MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 4

In Youngs expt., the distance between two slits is d/3 and the distance between the screen and the slits is 3D. The number of fringes in 1/3 m on the screen, formed by monochromatic light of wavelength

$3\lambda$ , will be?

0%
$\frac{d}{9D\lambda }$
0%
$\frac{d}{27D\lambda }$
0%
$\frac{d}{81D\lambda }$
0%
$\frac{d}{D\lambda }$

Explanation

Triangle width =

$\frac{\lambda D}{d}$
=

$\frac{3\lambda (3D)}{\frac{d}{3}}$
=

$\frac{27\lambda D}{d}$
no. of fringes =

$\frac{1}{3} \times \frac{27\lambda D}{d}$
=

$\frac{d}{81\lambda D}$

The Poissons ratio for inert gases is:

0%
$1.40$
0%
$1.66$
0%
$1.34$
0%
$None \ of \ these$

Explanation

For mono atomic molecule

$C_v = \dfrac {3R}{2}$ (as can be derived from standard results)

$C_p = \dfrac {5R}{2}$

Poissons ratio for any gas is given by:

$\text{poissons ratio} = \dfrac {C_p}{C_v}$

$= \dfrac {\dfrac {5R}{2}}{\dfrac {3R}{2}}$

$= \dfrac {5}{3}$

$\simeq 1.66$

One end of a string of length L and cross-sectional area A is fixed to a support and the other end is fixed to a bob of mass m. The bob is revolved in a horizontal circle of radius r, with an angular velocity

$\omega$ , such that the string makes an angle

$\theta$ with the vertical. The increase

$\Delta L$ in length of the string is

0%
$\dfrac{ML}{AY}$
0%
$\dfrac{MgL}{AY cos\theta }$
0%
$\dfrac{MgL}{AY sin\theta }$
0%
$\dfrac{MgL}{AY}$

Explanation

$\Delta L=\dfrac{T L}{AY}$
Substituting,

$T=mg/cos\theta$

$\Delta L=\dfrac{mg L}{AY cos\theta }$

One end of a string of length

$L$ and cross-sectional area

$A$ is fixed to a support and the other end is fixed to a bob of mass

$m$ . The bob is revolved in a horizontal circle of radius

$r$ with an angular velocity

$\omega$ such that the string makes an angle

$\theta$ with the vertical. The stress in the string is :

0%
$\dfrac{mg}{A}$
0%
$\dfrac{mg}{A} \left ( 1-\dfrac{r}{L} \right )$
0%
$\dfrac{mg}{A} \left ( 1+\dfrac{r}{L} \right )$
0%
none of these

Explanation

Stress

$=\dfrac{T}{A}$

$=\dfrac{mg}{A cos\theta }$ (on taking the component in the string direction)
where

$cos \theta =\dfrac{\sqrt{L^{2}-r^{2}}}{L}$

$=\sqrt{1-\dfrac{r^{2}}{L^{2}}}$
Stress

$=\dfrac{mg}{A\sqrt{1-\dfrac{r^{2}}{L^{2}}}}$

In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference $\dfrac{\pi}{3}$ .If maximum intensity on the screen is $I_0$ , the intensity at the point on the screen equidistant from the slits is____?

0%
$I_0$
0%
$\dfrac{I_0}{2}$
0%
$\dfrac{I_0}{4}$
0%
$\dfrac{3 I_0}{4}$

Explanation

$[I = I_o sin^2 \theta]$

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

$= \dfrac{3 I_o}{4}$

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, $\frac{\Delta V}{V}$ of water at the bottom of the ocean, given that the bulk modulus of water is $2.2 \times 10^9Nm^{-2}$ (consider $g=10 ms^{-2}$ )

0%
0.82%
0%
0.91%
0%
1.24%
0%
1.36%

Four uniform wires of the same material are stretched by the same force. The dimensions of wire are as given below. The one which has the minimum elongation has :

0%
radius $3\ mm$ , length $3\ m$
0%
radius $0.5\ mm$ , length $0.5\ m$
0%
radius $2 \ mm$ , length $2\ m$
0%
radius $3 \ mm$ , length $2\ m$

Explanation

According to Young's modulus of elasticity:

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

$\Delta l \propto 1/ r^{2}$
Hence only option "D" is satisfying the relation.

A rubber ball is brought into 200 m deep water, its volume is decreased by 0.1% then volume elasticity coefficient of the material of ball will be:

$(Given\ \rho = 10^3 kg/m^3$ and

$g = 9.8 ms^{-2})$

0%
$19.6 \times 10^8 N/m^2$
0%
$19.6 \times 10^{-10} N/m^2$
0%
$19.6 \times 10^{10} N/m^2$
0%
$19.6 \times 10^{-8} N/m^2$

Explanation

According to bulk modulus of elasticity,

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

According to pressure depth relation,

$P = h \rho g$

$B = -\dfrac{h \rho g}{\Delta V/V}$

or,

$B = \dfrac{200 \times 10^{3} \times 9.8}{ \dfrac{1}{1000} }$

or,

$B = 19.6 \times 10^{8} N / m^{2}$

(i) For a Searle's experiment, in the graph shown, there are two readings a and b that are not lying on the straight line
(ii) Experiment is not performed precisely

0%
Both (i) and (ii) are true and (ii) is reason for (i)
0%
Both (i) and (ii) are true but (ii) is not reason for (i)
0%
Only (i) is true
0%
Only (ii) is true

The velocity of sound in glass in0m/sec, it's density is 2,927 gm c.c. Calculate Youngs modules of glass in dynes/cm

$^{2}$

0%
$65 \times 10^{12}$
0%
$65 \times 10^{16}$
0%
$73.17 \times 10^{9}$
0%
$65 \times 10^{11}$

Explanation

Given

Velocity of sound in glass= 50 m/s

Density of glass = 2927g/cc = 2927000

$kg/m^{3}$

Young modulus of glass =?

Solution

We know that

$v=(\dfrac{Y}{D})^{\dfrac{1}{2}}$

$Y=v^{2} \times D$

$Y=50^{2} \times 2927000$

$Y=73.17 \times 10^{9} dyne /cm^{2}$

A sample of a liquid has an initial volume of

$1.5 L$ . The volume is reduced by

$0.2 mL,$ when the pressure increases by

$140 kPa$ . What is the bulk modulus of the liquid.

0%
$1.05 \times 10^{9}$ Pa
0%
$1.05 \times 10^{5}$ Pa
0%
$3.05 \times 10^{9}$ Pa
0%
$3.05 \times 10^{5}$ Pa

Explanation

bulk modulus of elasticity,

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

$B = \dfrac{-1.5 \times140 \times 10^{3} }{-0.2 \times 10^{-3} }$

$B = 1.05 \times 10^{9}Pa$

Velocity of sound, (Y) youngs modules, and p density of solid are related as

0%
$v=\sqrt{Y p}$
0%
$v=\sqrt{Y/p}$
0%
$v^2 = Yp$
0%
Both (a) and (b)

Explanation

$\begin{array}{l}\text { The speed of sound in a solid the depends } \\\text { on the youngls modulus of the medium and Density.}\end{array}$

$V=\sqrt{\frac{Y}{\rho}}$

$\begin{aligned}Y &=\text { Young's modulus } \\\rho &=\text { density of } \\& \text { medium }\end{aligned}$

Hence option B is correct

A spherical ball is compressed by

$0.01$ % when a pressure of

$100$ atmosphere is applied on it. Its bulk modulus of elasticity in

$dyne/cm^{2}$ will be approximately

0%
$10^{12}$
0%
$10^{14}$
0%
$10^{6}$
0%
$10^{24}$

Explanation

The pressure is 100atm i.e.

$100\times 10^5 Pa=10^7 N/m^2$
Now as

$1 N/m^2=10 dyne/cm^2$
we get
pressure as

$10^8 dyne/cm^2$
Now as

$\Delta V/V$ is given as

$\dfrac{0.01}{100}=10^{-4}$
Thus we get Bulk modulus as

$\dfrac{10^8}{10^{-4}}=10^{12}$

(i) In Searle's experiment, after every step of loading, one waits for sometime (2 or 3 min) before taking reading.
(ii) In this duration, the wire becomes free from kinks.

0%
Both (i) and (ii) are true and (ii) is reason for (i)
0%
Both (i) and (ii) are true but (ii) is not reason for (i)
0%
Only (i) is true
0%
Only (ii) is true

The modulus of elasticity of a material does not depend upon

0%
shape
0%
temperature
0%
nature of material
0%
impurities mixed

The ratio of change in dimension at right angles to applied force to the initial dimension is defined as

0%
$Y$
0%
$\eta$
0%
$\beta$
0%
$K$

Explanation

This is a factual question. The ratio is labelled

$\beta$ .

A wire is stretched through

$1 mm$ by certain load. The extension produced in the wire of same material with double the length and radius will be

0%
$4 mm$
0%
$3 mm$
0%
$1 mm$
0%
$0.5 mm$

Explanation

$\Delta l=\dfrac { FL }{ AE } \\ Double\quad length\quad and\quad radius\quad means,\quad L'=2L\quad and\quad A'={ 2 }^{ 2 }A=4A\\ Thus\quad \Delta l'=\dfrac { F.2L }{ 4A.E } =\dfrac { \Delta l }{ 2 } =\dfrac { 1 }{ 2 } =0.5mm$

Two identical wires of different materials have Young's moduli of elasticity as

$22\times10^{10}N/m^2$ and

$11\times10^{10}N/m^2$ respectively. If these are stretched by equal loads then the ratio of extensions produced in them will be:

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

Explanation

We know that

$\Delta l=\dfrac{FL}{AY}$ . Thus, for same load and same length and area (because similar wires), extension is inversely proportional to Young's modulus.
Thus, for Y in ratio

$2:1$ , extension will be in ratio

$1:2$

Which of the following pairs is not correct?

0%
strain-dimensionless
0%
stress- $N/m^{2}$
0%
modulus of elasticity- $N/m^{2}$
0%
poisson's ratio- $N/m^{2}$

Explanation

stress is

$\dfrac{F}{A}$ hence unit

$N/m^2$

strain is

$\dfrac{\Delta l}{L}$ so unit

$m/m$ therefore dimensionless

modulus of elasticity is

$\dfrac{stress}{strain}$ hence same unit as stress as the denominator is dimensionless

poisson's ratio

$\dfrac{-\epsilon_t}{\epsilon_l}$ so its also going to be dimensionless

Equal weights are suspended from the wires of same material and same lengths but with radii in the ratio

$1 : 2$ . The ratio of extensions produced in them will be

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

Explanation

We know that

$\Delta l=\dfrac{FL}{AY}$ . Thus, for same force, same material (Y) and same length, extension is inversely proportional to area, or

${r}^{2}$
Thus, for radii in ratio

$1:2$ , extension will be in ratio

$4:1$

The modulus of elasticity at constant temperature is

0%
$\gamma P$
0%
$\frac{P}{\gamma}$
0%
$P$
0%
$\frac{P}{V}$

Explanation

Bulk modulus of elasticity

$B=-V\frac { dP }{ dV }$
Use Perfect gas equation,

$PV=RT$
From this, we get

$\frac { dP }{ dV } =-\frac { RT }{ { V }^{ 2 } }$
Put this in the expression for

$B$ ,

$B=-V\left( \frac { -RT }{ { V }^{ 2 } } \right)$ =

$\frac { RT }{ { V } }$
At constant temperature,

$B=\frac { RT }{ { V } }= P$

A stress of

$2\ kg/mm^{2}$ is applied on a wire. If

$Y=10^{12}\ dyne/cm^2$ then the percentage increase in its length will be

0%
$0.196\%$
0%
$19.6\%$
0%
$1.96\%$
0%
$0.0196\%$

Explanation

Given

$:Stress=2\ kg/mm^2= 2\times\dfrac{9.8}{{ 10 }^{ -6 }} Pa= 19.6\times{10}^{6}Pa$

Young's Modulus of elasticity Y

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

$,\ 1\ dyne= 10^{-5}\ N$
Y

$= 10^{ 11 }\ N/m^2$
Now, from Hooke's law, we know that

$Stress=Y Strain$
Percentage change in length is strain

$\%$
Strain

$=\dfrac { \Delta L }{ L }$
Strain

$\%=\dfrac { Stress }{ Y } \times 100$
So, percentage change in length

$=\dfrac{ 19.6\times { 10 }^{ 6 }\times 100 }{ { 10 }^{ 11 } }=0.0196\%$

The expression for the determination of Poisson's ratio for rubber is

0%
$\displaystyle \sigma=\frac{1}{2}\left[1-\frac{dV}{AdL}\right]$
0%
$\displaystyle \sigma=\frac{1}{2}\left[1+\frac{dV}{AdL}\right]$
0%
$\displaystyle \sigma=\frac{1}{2}\frac{dV}{AdL}$
0%
$\displaystyle \sigma=\frac{dV}{AdL}$

Explanation

Volume of rubber,

$V=AL$

$\dfrac{dV}{dL}=- L\dfrac{dA}{dL}+A$ , negative sign because the are decreases as length increases
Area,

$A=\pi {R}^{2}$
So,

$\dfrac{dA}{A}=2\dfrac{dR}{R}$

$\therefore \quad \dfrac { dV }{ dL } = \dfrac { -2A\dfrac { dR }{ R } }{ \dfrac { dL }{ L } } +A=(-2\sigma +1)A$
Hence,

$\sigma =\dfrac { 1 }{ 2 } \left( 1-\dfrac { dV }{ AdL } \right)$

$10 \ meter$ long thick rubber pipe is suspended from one of its ends. The extension produced in the pipe under its own weight will be :

$(Y=5\times10^{6}N/m^{2}$ and density of rubber =

$1500 \ kg/m^{3}$ )

0%
$1.5\ m$
0%
$0.15\ m$
0%
$0.015 \ m$
0%
$0.0015 \ m$

Explanation

$\Delta l=\dfrac { FL }{ AY } =\dfrac { \dfrac { mg }{ 2 } L }{ AY } \quad (average\quad force\quad due\quad to\quad weight\quad at\quad centre\quad of\quad mass=\dfrac { mg }{ 2 } )\\ \\ =\dfrac { (1500\times A\times L\times g).L }{ 2AY } =\dfrac { 1500\times 10\times 10\times 10 }{ 2\times 5\times 1{ 0 }^{ 6 } } =0.15m$

Two identical wires are suspended from a roof, but one is of copper and other is of iron. Young's modulus of iron is thrice that of copper. The weights to be added on copper and iron wires so that the ends are on the same level must be in the ratio of

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

Explanation

We know that

$\Delta l=\dfrac{FL}{AY}$ . Thus, for same length and same length, extension is inversely proportional to Y and directly proportional to force, or

$\dfrac{F}{Y}$
Thus, for

$Y$ in ratio

$3:1$ for iron and copper respectively, extension will be in ratio

$1:1$ only if Forces are in ratio

$1:3.$

For a material

$\sigma=-0.25$ under an external stress, the longitudinal strain is

$10^{-2}$ . The percentage change in the diameter of the wire is

0%
$+1$ %
0%
$-1$ %
0%
$+0.25$ %
0%
$-0.25$ %

Explanation

$=\dfrac { \Delta d }{ d } \times 100$

$=-\sigma \dfrac { \Delta l }{ l } \times 100\quad$

$=-(-0.25)\times 1{ 0 }^{ -2 }\times 100$

$= +0.25$ %
(Negative sign is there because increase in length causes decrease in diameter)

A wire of density

$9\times10\space kg\space m^{-3}$ stretched between two clamps

$1\space m$ apart is subjected to an extension of

$4.9\times10^{-4}\space m$ . If Young's modulus of the wire is

$9\times10^{-4}\space m$ . The lowest frequency of transverse vibrations in the wire is

0%
$35\space Hz$
0%
$70\space Hz$
0%
$105\space Hz$
0%
$140\space Hz$

Explanation

Fundamental frequency

$\nu=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$ .
Now, Young's modulus

$Y=\dfrac{T/A}{\Delta l/l}=\dfrac{Tl}{A\Delta l}\Rightarrow T= \dfrac{YA\Delta l}{l}$ .
So,

$\nu=\dfrac{1}{2l}\sqrt{\dfrac{\dfrac{YA\Delta l}{l}}{A\rho}}=\dfrac{1}{2l}\sqrt{\dfrac{YA\Delta l}{lA\rho}}=\dfrac{1}{2}\sqrt{\dfrac{9\times10^{10}\times 4.9\times 10^{-4}}{9\times10^3}}=35$ Hz.
Ans: A

A wire made of the material of Young's modulus

$Y$ has an stress

$S$ applied to it. If Poisson's ratio of the wire is

$\sigma$ , the lateral strain is:

0%
$\displaystyle \sigma\frac{S}{Y}$
0%
$\displaystyle \sigma\frac{Y}{S}$
0%
$\displaystyle \sigma Y\times S$
0%
$\displaystyle \frac{S}{\sigma Y}$

Explanation

Longitudinal strain =

$S/Y$ ;
Poisson's ratio is given as

$\dfrac{Lateral \ strain}{Longitudinal\ Strain}$

$Lateral \ Strain = Poisson's\ ratio \times Longitudinal\ Strain=\sigma \dfrac{S}{Y}$

A steel girder can bear a load of

$20 tons$ . If the thickness of girder is double, then for the same depression it can bear a load of :

0%
40 ton
0%
80 ton
0%
160 ton
0%
5 ton

Explanation

Let the load on the girder be

$W_{ 1 }$ and the thickness be

$t_{1}$ .
Now, when the thickness is doubled, i.e.

$t_{2}=2t_{1}$ , the load on the girder is

$W_{2}$ .
Depression

$\delta =\dfrac { W{ l }^{ 3 } }{ 4b{ t }^{ 3 }Y }$
Since depression is same in both cases,

$\delta_{1}=\delta_{2}$

$\dfrac { { W }_{ 1 } }{ { t }_{ 1 }^{ 3 } } =\dfrac { { W }_{ 2 } }{ { t }_{ 2 }^{ 3 } }$
Hence,

$W_{2}= 160 N$

A spiral spring is stretched by a weight attached to it. The strain is

0%
shear
0%
elastic
0%
tensile
0%
bulk

A gas undergoes a process in which its pressure P and volume V are related as

$VP^n=$ constant. The bulk modulus for the gas in this process is:

0%
np
0%
p $^{1/n}$
0%
$\dfrac{p}{n}$
0%
p $^n$

Explanation

$Vp^n=$ constant

$B=\dfrac{\Delta p}{(\dfrac{\Delta V}{V})}$ where B is Bulk modulus

Differentiate on both sides w.r.t,

$p^n+np^{n-1}\dfrac{\Delta p .V}{\Delta V}=0$

$p^{n-1}(p+\dfrac{n\Delta p}{\Delta V}.V)=0$

$p=0$

$p+n\dfrac{\Delta p}{\Delta V/V}=0$

$p-nB=0$

$B=\dfrac{p}{n}$

A vertical steel post of diameter

$25 cm$ and length

$2.5 m$ supports a weight of

$8000 kg$ . Find the change in length produced.

(Given

$Y=2\times 10^{11}Pa$ )

0%
2.1 cm
0%
0.21 cm
0%
0.21 mm
0%
0.021 mm

Explanation

$\displaystyle \Delta l=\frac{Fl}{AY}=\frac{4\times8000\times10\times2.5}{\pi (.25)^{2}\times2\times10^{11}}=\frac{4\times16\times10^{-6}}{\pi}=0.021mm$

Which one of the following substance possesses the highest elasticity?

0%
rubber
0%
glass
0%
steel
0%
copper

Explanation

Modulus of elasticity gives the idea of elasticity. For above materials, values of E are:
Glass:

$50-90 GPa$
Rubber:

$0.01-0.1 GPa$
Steel:

$200 GPa$
Copper:

$117 GPa$

A spiral spring is stretched to 20.5 cm gradation on a metre scale when loaded with a 100 g load and to the 23.3 cm gradation by 200 g load. The spring is used to support a lump of metal in air and the reading now is 24.0 cm. The mass of metal lump is :

0%
250 gm
0%
225 gm
0%
145 gm
0%
750 gm

Explanation

Let original length of spring be

$L$ .

Thus,

$20.5-L=\dfrac{0.1kg\times g\times L}{AY}$ or

$\dfrac{L}{AY}=20.5-L$

Also, from second loading, we get similarly,

$\dfrac{0.2\times g\times L}{AY}=23.3-L$ or,

$\dfrac{L}{AY}=\dfrac{23.3-L}{2}$

Solve these to get,

$41-2L=23.3-L$ or

$L=17.7$

Also,

$\dfrac{L}{AY}=20.5-17.7=2.8$

Thus, for

$24\ cm$ ,

$\dfrac{mgL}{AY}=24-L$ or

$mg=\dfrac{(24-17.7)}{2.8}=2.25$ or

$m=225\ g$

A petite young woman of

$50\ kg$ distributes her weight equally over her high-heeled shoes. Each heel has an area of

$0.75cm^2$ . Find the pressure exerted by each heel? Take

$g=10 m/s^2$

0%
$6.66\times10^{6}Pa$
0%
$3.33\times10^{6}Pa$
0%
$1.67\times10^{6}Pa$
0%
$4.44\times10^{6}Pa$

Explanation

Pressure on each heel is :

$P=\dfrac {W/2}{A}$

$P=\dfrac {25\times 10}{0.75\times 10^{-4}}=3.33\times 10^6\ Pa$

A stone of mass m tied to one end of a thread of length l. The diameter of the thread is d and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle

$\theta$ with the vertical. Find the increase in length of the wire. Young's modulus of the wire is Y.

0%
$\displaystyle \frac{4mgl}{\pi d^{2}Y\cos\theta}$
0%
$\displaystyle \frac{4mgl}{\pi d^{2}Y\sin\theta}$
0%
$\displaystyle \frac{4mgl}{\pi d^{2}Y}$
0%
$\displaystyle \frac{4mgl}{\pi d^{2}Y\sec\theta}$

Explanation

The Tension generated in string can be calculated as

$Tcos\theta =mg$

$Thus\quad T=\dfrac { mg }{ cos\theta }$
Now,

$\Delta l=\dfrac { Tl }{ AY } =\dfrac { mg\times l }{ cos\theta \times \dfrac { \pi { d }^{ 2 } }{ 4 } \times Y } =\dfrac { 4mgl }{ \pi { d }^{ 2 }Ycos\theta }$

Outside a house,

$1 km$ from ground zero of a

$100$ kiloton nuclear bomb explosion, the pressure will rapidly rise to as high as

$2.8$ atm while the pressure inside the house is

$1$ atm. If the area outside the house is

$50\:m^2$ . The resulting net force exerted by the air in front of the house is :

0%
$9.1\times10^{6}N$
0%
$2.1\times10^{7}N$
0%
$1.41\times10^{7}N$
0%
$1.63\times10^{7}N$

Explanation

Pressure difference at the boundary of house

$= 2.8$ atm-

$1$ atm

$=1.8$ atm

So force exerted

$=PA=$ (

$1.8$ atm)(

$50$ )

$=1.8\times 101325 \times50 = 9.1\times 10^6N$

A uniform heavy rod of length $L$ , weight $W$ and cross-sectional area $A$ is hanging from a fixed support. Young's modulus of the material is $Y$ . Find the elongation of the rod.

0%
$\displaystyle \dfrac{WL}{AY}$
0%
$\displaystyle \dfrac{WL}{2AY}$
0%
$\displaystyle \dfrac{WL}{4AY}$
0%
$\displaystyle \dfrac{WL}{3AY}$

Explanation

$\displaystyle T=(L-x)\frac{W}{L};\:elongation=\frac{Tdx}{AY}$

$\displaystyle =\frac{(L-x)Wdx}{LAY};\:Total\:elongation=\frac{W}{LAY}\int_{0}^{L}(L-x)dx$
$\displaystyle =\frac{WL}{2AY}$

Poisson's ratio cannot exceed

0%
0.25
0%
1.0
0%
0.75
0%
0.5

Explanation

Poisson's ratio = Lateral strain/Longitudinal strain

$Y=3K(1-2\mu)\Rightarrow \mu=0.5-Y/6K$

$Y$ is young's modulus.

$\mu$ is poisson ratio

$K$ is compressibility of the substance which is inverse of Bulk's modulus. Maximum value of

$K$ is

$\infty$

So maximum value of Poisson's ratio

$\mu=0.5$

A student plots graph from his readings on the determination of Young's modulus of a metal wire but forgets to put the labels (figure). The quantities on

$X$ and

$Y$ -axes respectively may be

0%
weight hung and length increased.
0%
stress applied and length increased.
0%
stress applied and strain developed.
0%
length increased and the weight hung

The length of a wire is

$l_1$ when tension is

$T_1$ and

$l_2$ when tension is

$T_2$ . The natural length of the wire is

0%
$\displaystyle \frac{l_1+l_2}{2}$
0%
$\sqrt{l_1l_2}$
0%
$\displaystyle \frac{l_1T_2-l_2T_1}{T_2-T_1}$
0%
$\displaystyle \frac{l_1T_2+l_2T_1}{T_2+T_1}$
0%
$\displaystyle \frac{l_1T_1-l_2T_2}{T_1-T_2}$

Explanation

$\displaystyle \frac{T_1}{T_2}=\frac{l_1-l}{l_2-l}\:or\:T_1(l_2-l)=T_2(l_1-l)\:or\:l=\frac{l_1T_2-l_2T_1}{T_2-T_1}$

A sphere of mass M kg is suspended by a metal wire of length L and diameter d. When in equilibrium, there is a gap of

$\Delta l$ between the sphere and the floor. The sphere is gently pushed aside so that it makes an angle

$\theta$ with the vertical. Find

$\theta_{max}$ so that sphere fails to rub the Floor, Young's modulus of the wire is Y.

0%
$\displaystyle \sin^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$
0%
$\displaystyle \tan^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$
0%
$\displaystyle \cos^{-1}\left(1-\frac{Y\pi d^{2}\Delta l}{8MgL}\right)$
0%
none

Explanation

or $\displaystyle 1-\cos\theta=\dfrac{Y\pi d^{2}\Delta l}{8MgL}$ or $\displaystyle \cos\theta=1-\dfrac{Y\pi d^{2}\Delta l}{8MgL}$

Here we have used the energy conservation

$KE=PE$

$\displaystyle \dfrac{mv^{2}}{2}=mgl(1-\cos\theta)$

or $\displaystyle \dfrac{mv^{2}}{l}=2mg(1-\cos\theta)$

Thus

$\displaystyle \theta=\cos^{-1}\left(1-\dfrac{Y\pi d^{2}\Delta l}{8MgL}\right)$

A copper wire of cross-section A is under a tension T. Find the decrease in the cross-section area. Young's modulus is Y and Poisson's ratio is

$\sigma$

0%
$\displaystyle \frac{\sigma T}{2AY}$
0%
$\displaystyle \frac{\sigma T}{AY}$
0%
$\displaystyle \frac{2\sigma T}{AY}$
0%
$\displaystyle \frac{4\sigma T}{AY}$

Explanation

$\dfrac { \Delta d }{ d } =-\sigma \dfrac { \Delta l }{ l } =-\sigma \dfrac { T }{ AY } \\ Thus\quad \Delta d=-\sigma d\dfrac { T }{ AY } ,\quad hence\quad \Delta A=\Delta (\pi /4{ d }^{ 2 })=2.\pi /4{ .d }\Delta d\\ Hence\quad \Delta A=2.\pi /4{ .d }.d.(\dfrac { -\sigma T }{ AY } )=2\dfrac { -\sigma T }{ Y } \\ or\quad Decrease\quad =\quad -\dfrac { \Delta A }{ A } =\dfrac { 2\sigma T }{ AY }$

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio

$3 : 2$ . Their elongations are in the ratio

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

Explanation

HINT: Using Hooke's law we can find elongation in wire for a given load.

STEP 1: Use elongation formula

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

where

$F =$ force

$A =$ Area

$l =$ actual length

$Y =$ Young's modulus

Given same force, same length and same material means the Young's modulus is also same.

Volume = Area

$\times$ length

Given,

$m_1 : m_2 = 3:2$

mass = volume

$\times$ density

= Area

$\times$ length

$\times$ density (length & density are same)

$A_1 : A_2\;=\;3:2$

From eq.(1), we set

$\dfrac{\Delta l_1}{\Delta l_2} = \dfrac{A_2}{A_1} = \dfrac 23$

Thus, option C is correct.

Two identical wires of iron and copper with their Young's modulus in the ratio

$3:1$ are suspended at same level. They are to be loaded so as to have the same extension and hence level. Ratio of the weight is

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

Explanation

$\delta l\quad =\quad \dfrac { FL }{ AE } \\ \Rightarrow \quad \dfrac { { F }_{ c }.L }{ { A }.{ E }_{ c } } \quad =\quad \dfrac { { F }_{ i }.L }{ { A }.{ 3E }_{ c } } \quad or\quad { 3F }_{ c }\quad =\quad { F }_{ i }\\ Thus,\quad 3\quad { W }_{ c }\quad =\quad { W }_{ i }$

The length of a wire is increased by

$1\ mm$ on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

0%
$0.25\ mm$
0%
$0.5\ mm$
0%
$2\ mm$
0%
$4\ mm$

Explanation

Young's Modulus of elasticity =stress/strain

$Y = \dfrac {F/a}{\triangle l/l}$ or

$Y = \dfrac {Fl}{a\triangle l}$
or

$\triangle l = \dfrac {Fl}{aY} = \dfrac {Fl}{\pi r^{2} Y}$
In the given problem,

$\triangle l \propto \dfrac {l}{r^{2}}$
When both

$l$ and

$r$ are doubled,

$\triangle l$ is halved.

Young's modulus of rubber is

$10^{4} N/m^{2}$ and area of cross section is

$2\ cm^{-2}$ . If force of

$2\times 10^{5}$ dyn is applied along its length, then its initial

$l$ becomes.

0%
$3l$
0%
$4l$
0%
$2l$
0%
None of these

Explanation

We have

$1\ dyne=10^{ -5 }N$

$\Delta l=\dfrac { F.L }{ A.Y } =\dfrac { 2\times 10^{ 5 }\times 10^{ -5 }\times l }{ 2\times { 10 }^{ -4 }\times { 10 }^{ 4 } } =l\\ Thus\quad new\quad length\quad =\quad l+\Delta l=2l$

A cube is shifted to a depth of

$100m$ is alake. The change in volume is

$0.1$ %. The bulk modules of the material is nearly

0%
$10Pa$
0%
${10}^{4}Pa$
0%
${10}^{7}Pa$
0%
${10}^{9}Pa$

Explanation

$B=-V\dfrac{dP}{dV}$
Now, change in pressure,

$dP=\rho gh=1000 \times 10\times 100={10}^{6}$
Also, given

$\dfrac{dV}{V}=-0.001$
Thus,

$B=-\dfrac{{10}^{6}}{-0.001}={10}^{9}$

Two wires of the same length and same material but radii in the ratio of

$1:2$ are stretched by unequal forces to produce equal elongation. The ratio of the two forces is

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

Explanation

$Y = \dfrac {Fl}{a\triangle l}$
In the given problem,

$Y, l$ and

$\triangle l$ are constants.

$\triangle F\propto a$
or

$F \propto \pi r^{2}$ or

$F\propto r^{2}$ or

$\dfrac {F_{1}}{F_{2}} = \dfrac {r_{1}^{2}}{r_{2}^{2}} = \dfrac {1}{4}$ .

A long wire hangs vertically with its upper end clamped. A torque of

$8\ Nm$ applied to the free end twists it through

$45^{\circ}$ . The potential energy of the twisted wire is

0%
$\pi J$
0%
$\dfrac {\pi}{2} J$
0%
$\dfrac {\pi}{4} J$
0%
$\dfrac {\pi}{8} J$

Explanation

the elastic potential energy twisted by a torque T by an angle

$\theta$

$U = \dfrac{T\theta}{2}$

$U = \dfrac{8\times \dfrac{\pi}{4}}{2} = \pi J$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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