omega ^{2}L^{2}}{2AY}$
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$\frac{m\omega ^{2}L^{2}}{3AY}$
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$\frac{2m\omega ^{2}L^{2}}{AY}$

Explanation

$\begin{array}{l}\text { Consider a Small element } \\\text { of length } d x \text { at a distance } \\x \text { from the axis of rotation. } \\\text { Mass of length } L=M \\\qquad d x=\left(\frac{M}{L}\right) d x\end{array}$

$\begin{array}{l}\text { let Tension in the de part be } d\tau \\\qquad d T=(d m) \omega^{2} x=\text { Cenlripelal force }\end{array}$

$\left(\because \text { centripelal force }=m \omega^{2}r\right)$

$\begin{aligned}\therefore T &=\int_{x}^{L}\left(\frac{M}{L}\right) d x \omega^{2} x \\T &=\int_{L}\left(\frac{M}{L}\right) \omega^{2} x d x \\T &=\frac{M}{L}\frac{\omega^{2}}{2}\left(L^{2}-x^{2}\right)\end{aligned}$

$\begin{array}{l}\text { let } d l \text { be increase in length of the elemant } \\\text { Then } Y=\frac{T / A}{d 1 / d x} \Rightarrow d l=\frac{T d x}{A Y}\end{array}$

$\begin{array}{l}\qquad d l=\frac{M \omega^{2}}{2 L A Y}\left[L^{2}-x^{2}\right] d x \\\text { So, total elongation of the whole rod is } \\\qquad l=\int_{0}^{L} \frac{M \omega^{2}}{2 L A Y}\left[L^{2}-x^{2}\right] d x\end{array}$

$=\frac{1}{3} \frac{m \omega^{2} L^{2}}{A Y}$

The equation of a stationary wave in a metal rod is given by

$y=0.002\displaystyle \sin\frac{\pi x}{3}\sin 1000t$ , where

$\mathrm{x}$ is in cm and

$\mathrm{t}$ is in second. The maximum tensile stress at a point

$\mathrm{x}=2\mathrm{c}\mathrm{m}$ (Young's modulus

$\mathrm{Y}$ of material of rod

$=8\times 10^{11}dyne/cm^{2})$ will be

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$\displaystyle \frac{\pi}{3}\times 10^{8}dyne/cm^{2}$
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$\displaystyle \frac{4\pi}{3}\times 10^{8}dyne/cm^{2}$
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$\displaystyle \frac{8\pi}{3}\times 10^{8}dyne/cm^{2}$
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$\displaystyle \frac{2\pi}{3}\times 10^{8}dyne/cm^{2}$

Explanation

The equation of stationary wave in the metal rod is:

$y=0.002Sin\dfrac{\pi x}{3} sin 1000t$

The maximum tensile stress at the point

$x=2\ cm$ will be:

$T= Y \times (\dfrac{dy}{dx})_{max}=Y \times 2 \times 10^{-3} \times \dfrac{\pi}{3}Cos \dfrac{2 \pi}{3}= \dfrac{8 \pi}{3} \times 10^8 dyne/cm^2$

A copper wire is held at the two ends between two rigid supports. At

$30^{\mathrm{o}}\mathrm{C}$ , the wire is just taut,with negligible tension. If

$Y=13\times 10^{11}Nm^{-2}, \alpha =1.7\times 10^{-5}(^{\circ}C)^{-1}$ and density

$\rho=9\times 10^{3}kgm^{-3}$ , then the speed of transverse wave in this wire at

$10^{o}$ C is:

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$90 ms^{-1}$
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$70 ms^{-1}$
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$60 ms^{-1}$
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$100 ms^{-1}$

Explanation

$V=\sqrt{\dfrac{T}{m}}=\sqrt{\dfrac{Y\alpha \Delta t}{\rho}}$

where Thermal force

$=$ Tension in wire

$=T =YA\alpha\Delta t$

and linear density

$= m =A\rho$

So, now putting in the derived expression,

$V=\sqrt{\dfrac{13\times10^{11}\times1.7\times10^{-5}\times10}{9\times10^3}}=70 m/s$

In performing an experiment to determine the Young's modulus Y of steel, a student can record the following values:
length of wire l

$=(\ell_{0}\pm\Delta$ l

$){m}$
diameter of wire

${d}=({d}_{0}\pm\Delta {d})$ mm
force applied to wire

${F}$ =

$({F}_{0}\pm\Delta {F}){N}$
extension of wire

${e}=({e}_{0}\neq\Delta {e})$ mm
In order to obtain more reliable value for Y, the followlng three techniques are suggested.
Technique (i) A shorter wire ls to be used.
Technique (ii) The diameter shall be measured at several places with a micrometer screw gauge.
Technique (iii) Two wires are made irom the same ntaterial and of same length. One is loaded at a fixed weight and acts as a reference for the extension of the other which is load- tested
Which of the above techniques is/are useful?

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i and ii only
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ii and iii only
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i only
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iii only

One end of a horizontal thick copper wire of length

$2L$ and radius

$2R$ is welded to an end of another horizontal thin copper wire of length

$L$ and radius

$R$ . When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:

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$0.25$
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$0.50$
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$2.00$
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$4.00$

Explanation

We have change in length

$\triangle l=\cfrac { F.l }{ YA }$
Since the two rods are in series

${ \left( { F }_{ 1 } \right) }_{ rest }={ \left( { F }_{ 2 } \right) }_{ rest }$

$\triangle l\propto \cfrac { l }{ { R }^{ 2 } }$

$\left( \because A=\pi { R }^{ 2 } \right)$

$\therefore$

$\cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } }$

$=\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \times \cfrac { { { R }_{ 2 } }^{ 2 } }{ { { R }_{ 1 } }^{ 2 } }$

$\cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } }$

$=\cfrac { 2L }{ L } \times \cfrac { { { R }_{ 1 } }^{ 2 } }{ 4{ { R }_{ 1 } }^{ 2 } }$

$\cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } } =\cfrac { 1 }{ 2 }$

$\therefore \cfrac { \triangle { l }_{ 2 } }{ \triangle { l }_{ 1 } } =\cfrac { 2 }{ 1 } =2$

$32 g$ of

$O_{2}$ is contained in a cubical container of side

$1 m$ and maintained at a temperature of

$127 ^{0} C$ . The isothermal bulk modulus of elasticity of the gas in terms of universal gas constant

$R$ is

0%
$127 R$
0%
$400 R$
0%
$200 R$
0%
$560 R$

Explanation

Bulk modulus of elasticity,

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

"B" is directly proportional to "p"

Now, Ideal gas equation

$PV = n RT$

$P = nRT / V$

$B = P = nRT /V$

$B = \dfrac{1 \times R \times 400}{ 1^{3} }$

$B =400 R$

A uniform rod of mass m, length L, area of cross-section A and Youngs modulus Y hangs from a rigid support. Its elongation under its own weight will be:

0%
zero
0%
mgL/2YA
0%
mgL/YA
0%
2mgL/YA.

Explanation

$\begin{array}{l}M=\text { Mass } \\A=\text { Area of cross Sechon } \\L=\text { Length of Rod }\end{array}$

$Y=\text { Yourgls modulus }$

$\text { mass of } \operatorname{leng}{th } \ L=M$

$\text { Mass of length } x=\left(\frac{M}{L}\right) x$

$\begin{array}{l}\text { Tension at point } B=\left(\frac{M}{L} x\right) g \\Y=\frac{\text { Tension/Area }}{\frac{\Delta L}{L}}=\text { strain }\end{array}$

$\begin{array}{l}\therefore Y=\frac{\frac{I}{A}}{\frac{d c}{d x}} \quad\left(d c=\begin{array}{l}\text { incressed } \\\text { length }\end{array}\right) \\d c=\int \frac{T}{y_{A}} d x\end{array}$

$\begin{array}{l}\text { integrating botu sides } \\\Delta L=\int_{0}^{L} \frac{\left(\frac{M}{L} x\right) g}{Y_{A}} d x\end{array}$

$\Delta L=\frac{M g}{A Y L} \int_{0}^{L } x d\quad\therefore \Delta L=\frac{M g L}{2Y A}$

Hence option B is correct

A wire elongates by

$1 mm$ when a load W is hanged from it. lf the wire goes over a pulley and two weights

$\mathrm{W}$ each are hung at the two ends, the elongation of the wire will be (in mm):

0%
$1/2$
0%
$1$
0%
$2$
0%
zero

Explanation

Initially the length of wire is

$L$

$\therefore$

$Y = \dfrac{W L}{A \delta L}$ where

$\delta L = 1 mm$

$\implies$

$\dfrac{WL}{AY} =1 mm$ ..................(1)

Now the wire goes over the pulley.

Thus the length of wire on each side of the pulley

$L' = \dfrac{L}{2}$

Let the extensions in the wires on the two sides of the pulley be

$\delta L_1$ and

$\delta L_2$ respectively.

For one side

$Y = \dfrac{W L/2}{A \delta L_1}$

$\implies$

$\delta L_1=\dfrac{WL}{2AY} =0.5 mm$ (using 1)

For another side

$Y = \dfrac{W L/2}{A \delta L_2}$

$\implies$

$\delta L_2=\dfrac{WL}{2AY} =0.5 mm$ (using 1)

Thus total elongation

$\delta L_T = 0.5 + 0.5 =1 mm$

A steel rope has length

$L$ , area of cross-section

$A$ , Young's modulus

$Y$ . [

$Density = d$ ]. If the steel rope is vertical and moving with the force acting vertically up at the upper end, find the strain at a point

$\displaystyle \frac{L}{3}$ from lower end.

0%
$(dgL)/2Y$
0%
$(dgL)/4Y$
0%
$(dgL)/6Y$
0%
$(dgL)/8Y$

Explanation

Since steel rope has mass therefore, tension in it will not remain constant
and it will proportional to the length of it from the end with which it is tied up.

$T\alpha \ l$

Therefore tension at

$l=L/3$ will be .

$T_{l=L/3}=T/3$

$T_{l=L/3}=[dALg]/6$

We know that,

$stress=F/A=T/A$

$stress= [dALg]/6A$

$strain=stress/Y$

$strain= [dALg]/6AY$

Therefore, option(C) is correct

A uniform cylindrical wire is subjected to a longitudinal tensile stress of

$5\times 10^{7} N/m^{2}$ . Young's modulus of the material of the wire is

$2\times 10^{11} N/m^{2}$ . The volume change in the wire is

$0.02\%$ . The fractional change in the radius is

0%
$0.25\times 10^{-4}$
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$0.5\times 10^{-4}$
0%
$0.1\times 10^{-4}$
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$1.5\times 10^{-4}$

Explanation

$Y=\dfrac{Stress}{\Delta L/L}$

$\dfrac{\Delta L}{L}=\dfrac{stress}{Y}=\dfrac{5\times 10^7}{2\times 10^{11}}=2.5\times 10^{-4}$

Now

$V=\pi r^2L$

$\dfrac{\Delta V}{V}=\dfrac{\pi\Delta (r^2L)}{\pi r^2L}$

$=\dfrac{r^2\Delta L+L\times 2r\Delta r}{r^2L}$

$\dfrac{\Delta L}{L}+2\dfrac{\Delta r}{r}$

$2\dfrac{\Delta r}{r}=\dfrac{\Delta V}{V}-\dfrac{\Delta L}{L}$

$=\dfrac{0.02}{100}-2.5\times 10^{-4}$

$\dfrac{\Delta r}{r}=1\times 10^{-4}-\dfrac{2.5}{2}\times 10^{-4}=-0.25\times 10^{-4}$

A slightly conical wire of length L and end radii

$r_1$ and

$r_2$ is stretched by two forces F and F applied parallel to the length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the extension produced is:

0%
$\displaystyle \frac{FL}{\pi r_1 r_2Y}$
0%
$\displaystyle \frac{FLY}{\pi r_1 r_2}$
0%
$\displaystyle \frac{FL}{\pi r_1 Y}$
0%
$\displaystyle \frac{FL}{\pi r_1^{2}Y}$

Explanation

$R$ be the rate of increase of radius per length,

$R=\dfrac{r_2-r_1}{L}$
Consider a element of thickness

$dx$ at a distance

$x$ from narrow end of the wire.
The radius of the element is

$y=r_1+Rx$ .
The extension in the element is

$\displaystyle dl=\dfrac{F dx}{(\pi y^2)Y}$
thus,

$\displaystyle l=\frac{F}{\pi Y}\int_0^L\frac{dx}{r_1+Rx}$
let

$r_1+Rx=p, Rdx=dp$
or

$\displaystyle l=\dfrac{F}{\pi Y}\int^{r_1+RL}_0\dfrac{dp}{R p^2}=\dfrac{F}{\pi YR}\left[\dfrac{1}{r_1}-\dfrac{1}{r_1+RL}\right]=\dfrac{F}{\pi YR}\left[\dfrac{r_1+RL-r_1}{r_1(r_1+RL)}\right]$
or

$\displaystyle l=\dfrac{FL}{\pi Y(r_2-r_1)}.\frac{r_2-r_1}{r_1(r_1+r_2-r_1)}=\dfrac{FL}{\pi r_1r_2 Y}$

The speed of a traverse wave travelling on a wire having a length

$50\space cm$ and mass

$50\space g$ is

$80\space ms^{-1}$ . The area of cross-section of the wire is

$1\space mm^2$ and its Young's modulus is

$16\times10^{11}\space Nm^{-2}$ . Find the extension of the wire over natural length.

0%
$2\space cm$
0%
$2\space mm$
0%
$0.2\space mm$
0%
$0.02\space mm$

Explanation

$\quad v = \sqrt{\displaystyle\frac{T}{\mu}}$

$or \quad T = v^2\mu = (80)^2\left(\displaystyle\frac{50}{0.5}\times10^{-3}\right) = 640\space N$

$and \quad Y = \displaystyle\frac{Fl}{A\space\triangle l}$

$or \quad \triangle l = \displaystyle\frac{Fl}{AY} = \displaystyle\frac{640\times0.5}{10^{-6}\times16\times10^{11}} = 20\times10^{-5}\space m$

A copper rod of length

$l$ is suspended from the ceiling by one of its ends. Find the elongation

$\Delta l$ of the rod due to its own weight.

0%
$\displaystyle\Delta l=\frac{1}{2}\frac{\rho gl^2}{E}$
0%
$\displaystyle\Delta l=\frac{1}{3}\frac{\rho gl^2}{E}$
0%
$\displaystyle\Delta l=\frac{1}{4}\frac{\rho gl^2}{E}$
0%
$\displaystyle\Delta l=\frac{1}{5}\frac{\rho gl^2}{E}$

Explanation

Let the tension acting on

$dx$ element of rod be

$T_x$ and due to this the elongation in that element be

$\delta x$

$T_x = M_x g$

$T_x = \rho (S x)g$ ...........(1) Where

$S$ is the cross sectional area of the rod

From Hooke's law,

$E = \dfrac{T_x dx}{S \delta x}$

$\implies \delta x = \dfrac{T_x dx}{SE}$

$\therefore$ Total elongation in the rod

$\int \delta x = \int_0^l \dfrac{T_x }{SE} dx$

$\therefore$

$\Delta l = \int_0^l \dfrac{\rho g}{E} xdx$

$\Delta l = \dfrac{\rho g}{E} \times \dfrac{x^2}{2}\bigg|_0^l$

$\implies$

$\Delta l = \dfrac{\rho g l^2}{2 E}$

Six identical uniform rods

$PQ, QR, RS, ST, TU$ and

$UP$ each weighing W are freely joined at their ends to form a hexagon. The rod

$PQ$ is fixed in a horizontal position and middle points of

$PQ$ and

$ST$ are connected by a vertical string. The tension in string is

0%
$W$
0%
$3W$
0%
$2W$
0%
$4W$

Explanation

Let a small displacement be given to the system in vertical plane in frame such that ST remains horizontal then let vertical displacement of centres of rods UP and QR be y then vertical displacement of centres of UT and RS will be 3y and that of TS will by 4y. Equating the total virtual work to zero we get.

$(W + W) \delta y + (W + W) 3 \delta y + W (4 \delta y) - T (4 \delta y) = 0$
(where T is the tension in thread)

$\Rightarrow 2W + 6 W + 4W = 4 T$

$\Rightarrow T = 3 W$

Two opposite forces

$F_1 = 120N \:and \:F_2 = 80N$ act on an heavy elastic plank of modulus of elasticity

$y=2\times 10^{11}N/m^2$ and length

$L = 1m$ placed over a smooth horizontal surface. The cross-sectional area of plank is

$A = 0.5m^2$ . If the change in the length of plank is (in nm )

0%
1
0%
0.5
0%
5
0%
2

Explanation

We use the formula

$\Delta L= \dfrac{FL}{Ay}$
The average force acting on the plank is

$\dfrac{120+80}{2}=100 N$ , Thus

$\Delta L=\dfrac{100\times 1}{0.5\times 2\times 10^11}$
or

$\Delta L=1\times 10^{-9}=1 nm$

A metal wire of length L, area of cross section A and Young's modulus Y is stretched by a variable force F such that F is always slightly greater than the elastic forces of resistance in the wire. Then the elongation of the wire is

$l$ .

0%
The work done by F is $\displaystyle \frac{YAl^2}{2L}$
0%
The work done by F is $\displaystyle \frac{YAl^2}{L}$
0%
The elastic potential energy stored in the wire is $\displaystyle \frac{YAl^2}{2L}$
0%
No heat is produced during the elongation

Explanation

We have

$l=\dfrac{FL}{AY}$
or

$F=\dfrac{AY}{L}l$
Work done is given using the relation

$dW=Fdx$
or

$dW=\dfrac{AY}{L}xdx$
or

$W=\dfrac{AY}{L}\int^l_0xdx$
or

$W=\dfrac{AYl^2}{2L}$
The elastic potential energy is given as

$\Delta U=W$
or

$\Delta U=\dfrac{AYl^2}{2L}$ (This is the energy stored in the wire).
All the work done on the wire is stored as its potential energy, thus there is no heat produced during elongation.

The length of a metal is

$l_1$ when the tension in it is

$T_1$ and is

$l_2$ when the tension is

$T_2$ . The original length of the wire is :

0%
$\displaystyle\frac{l_1+l_2}{2}$
0%
$\displaystyle\frac{l_1T_2+l_2T_1}{T_1+T_2}$
0%
$\displaystyle\frac{l_1T_2-l_2T_1}{T_2-T_1}$
0%
$\displaystyle \sqrt{T_1T_2l_1l_2}$

Explanation

If l is the original length of wire, then change in length of first wire,

$\displaystyle\displaystyle\displaystyle \Delta l_1=(l_1-l)$
Change in length of second wire,

$\Delta l_{ 2 }=(l_{ 2 }-l)$
Now,

$Y=\displaystyle\frac{T_1}{A}\times \frac{l}{\Delta l_1}=\frac{T_2}{A}\times \frac{l}{\Delta l_2}$
or,

$\displaystyle\frac{T_1}{\Delta l_1}=\frac{T_2}{\Delta l_2}$ , or,

$\displaystyle\displaystyle\frac{T_1}{l_2-l}=\frac{T_2}{l_2-l}$

Thus,

$T_{ 1 }l_{ 2 }-T_{ 1 }l=T_{ 2 }l_{ 1 }-lT_{ 2 }$

$\Rightarrow l=\displaystyle\frac{T_2l_1-T_1l_2}{T_2-T_1}$

In Searle's experiment to find Young's modulus the diameter of wire is measured as

$d=0.05cm$ , length of wire is

$l=125cm$ and when a weight,

$m=20.0kg$ is put, extension in wire was found to be

$0.100cm$ . Find the maximum permissible error in Young's modulus

$(Y)$ . Use:

$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$ .

0%
$6.3\%$
0%
$5.3\%$
0%
$2.3\%$
0%
$1\%$

Explanation

$Y=\displaystyle\frac{mgl}{(\pi/4)d^2x}$ .....

$(1)$

$\displaystyle\left[\frac{dY}{Y}\right]_{max}$

$=\displaystyle \frac{\Delta m}{m}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}+\frac{\Delta x}{x}$

$m=20.0kg\Rightarrow \Delta m=0.1kg$

$l=125cm\Rightarrow \Delta l=1cm$

$d=0.050cm\Rightarrow \Delta d=0.001cm$

$x=0.100cm\Rightarrow \Delta x=0.001cm$

$\displaystyle\left[\frac{dY}{Y}\right]_{max}=\displaystyle\left[\frac{0.1kg}{20.0kg}+\frac{1cm}{125cm}+2\times \frac{0.001cm}{0.05cm}+\frac{0.001cm}{0.100cm}\right]\times 100\%=6.3\%$

Young's modulis of brass and steel are

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

$2\times 10^{11}\ N/m^2$ , respectively. A brass wire and a steel wire of the same length are extended by

$1mm$ under the same force. The radii of brass and steel wires are

$R_B$ and

$R_S$ respectively. Then

0%
$R_S=\sqrt 2R_B$
0%
$R_S=\dfrac{R_B}{\sqrt 2}$
0%
$R_S=4R_B$
0%
$R_S=\dfrac{R_B}{4}$

Explanation

We know that

$Y=\dfrac{FL}{\pi r^2l}$ or

$r^2=\dfrac{FL}{(Y\pi l)}$

$\therefore R^2_B=\dfrac{FL}{(Y_B\pi l)}$ and

$R^2_S=\dfrac{FL}{(Y_S\pi l)}$

$\dfrac{R^2_B}{R^2_S}=\dfrac{Y_S}{Y_B}=\dfrac{2\times 10^{10}}{10^{10}}=2$

$R^2_B=2R^2_S$

Thus, we get

$R_S=\dfrac{R_B}{\sqrt{2}}$

Two wire A and B have equal lengths and are made of the same material, but the diameter of A is twice that of wire B. Then, for given load

0%
The extension of B will be four times that of A
0%
the extension of A and B will be equal
0%
the strain in B is four times that in A
0%
the strains in A and B will be equal

Explanation

$\Delta l=\dfrac { FL }{ AY }$
Thus, extension is inversely proportional to area, and also Strain=

$\dfrac{\Delta l}{l}$ is inversely proportional to area. Thus strain in B will be 4 times that in A since diameter of B is half of A, or area of B is one-fourth of A; and similar for extensions.

If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be :

0%
$\dfrac{{ b }^{ 2 }}{2c}$
0%
$\dfrac{bc}{{ 2a }^{ 2 }}$
0%
$\dfrac{{ ba }^{ 2 }}{2c}$
0%
$\dfrac{{ 2b }^{ 2 }c}{a}$

Explanation

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

Force on the Steel wire

$=20N$

Elongation in the steel wire

$=\dfrac{20\times l_s}{A_s Y_s}$

Force on the brass wire

$=20N+20N=40N$

Elongation in the brass wire

$=\dfrac{40\times l_b}{A_b Y_b}$

$\Rightarrow \dfrac{\Delta L_b}{\Delta L_s}=\dfrac{40}{20}\dfrac{L_b}{L_s}\dfrac{r_s^2}{r_b^2}\dfrac{Y_s}{Y_b}=\dfrac{2b^2c}{a}$

The length of a metal wire is

$l_1$ when the tension in it is

$F_1$ and

$l_2$ when the tension is

$F_2$ . Then original length of the wire is:

0%
$\dfrac{l_1F_1+l_2F_2}{F_1+F_2}$
0%
$\dfrac{l_2-l_1}{F_2-F_1}$
0%
$\dfrac{l_1F_2-l_2F_1}{F_2-F_1}$
0%
$\dfrac{l_1F_1-l_2F_2}{F_2-F_1}$

Explanation

$F\propto \Delta l$

$F_1\propto l_1-l$

Where $l$ , is the original length of the wire

$F_2\propto l_2-l$

$\dfrac{F_1}{F_2}=\dfrac{l_1-l}{l_2-l}$

$F_1l_2-F_1l=F_2l_1-F_2l$

$(F_2-F_1)l=F_2l_1-F_1l_2$

$l=\dfrac{F_2l_1-F_1l_2}{F_2-F_1}$

A copper wire and a steel wire of the same length and same cross section are joined end to end to form a composite wire. The composite wire is hung from a rigid support and a load is suspended from the other end. If the increase in length of the composite wire is

$2.4\ mm$ , then the increase in lengths of steel and copper wires are:

$(Y_{cu} = \, 10 \times \, 10^{10} \, N/m^2, \, Y_{steel} = \, 2 \times \, 10^{11} \, N /m^2)$

0%
$1.2\ mm$ , $1.2\ mm$
0%
$0.6\ mm$ , $1.8\ mm$
0%
$0.8\ mm$ , $1.6\ mm$
0%
$0.4\ mm$ , $2.0\ mm$

Explanation

Given :

$L_{cu} = L_{steel} = L$

$A_{cu} = A_{steel} = A$

Let the increase in the lengths of individual wires be

$\Delta l_{cu}$ and

$\Delta l_{steel}$ .

Total increase in the length of composite wire

$\Delta l_{total} = 2.4\ mm$

$\therefore$

$\Delta l_{cu} + \Delta l_{steel} = 2.4$ ...........(1)

Also

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

$\implies \Delta l = \dfrac{FL}{AY}$

$\therefore$

$\Delta l_{cu} = \dfrac{F L_{cu}}{A_{cu} Y_{cu}} = \dfrac{FL}{A Y_{cu}}$

Similarly

$\Delta l_{steel} = \dfrac{F L_{steel}}{A_{steel} Y_{steel}} = \dfrac{FL}{A Y_{steel}}$

Dividing these two equations we get

$\dfrac{\Delta l_{cu}}{\Delta l_{steel}} = \dfrac{Y_{steel}}{Y_{cu}}$

$\therefore$

$\dfrac{\Delta l_{cu}}{\Delta l_{steel}} = \dfrac{2\times 10^{11}}{10 \times 10^{10} } = 2$

$\implies$

$\Delta l_{cu} = 2\Delta l_{steel}$ ............(2)

From (1) and (2),

$2\Delta l_{steel} + \Delta l_{steel} = 2.4$

$\implies$

$\Delta l_{steel} =0.8\ mm$

$\therefore$

$\Delta l_{cu } = 2 \times 0.8 =1.6\ mm$

A tension of

$20\ N$ is applied to a copper wire of cross sectional area

$0.01 cm^2$ , Young's Modulus of copper is

$1.1\times 10^{11} N/m^2$ and Poisson's ratio is 0.The decrease in cross sectional area of the wire is:

0%
$1.16\times 10^{-6}cm^2$
0%
$1.16\times 10^{-5}m^2$
0%
$1.16\times 10^{-4}m^2$
0%
$1.16\times 10^{-3}cm^2$

Explanation

In case Poisson ratio (

$\nu$ ) is given, the relation between stress and strain contains factor of 2,

Hence,

$\dfrac { \Delta A }{ A } =\dfrac { 2 }{ Y } \dfrac { F }{ A } \nu$

$\Delta A=\dfrac { 2F }{ Y } \nu =2\dfrac { 20N }{ 1.1 \times { 10 }^{ 11 }N/{ m }^{ 2 } } (0.32)=1.16 \times { 10 }^{ -6 }{ cm }^{ 2 }$

The length of a metal wire is

$l_{1}$ when the tension in it is

$T_{1}$ and is

$l_{2}$ when the tension is

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

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

Explanation

Let the natural length of wire be

$l_o$ .

Using Hooke's law,

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

where

$\Delta l = l-l_o$

We get

$l-l_o = \dfrac{Tl_o}{AY}$

Case 1 : Tension is

$T_1$ and length of wire

$l=l_1$

$\therefore$

$l_1-l_o = \dfrac{T_1l_o}{AY}$ .....(1)

Case 2 : Tension is

$T_2$ and length of wire

$l=l_2$

$\therefore$

$l_2-l_o = \dfrac{T_2l_o}{AY}$ .....(2)

Dividing both equations

$\dfrac{l_1 - l_o}{l_2 - l_o} = \dfrac{T_1 }{T_2}$

$\implies$

$l_o = \dfrac{l_1T_2 - l_2T_1}{T_2-T_1}$

An Aluminium and Copper wire of same cross sectional area but having lengths in the ratio

$2 : 3$ are joined end to end. This composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is

$2.1 \ mm$ , the increase in lengths of Aluminium and Copper wires are : [

$\displaystyle { Y }_{ Al }=20\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ and

$\displaystyle { Y }_{ Cu }=12\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$ ]

0%
$0.7 \ mm; 1.4\ mm$
0%
$0.9\ mm; 1.2\ mm$
0%
$1.0\ mm; 1.1\ mm$
0%
$0.6 \ mm; 1.5\ mm$

Explanation

Since the tension and cross section area of both the wires are same, stress in both the wires is same.

Since,

$\dfrac { T }{ A } =Y\times \dfrac { \Delta l }{ l }$

Hence,

$\Delta l=k\dfrac { l }{ Y }$

$\dfrac { \Delta { l }_{ 1 } }{ \Delta { l }_{ 2 } } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } }\times \dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } } =\dfrac { 2 }{ 3 }\times \dfrac { 12 }{ 20 } =\dfrac { 2 }{ 5 }$

Hence,

$2x+5x=2.1\ mm$ (Given)

$x=0.3\ mm$

Hence the lengths are

$2x=0.6\ mm$ and

$5x=1.5\ mm$

The load versus elongation graph of four wires of same length and of the same material is shown in figure. The thinnest wire is represented by the line :

0%
$OA$
0%
$OB$
0%
$OC$
0%
$OD$

Explanation

Hint: Young's Modulus is the ratio of stress and strain.

Solution:

Step1: Observe the given diagram,

Step:2 Use Young’s modulus equation

Using Young’s modulus equation we have,

$Y = \dfrac{{F \times l}}{{A \times \Delta l}}.....(1)$

Where, Y=Young’s modulus, F=Applied force, l=Initial length, A=Square area & ΔL= Elongation.

So, we have slope, $\dfrac{{F(load)}}{{\Delta l(elongation)}} = \dfrac{{Y \times A}}{l}$

Hence, $slope \propto A$ or $slope \propto {r^2}$

It is clear that, for thinnest wire slope is minimum i.e. wire ‘OA’ is thinnest.

Hence, option (A) is correct.

The Young's modulus of a material is

$2\times { 10 }^{ 11 }N/{ m }^{ 2 }$ and its elastic limit is

$1.8\times { 10 }^{ 8 }N/{ m }^{ 2 }$ . For a wire of

$1m$ length of this material, the maximum elongation achievable is

0%
$0.2mm$
0%
$0.3mm$
0%
$0.4mm$
0%
$0.5mm$

Explanation

From Hooke's law

$strain=\cfrac { stress }{ Y }$
where

$strain = \dfrac{e}{l}$
Given :

$stress = 1.8\times 10^8 \ N/m^2$

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

$l=1 \ m$

$\therefore$

$e=\cfrac { stress }{ Y } l=\cfrac { { 10 }^{ 8 } }{ 2\times { 10 }^{ 11 } } \times 1=5\times { 10 }^{ -4 }m=0.5mm$

A rigid bar of mass

$15 kg$ is supported symmetrically by three wires each

$2 m$ long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the tension? (Given E for copper =

$110\times 10^{9} N/m^{2}$ and E for iron =

$190\times 10^{9} N/m^{2}$ ).

0%
$12.6 : 2$
0%
$1.31 : 1$
0%
$4.65 : 3$
0%
$2.69 : 4$

Explanation

Let

${ A }_{ 1 }$ and

${ A }_{ 2 }$ be the areas of cross-section of copper and iron wires respectively. If

${ d }_{ 1 }$ and

${ d }_{ 2 }$ be their respective diameters, then

${ A }_{ 1 }=\dfrac { \pi { d }_{ 1 }^{ 2 } }{ 4 }$
and

${ A }_{ 2 }=\dfrac { \pi { d }_{ 2 }^{ 2 } }{ 4 }$

$\therefore \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { { d }_{ 1 }^{ 2 } }{ { d }_{ 2 }^{ 2 } } ={ \left( \dfrac { { d }_{ 1 } }{ { d }_{ 2 } } \right) }^{ 2 }$

$L=2m$
Let

$\Delta l$ be the extension produced in each wire.
Let

$F=$ tension produced in each wire.

$\therefore$ From the relation,

$Y=\dfrac { stress }{ strain }$ , we get
Strain for copper wire

$=\dfrac { { F }/{ { A }_{ 1 } } }{ { Y }_{ 1 } }$ and
Strain for iron wire

$=\dfrac { { F }/{ { A }_{ 2 } } }{ { Y }_{ 2 } }$
As the bar is supported symmetrically, the two strains are equal

$\dfrac { F }{ { A }_{ 1 }{ Y }_{ 1 } } =\dfrac { F }{ { A }_{ 2 }{ Y }_{ 2 } }$

$\Rightarrow { A }_{ 1 }{ Y }_{ 1 }={ A }_{ 2 }{ Y }_{ 2 }\Rightarrow \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } }$

$\Rightarrow \dfrac { { \pi { d }_{ 1 }^{ 2 } }/{ 4 } }{ { \pi { d }_{ 2 }^{ 2 } }/{ 4 } } =\dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } }$

$\Rightarrow \dfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } } } =\sqrt { \dfrac { 190\times { 10 }^{ 9 } }{ 110\times { 10 }^{ 9 } } }$

$\dfrac { { d }_{ 1 } }{ { d }_{ 2 } } =1.31$

$\Rightarrow { d }_{ 1 }:{ d }_{ 2 }=1.31:1$

Two wires having same length and material are stretched by same force. Their diameters are in the ratio 1 :The ratio of strain energy per unit volume for these two wires (smaller to larger diameter) when stretched is

0%
3 : 1
0%
9 : 1
0%
27 : 1
0%
81 : 1

Explanation

Strain energy per unit volume

$U = \dfrac{1}{2}\times Stress \times Strain$

From Hooke's law,

$Stress = Y \times Strain$

$\implies$

$U = \dfrac{1}{2Y} \times (Stress)^2$

where

$Stress = \dfrac{F}{A} = \dfrac{F}{\pi d^2/4}$

$\implies$

$U \propto \dfrac{1}{d^4}$

Given :

$\dfrac{d_s}{d_l} = \dfrac{1}{3}$

Thus ratio of strain energy per unit volume

$\dfrac{U_s}{U_l} = \bigg(\dfrac{d_l}{d_s}\bigg)^4$

$\implies$

$\dfrac{U_s}{U_l} = \bigg(\dfrac{3}{1}\bigg)^4 = 81$

When a metallic wire is stretched with a tension

$\displaystyle { T }_{ 1 }$ its length is

$\displaystyle { l }_{ 1 }$ and with a tension

$\displaystyle { T }_{ 2 }$ its length is

$\displaystyle { l }_{ 2 }$ . The original length of the wire is:

0%
$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }-{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } }$
0%
$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }+{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } }$
0%
$\displaystyle \sqrt { { l }_{ 1 }{ l }_{ 2 } }$
0%
$\displaystyle \frac { { l }_{ 1 }{ l }_{ 2 } }{ 2 }$

Explanation

From the relation between stress and strain, we get,

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

Thus

$\dfrac { { T }_{ 1 } }{ A } =Y\dfrac { { l }_{ 1 }-l }{ l }$ and

$\dfrac { { T }_{ 2 } }{ A } =Y\dfrac { { l }_{ 2 }-l }{ l }$

Dividing the two equations, we get,

$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { l-{ l }_{ 1 } }{ l-{ l }_{ 2 } }$

$l=\dfrac { { l }_{ 1 }{ T }_{ 2 }-{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } }$

A thick uniform rubber rope of density

$1.5\ g\ cm^{-3}$ and Young's modulus

$5 \times 10^{6}$

$N m^{-2}$ has a length of

$8 m$ . When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

0%
$9.6 \times 10^{-2}$
0%
$19.2 \times 10^{-2}$
0%
$9.6 \times 10^{-3}$
0%
$9.6$

Explanation

The youngs modulus

$Y$ is defined as

$Y=\frac{FL}{Ax}$ , where

$F$ is force apllied ,

$L$ is original length

$x$ is change in length and

$\sigma=$ density of the wire.

$\Rightarrow x=\frac{\sigma L^{2}g}{AY} (\therefore F=mg=\sigma \times A\times L \times g)$

$\Rightarrow x=\frac{\sigma L^{2}g}{Y}=\frac{1.5 \times 10^{-3} \times 64 \times 10}{5 \times 10^{6}}=19.2 \times 10^{-2}$

Therefore option

$B$ is correct

A steel wire of length

$4.5m$ and cross-sectional area

$3\times { 10 }^{ -5 }{ m }^{ 2 }$ stretches by the same amount as a copper wire of length

$3.5m$ and cross-sectional area of

$4\times { 10 }^{ -5 }{ m }^{ 2 }$ under a given load. The ratio of the Young's modulus of steel to that of copper is:

0%
$1.3$
0%
$1.5$
0%
$1.7$
0%
$1.9$

Explanation

For copper wire,

${ L }_{ C }=3.5m,{ A }_{ C }=4\times { 10 }^{ -5 }{ m }^{ 2 }\quad$
for steel wire,

${ L }_{ S }=4.5m,{ A }_{ S }=3\times { 10 }^{ -5 }{ m }^{ 2 }$

As Young's modulus,

$\quad Y=\cfrac { (F/A) }{ (\Delta L/L) }$

As applied force

$F$ and extension

$\Delta L$ are same for steel and copper wire

$\therefore \cfrac { F }{ \Delta L } =\cfrac { { Y }_{ S }{ A }_{ S } }{ { L }_{ S } } =\cfrac { { Y }_{ C }{ A }_{ C } }{ { L }_{ C } }$

where the subscripts

$C$ and

$S$ refers to copper and steel respectively

$\therefore \cfrac { { Y }_{ S } }{ { Y }_{ C } } =\cfrac { { L }_{ S } }{ { L }_{ C } } \times \cfrac { { A }_{ C } }{ { A }_{ S } } =\cfrac { \left( 4.5m \right) \left( 4\times { 10 }^{ -5 }{ m }^{ 2 } \right) }{ \left( 3.5m \right) \left( 3\times { 10 }^{ -5 }{ m }^{ 2 } \right) } =1.7$

1028171_936996_ans_a73d8d5fa998480496a2d9a635784daa.PNG

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere,

$\left(\displaystyle\frac{dr}{r}\right)$ , is?

0%
$\displaystyle\frac{mg}{3 Ka}$
0%
$\displaystyle\frac{mg}{Ka}$
0%
$\displaystyle\frac{Ka}{mg}$
0%
$\displaystyle\frac{Ka}{3 mg}$

Explanation

Volume of sphere

$V = \dfrac{4\pi}{3}r^3$
Decrease in volume of sphere

$-dV = \dfrac{4\pi}{3} 3r^2 dr$
Bulk modulus

$K = \dfrac{P V}{-d V}$

$\implies \$

$-dV = \dfrac{PV}{K}$ ........(1)
Pressure

$P = \dfrac{Force}{Area} = \dfrac{mg}{a}$ ......(2)
Using (2) in (1), we get

$\dfrac{4\pi }{3}3r^2 dr = \dfrac{mg}{a}.\dfrac{4\pi}{3}r^3.\dfrac{1}{K}$

$\implies \ \dfrac{dr}{r} = \dfrac{mg}{3Ka}$

Two metal wire 'P' and 'Q' of same length and material are stretched by same load. Their masses are in the ratio

$m_1 : m_2$ . The ratio of elongations of wire 'P' to that of 'Q' is

0%
$m_{1}^{2} : m_{2}^{2}$
0%
$m_{2}^{2} : m_{2}^{1}$
0%
$m_2 : m_1$
0%
$m_1 : m_2$

Explanation

We know that:-

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

$Y$ =Young's Modulus

$\implies \Delta l=\dfrac{Fl}{AY}$

$\implies \Delta l\propto \dfrac{1}{A}$

Now,

$m=Al\rho\implies m\propto A$

Hence,

$\dfrac{\Delta l_P}{\Delta l_Q}=\dfrac{A_Q}{A_P}$

$\implies \dfrac{\Delta l_P}{\Delta l_Q}=\dfrac{m_2}{m_1}$

Hence, answer is option-(C).

An amusement park ride consists of airplane shaped cars attached to steel rods. Each rod has a length of 20.0 m and a cross-sectional area of 8.00

$cm^2$ . Young's modulus for steel is

$2 \, \times \, 10^{11} \, N/m^2.$
b. When operating, the ride has a maximum angular speed of

$\sqrt{1}$ 9/5 rad/s. How much is the rod stretched (in mm) then?

0%
0.38 mm
0%
0.55 mm
0%
0.45 mm
0%
0.34 mm

Explanation

0.38 mm
Sln :
The force that each car exerts on the cable is F =

$m\omega^2l_0 \, = \, \dfrac{W}{g} \omega^2 l_0.$

$\Delta l \, = \, \dfrac{Fl_0}{YA} \, = \, \dfrac{W\omega^2l_0^2}{gYA}$

The rubber cord of a catapult has cross-section area

$2m{m}^{2}$ and a total unstretched length

$15cm$ . It is stretched to

$18cm$ and then released to project a particle of mass

$3g$ . Calculate the velocity of projection if

$Y$ for rubber is

$8\times { 10 }^{ 8 }N/{ m }^{ 2 }$ .

0%
$56.5{ms}^{-1}$
0%
$10{ms}^{-1}$
0%
$20{ms}^{-1}$
0%
$40{ms}^{-1}$

Explanation

Given,

$Y=8\times { 10 }^{ 8 }N/{ m }^{ 2 }$

$A=2{ mm }^{ 2 }=\dfrac { 2 }{ { \left( 1000 \right) }^{ 2 } } { m }^{ 2 }$

$L=15cm=\dfrac { 15 }{ 100 } m$

$l=\left( 18-15 \right) =3cm=\dfrac { 3 }{ 100 } m$

Let, velocity will be V m/s.

$\therefore$ Stored potential energy = Kinetic energy

or,

$\dfrac { YA{ l }^{ 2 } }{ 2L } =\dfrac { 1 }{ 2 } m{ v }^{ 2 }$

or,

${ v }^{ 2 }=\dfrac { YA{ l }^{ 2 } }{ mL }$

or,

${ v }^{ 2 }=\dfrac { \left( 8\times { 10 }^{ 8 } \right) \times 2\times { 3 }^{ 2 }\times 100\times 1000 }{ { \left( 1000 \right) }^{ 2 }\times { \left( 100 \right) }^{ 2 }\times \left( 3\times 10 \right) \times 15 }$

$\therefore \quad v=56.5m/s$

A student did an experiment to determine the Youngs modulus

$(Y)$ of a nylon thread,

$2.00\ m$ in length, using Searles method. A Vernier callipers having a least count of

$0.01\ mm$ was used to measure the diameter of the thread. The length and extensions were measured by using a scale with a least count of

$1\ mm$ . The data obtained by the student are shown in the table below. Assume the uncertainty in force to be

$0.01\ N$ .

Length of the thread	$1994\ mm$
Force	$0.32\ N$
Extension of the thread	$95\ mm$
Diameter of the thread	$0.04\ mm$

Choose the correct alternative(s).

0%
The student reports $Y$ to be about $5GPa$
0%
The reported uncertainty in $Y$ is $54$ %
0%
Measuring the extension using a travelling microscope with a least count of $0,1\ mm$ would improve the accuracy of the result by at most $1$ %
0%
The stress is $1.8\ GPa$

The density of water at the surface of the ocean is

$\rho$ and atmospheric pressure is

${P}_{0}$ . If the bulk modulus of water is

$K$ , what is the density of ocean water at a depth where the pressure is then

$n{P}_{0}$ ?

0%
$\cfrac { \rho K }{ K-n{ P }_{ 0 } } \quad$
0%
$\cfrac { \rho K }{ K+n{ P }_{ 0 } }$
0%
$\cfrac { \rho K }{ K-(n-1){ P }_{ 0 } }$
0%
$\cfrac { \rho K }{ K+(n-1){ P }_{ 0 } }$ .

Explanation

Let the volume of certain mass

$(M)$ of water at surface be

$V$

Density of water,

$\rho= \dfrac MV$

Pressure at a depth,

$P'= nP_o$

Thus, change in pressure,

$\Delta P= nP_o- P_o= (n-1)P_o$

Given Bulk Modulus

$= K=\dfrac{\Delta P}{\frac{\Delta V}V }\implies \Delta V=\dfrac{(n-1)P_o V}{K}$

As at a depth, the pressure increases. Thus the volume of water will decrease at that depth.

$\therefore$ Volume of water of mass

$(M)$ at that depth,

$V'= V-\Delta V$

$\therefore V'= V- \dfrac{(n-1)P_o V}{K}= (K- (n-1)P_o)\dfrac VK$

$\therefore$ Density at that depth,

$\rho' =\dfrac M{V'}= \rho \dfrac V{V'}$

$\therefore \rho' =\dfrac{\rho K}{K- (n-1)P_o}$

A solid sphere of radius

$R$ , made up of a material of bulk modulus

$K$ is surrounded by a liquid in a cylindrical container. A massless piston of area

$A$ floats on the surface of the liquid. When a mass

$M$ is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is

0%
$\dfrac {Mg}{2AK}$
0%
$\dfrac {Mg}{3AK}$
0%
$\dfrac {Mg}{AK}$
0%
$\dfrac {2Mg}{3AK}$

Explanation

Step 1: Find change in pressure when mass $M$ is placed on massless piston.

It is given that a sphere of radius

$R$ made of material of bulk modulus

$K$ is kept in a cylindrical container filled with liquid.

A massless piston of area

$A$ is floating on the surface of liquid. Since the piston is massless, initial force by piston on liquid is zero,

$F_i = 0$

When mass

$M$ is kept on piston force exerted by piston on liquid becomes,

$F_f = Mg$

Thus, change in pressure on liquid is given as,

$dP = \dfrac{dF}{A}$

$\Rightarrow dP = \dfrac{Mg - 0}{A}$

$\Rightarrow dP = \dfrac{Mg}{A}$

Step 2: Use formula for Bulk Modulus for Sphere to find fractional change in radius of sphere.

As radius of sphere is

$R$ , its volume is given as,

$V = \dfrac{4}{3} \pi R^3$

Let after putting mass on piston small chabge in volume of sphere is

$dV$ , it is viven as,

$dV = \dfrac{4}{3} \pi 3R^2dR = 4\pi R^2dR$

Now by using formula for Bulk Modulus of sphere, we have

$K = \dfrac{dP}{\dfrac{dV}{V}}$

$\Rightarrow K = \dfrac{dP}{\dfrac{dV}{V}}$

$\Rightarrow K = \dfrac{\dfrac{Mg}{A}}{\dfrac{4\pi R^2dR}{\dfrac{4}{3} \pi R^3}}$

$\Rightarrow \dfrac{dR}{R} K = \dfrac{Mg}{3A}$

$\Rightarrow \dfrac{dR}{R} = \dfrac{Mg}{3KA}$

Thus, fractional change in radius is given as, $\dfrac{dR}{R} = \dfrac{Mg}{3KA}$ .

Option B is correct.

If the ratio of diameters, lengths and Young's moduli of steel and brass wires shown in the figure are

$p,q$ and

$r$ respectively. Then the corresponding ratio of increase in their lengths would be:

0%
$\cfrac { 3q }{ 5{ p }^{ 2 }r }$
0%
$\cfrac { 5q }{ 3{ p }^{ 2 }r }$
0%
$\cfrac { 3q }{ 5{ p }^{ }r }$
0%
$\cfrac { 5q }{ { p }^{ }r }$

Explanation

Given:

$\cfrac { { F }_{ S } }{ { F }_{ B } }=\dfrac{5}{3}$

$\cfrac { { D }_{ B } }{ { D }_{ S }}=p$

$\cfrac { { L }_{ S } }{ { L }_{ B } } =q$

$\cfrac { { Y }_{ S } }{ { Y }_{ B } }=r$

As Young's modulus

$Y=\cfrac { FL }{ A\Delta L } =\cfrac { 4FL }{ \pi { D }^{ 2 }\Delta L }$
where the symbols have their usual meanings

$\therefore \Delta L=\cfrac { 4FL }{ \pi { D }^{ 2 }Y }$

$\therefore \cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ B } } =\cfrac { { F }_{ S } }{ { F }_{ B } } \cfrac { { L }_{ S } }{ { L }_{ B } } \cfrac { { D }_{ B }^{ 2 } }{ { D }_{ S }^{ 2 } } \cfrac { { Y }_{ B } }{ { Y }_{ S } }$

$\quad \therefore \cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ B } } =\left( \cfrac { 5}{ 3 } \right) \left( q \right) { \left( \cfrac { 1 }{ p } \right) }^{ 2 }\left( \cfrac { 1 }{ r } \right) =\cfrac { 5q }{ 3{ p }^{ 2 }r }$

1028161_936991_ans_6d21253a2ca0400c8b75159cc7910531.png

A uniform pressure P is exerted on all sides of a solid cube at temperature t

$^oC$ . By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? The coefficient of volume expansion of cube is

$\alpha$ and the bulk modulus is K.

0%
$\dfrac{P}{\alpha K}$
0%
$\dfrac{P\alpha}{K}$
0%
$\dfrac{PK}{\alpha}$
0%
$\dfrac{2K}{P}$

Explanation

$K=\dfrac{-PV}{\Delta V_1}$

$-\Delta V_1=\dfrac{PV}{K}$

Let us suppose change in volume be

$\Delta V_2$

when change in temperature is

$\Delta T$

$\Delta V_2=V\alpha\Delta T$

For no change in volume after raising temperature,

$\Delta V_1+\Delta V_2=0$

$\dfrac{PV}{K}=V\alpha\Delta T$

$\Delta T=\dfrac{P}{\alpha K}$

A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is?

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

Explanation

$\begin{array}{l}\text { Given, an ideal gas at a pressure } \\\text { and atsolite temparature T. } \\\text { Volume Stress }=-\beta \text { ( Volume strain) }\end{array}$ \

$\begin{array}{l}\frac{F}{A}=-\beta\left(\frac{\Delta V}{V}\right) \\\therefore \quad \beta=-\frac{\Delta P}{\Delta V / V} \\\text { Now, in isethermal conditions } \\P V=n R T=\text { constant }\end{array}$

$\begin{array}{l}\text { Taking log on beth sides, } \log (P V)=\log (n R T) \\\qquad \begin{array}{c}\ln P+\ln V=\ln (n R T) \\\text { Differmtiating, } \frac{1}{P} d P+\frac{1}{V} d V=0 \\\therefore \quad \frac{d V}{V}=-\frac{d P}{P}\end{array}\end{array}$

$\begin{array}{l}\therefore \text { Bule Medulus ( } \beta \text { ) in iscthermal }\\\text { conditions is, P. } \\\beta=\frac{\Delta P}{-\Delta P / P}=P \text { ( using equation 1) }\end{array}$

A brass rod of length

$1 \ m$ is fixed to a vertical wall at one end, with the other end keeping free to expand. When the temperature of the rod is increased by

$120^{\circ}C$ , the length increases by

$3 \ cm$ . What is the strain?

0%
$0$
0%
$10$
0%
$20$
0%
$30$

Explanation

$\triangle L=\alpha \triangle T$

Strain

$=\cfrac { \triangle L }{ L } =\cfrac { 0.03m }{ 1m } \approx 0$

Two wires are made of the same material and have the same volume. However

$wire\ 1$ has cross-sectional area

$A$ and

$wire\ 2$ has cross-sectional area

$3A.$ If the length of

$wire\ 1$ is increases by

$\Delta x$ on applying force

$F,$ how much force is needed to stretch

$wire\ 2$ by the same amount?

0%
$6F$
0%
$9F$
0%
$F$
0%
$4F$

Explanation

$\begin{array}{l}\text { Wire 1: Cross sectional area = A } \\\text { Volume }=V \\\text { Wire 2: cross sectional area = } 3 A \\\text { Volume }=V\end{array}$

$\begin{array}{l}\text { Since both wires are made of same } \\\text { material, so young's modulus for both } \\\text { wires is same. } \\\qquad \begin{array}{ll}A_{1} L_{1} & =A_{2} L_{2} \quad(\text { as volume is sane })\\A_{1} L_{1} & =\left(3 A_{1}\right)\left(L_{2}\right) \\L_{2} & =\frac{L_{1}}{3}...........(1)\end{array}\end{array}$

$\begin{array}{l}F=Y A_{1}\left(\frac{\Delta x}{L_{1}}\right) \\F^{\prime}=y\left(3 A_{1}\right)\left(\frac{3 \Delta x}{L_{1}}\right)={9 F} \\\text { Force required to stretch wire } 2 \text { by a }\\\text { distance } \triangle x \text { is } 9 \mathrm{~F}\text { . }\end{array}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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