MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 16

A ball of mass m is suspended on a weightless elastic thread whose stiffness factor is k. The ball is lifted so that the thread is not stretched and let fall with zero initial velocity. The maximum stretch of the thread in the process of ball's motion is

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$\frac{mg}{k}$
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$\frac{2mg}{k}$
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$-\frac{GM^2}{R}$
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$-\frac{6}{5}\frac{GM^2}{R}$

Explanation

$\begin{array}{l}m g x=\frac{1}{2} k x^{2} \\m g=\frac{1}{2} k x \\x=\frac{2 m g}{x}\end{array}$

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

$1$ increases by

$\Delta x$ on applying forced

$F$ , how much force is needed to stretch wire

$2$ by the same amount?

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$4F$
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$6F$
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$9F$
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$F$

Explanation

$\begin{array}{l} \text { By Hooke's law - } \\ \qquad Y=\frac{F l}{A \Delta l}=\frac{F V}{A^{2} \Delta l} \\ \text { for wire } 1 \text { and } 2 \text { . } \\ Y=\frac{F . V}{A^{2} \Delta x} \\ Y=\frac{F^{\prime} V}{(3 A)^{2} \Delta x} \\ \text { from (1) and (2) } \\ \qquad F^{\prime}=9 \mathrm{~F} \end{array}$

The pressure on an object of bulk modulus

$B$ undergoing hydraulic compression due to a stress exerted by surrounding fluid having volume strain

$\left(\dfrac{\triangle V}{V}\right)^2$

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$B^2\left(\dfrac{\triangle V}{V}\right)$
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$B\left(\dfrac{\triangle V}{V}\right)^2$
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$\dfrac{1}{B}\left(\dfrac{\triangle V}{V}\right)$
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$B\left(\dfrac{\triangle V}{V}\right)$

Explanation

$\begin{array}{l} \text { We know that } \\ \text { Bulk modulus (B)= Normal stress/Bulk stress } \\ \text { Normal strain/volume strain } \end{array}$

$\begin{array}{l} \quad B=\left|\frac{P}{\left(\frac{\Delta V}{V}\right)^{2}}\right| \\ \text { Where, volume strain }=\left(\frac{\Delta V}{V}\right)^{2} \text { given } \\ S O, \Rightarrow \quad B\left(\frac{\Delta V}{V}\right)^{2}=P \\ \therefore P=B\left(\frac{\Delta V}{V}\right)^{2} \end{array}$

The graph is drawn between the applied force F and the strain (x) for a thin uniform wire the wire hehaves as a liquid in the part

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ab
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bc
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cd
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oa

Stress S is needed to break a copper wire having radius R. The stress needed to break a copper wire of radius 2R will be :-

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S
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2S
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4S
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S/2

A block of mass

$100 kg$ is suspended with a massless wire of length

$20 m .$ Wire is elongated by

$0.01\%.$ The amount of recoverable energy is

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$0.5J$
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$1 J$
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$2 J$
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$3 J$

A solid sphere of radius

$r$ made of a material of bulk modulus

$B$ is surrounded by a liquid in a liquid in a cylindrical container. A mass less piston of area a floats on the surface of the liquid. When a mass

$m$ is placed on the piston compress the liquid, the fraction change in the radius of the sphere

$(dr/r)$ is :

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$B a/mg$
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$B a/3mg$
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$mg/3B a$
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$mg/B a$

Explanation

$\Delta P=\frac{m g}{a}$

$\frac{\Delta V}{V}=\frac{\Delta p}{B}$

$\begin{aligned} v=\frac{4}{3} \pi g^{3} & \\ \frac{\Delta v}{v}=\frac{3 \Delta n}{r} \\ \frac{3 \Delta g}{r}=\frac{m g}{a \beta} \\ \frac{\Delta r}{r}=\frac{m g}{3 a \beta} \end{aligned}$

Young's moduli of two wires A and B are in the ratio

$7 : 4$ . Wire A is 2 m long and has radius R. Wire B is

$1.5 m$ long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to : :

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$1.9 mm$
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$1.7 mm$
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$1.5 mm$
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$1.3 mm$

Explanation

Given

$\dfrac{Y_A}{Y_B} = \dfrac{7}{4}$

$L_A = 2m$

$A_A = \pi R^2$

$L_B = 1.5 m$

$A_B = \pi (2 mm)^2$

$\dfrac{F}{A} = Y \left(\dfrac{\ell}{L} \right)$
given F and

$\ell$ are same

$\Rightarrow \dfrac{AY}{L}$ is same

$\dfrac{A_A Y_A}{L_A} = \dfrac{A_B Y_B}{L_B}$

$\Rightarrow \dfrac{(\pi R^2) \left(\dfrac{7}{4} Y_B \right)}{2} = \dfrac{\pi (2 mm)^2 . Y_B}{1.5}$

$R = 1.74 \, mm$

Three blocks system is shown in the figure. Each has mass

$3$ kg. String connected to P and Q are of equal cross-section and Young's modulus of

$0.005$

$cm^2$ and

$2\times 10^{11}N/m^2$ respectively, neglect friction. , the elastic potential energy stored per unit volume in the wire connecting blocks P in steady-state is: (Take

$g=10m/s^2$ )

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$500 J/m^3$
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$1000 J/m^3$
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$2000 J/m^3$
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$3000 J/m^3$

Now the hollow cylinder (of radius

$x$ ) is cut along its length parallel to axis

$OO'$ before twisting. It is found as a rectangle

$DEFG$ . Due to the twisting couple, its deforms to a parallelogram

$D'EFG'$
Then find the shearing force on the face area of the cylinder:

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$\eta xL$
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$2\eta \pi x dx$
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$\dfrac{\eta\theta}{L}2\pi x^{2}dx$
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$\dfrac{\eta\theta}{L}\pi x^{2}dx$

A boy's catapult is made of rubber cord which is

$42\ cm$ long, with

$6\ mm$ diameter of cross-section and of negligible mass. The boy keeps a stone weighing

$0.02\ kg$ on it and stretches the cord by

$20\ cm$ by applying a constant force. When released, the stone flies off with a velocity of

$20\ ms^{-1}$ . Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to

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$10^{4} Nm^{-2}$
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$10^{8} Nm^{-2}$
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$10^{6} Nm^{-2}$
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$10^{3} Nm^{-2}$

Explanation

Energy of catapult

$= \dfrac {1}{2}\times \left (\dfrac {\triangle l}{l}\right )^{2} \times Y\times A \times l$

$=$ Kinetic energy of the ball

$= \dfrac {1}{2} mv^{2}$
therefore,

$\dfrac {1}{2}\times \left (\dfrac {20}{42}\right )^{2}\times Y\times \pi \times 3^{2} \times 10^{-6}\times 42 \times 10^{-2} = \dfrac {1}{2} \times 2\times 10^{-2} \times (20)^{2}$

$Y\simeq 3\times 10^{6} Nm^{-2}$ .

Give a suitable relation between

$x,\theta L$ and

$\phi$
(

$\angle BAB'=\phi$ shown in figure

$b$ ):

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$\phi=x\theta/L$
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$\theta=x\phi/L$
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$\theta=(x^{2}/L^{2})\phi$
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$\theta=\phi$

The total elongation of the bar, if the bar is subjected to axial forces as shown in figure and the cross-sectional area of bar is

$10 cm^2$ , is: (Take

$E=8\times 10^{2}$ dyne/

$cm^2)$

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$0.01$ cm
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$0.5$ cm
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$0.0675$ cm
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$0.775$ cm

Find the restoring couple applied by the cylinder in equilibrium condition:

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$\dfrac{2\pi\eta R^{4}\theta}{2L}$
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$\dfrac{2}{3}\dfrac{\pi\eta R^{2}\theta}{L}$
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$\dfrac{\pi\eta R^{4}\theta}{4L}$
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$\dfrac{2\pi G\eta^{3}}{3L}$

Explanation

The angle of shear from figure (d) is

$\angle DED'=\phi$

also from figure (d)

$DD'=l\phi$

also from figure (d)

$DD'=x\theta$

$l\phi=x\theta \implies \phi=\dfrac{x\theta}{l}$

$\eta=\dfrac{F}{\phi}$

$F=\eta.\phi=\dfrac{\eta x\theta}{l}$

The surface area of this hollow cylinder =

$2\pi x dx$

Total shearing force on this area =

$2\pi x dx\dfrac{\eta x\theta}{l}$

$2\pi \eta\dfrac{\theta}{l}x^2dx$

The moment of this force =

$2\pi \eta\dfrac{\theta}{l}x^2dx.x$

The moment of this force =

$2\pi \eta\dfrac{\theta}{l}x^3dx$

Now, integrating between the limits x = 0 and x = r, We have, total twisting couple on the cylinder

$C_{total}=\int$

$(2\pi \eta\dfrac{\theta}{l}x^3)dx$

∴ Total twisting couple =

$\dfrac{\pi \eta\theta r^4}{2l}$

Restroring couple is double of twisting couple so restoring couple =

$2\dfrac{\pi \eta\theta r^4}{2l}$ .

Hence option A is correct.

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 16 - 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 16 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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