MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 6

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

$1:2$ are stretched by the same force. The potential energy per unit volume for the two wires when stretched will be in the ratio :-

0%
$16 :1$
0%
$4:1$
0%
$2:1$
0%
$1:1$

Explanation

$U = \dfrac{1}{2} F\Delta l = \dfrac{1}{2} F \dfrac{FL}{YA} = \dfrac{1}{2} \dfrac{F^2L}{YA}$

same material means they have same bulk modulus and length are also same.

therefore

$d_1:d_2 = {r_1}:{r_2} =\dfrac{1}{2}$

$\dfrac{U_1}{U_2} = \dfrac{\dfrac{1}{2}\dfrac{F^2L}{Y\pi r_1^4}}{\dfrac{1}{2}\dfrac{F^2L}{Y\pi(2r_1)^4}} = \dfrac{16}{1}$

The graph showing the extension of is wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. of the cross-sectional area of the wire is

$1{mm}^{2}$ , then the Young's modulus of the material of the wire .( In graph, X-axis 1 unit = 10mm) .

0%
$2\times {10}^{11}N{m}^{-1}$
0%
$2\times {10}^{10}N{m}^{-2}$
0%
$\dfrac {1}{2}\times {10}^{11}N{m}^{-2}$
0%
None of these

Explanation

Let

$Y$ denote the Young's Modulus.
Then

$Y\dfrac{\triangle L}{L_{0}}=\dfrac{F}{A}$

$L=1m$ ,

$A=10^{-6}m$

Hence

$Y(\dfrac{(5-1)\times 10^{-3}}m{1m}=\dfrac{100-20N}{10^{-6}m^{2}}$

$Y(4\times 10^{-3})=80\times 10^{6} N m^{-2}$

$Y=20\times 10^{9} N m^{-2}$

$=2\times 10^{10} N m^{-2}$

Copper wire of length

$3m$ and area of cross-section

$1\:mm^{2}$ , passes through an arrangement of two frictionless pulleys,

$P_{1}$ and

$P_{2}$ . One end of the wire is rigidly clamped and a mass of

$1 kg$ is hanged from the other end. If Young's modulus for copper is

$10\times 10^{10}N/m^{2}$ , the elongation in the wire is :

0%
$0.05 mm$
0%
$0.1 mm$
0%
$0.2 mm$
0%
$0.3 mm$

Explanation

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

$\Delta l = \dfrac{10 \times 3}{10\times 10^{10} \times 10^{-6}} = 3\times 10^{-4} =0.3mm$

An increase in pressure required to decreases the

$100\ liters$ volume of a liquid by

$0.004$ % in container is: (Bulk modulus of the liquid

$=2100\:MPa$ ):

0%
$188\ kPa$
0%
$8.4\ kPa$
0%
$18.8 \ kPa$
0%
$84 \ kPa$

Explanation

Bulk modulus

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

$\displaystyle B=2100\times 10^{6} \ Pa$

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

$\displaystyle \dfrac{\Delta P}{V}=\frac{0.004}{100}$

$\dfrac{\Delta V}{V}=4\times 10^{-15}$

$\displaystyle \Delta P=2100\times 10^{6}\times 4\times 10^{-5}$

$\Delta P=84000\:Pa$

$\Delta P=84\:kPa$

A ball falling in a lake of depth 200 m show 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball:-

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

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

$P = h \rho g = 200 \times 10^3 \times 9.8 = 19.6 \times 10^5$

$B =\dfrac{ 19.6 \times 10^5}{\dfrac{0.1}{100}} = 19.6 \times 10^8 N/m^2$

A steel wire

$1.5 m$ long and of radius

$1 mm$ is attached with a load

$3 kg$ at one end the other end of the wire is fixed it is whirled in a vertical circle with a frequency

$2Hz$ . Find the elongation of the wire when the weight is at the lowest position:
(

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

$g=10\:m/s^{2}$ )

0%
$1.77\times 10^{-3}m$
0%
$7.17\times 10^{-3}m$
0%
$3.17\times 10^{-7}m$
0%
$1.37\times 10^{-7}m$

Explanation

Forces at the lowest point are gravitational force and centrifugal force

$F = mg +m \omega^2r = mg + m \omega^2 L$

$\Delta l = \dfrac{FL}{YA} = \dfrac{(mg+m\omega^2L) L}{Y \times \pi r^2}$

$\omega = 2\pi f = 2\pi \times 2 =4\pi$

$\Delta l = \dfrac{(30+ 3 \times 16 \pi^2\times 1.5)\times 1.5}{2 \times 10^{11} \times \pi \times (10^{-3})^2} = 1.77 \times 10^{-3} m$

There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is :

0%
$+0.50$
0%
$-0.50$
0%
$0.25$
0%
$-0.25$

Explanation

Let the material of length

$l$ and side

$s$

If a material maintains constant volume during stretching

$V = l \times s^2$

Differentiate wrt

$dl$

$dV = s^2.dl+ l .2s.ds$

$dl .s = 2l .ds$

$\dfrac{ds}{dl} = -\dfrac{1}{2}\dfrac{s}{l}$

$\eta = -\dfrac{\dfrac{ds}{s}}{\dfrac{dl}{l}} = \dfrac{1}{2}$

A copper wire having

$Y=1\times 10^{11}N/m^{2}$ with length

$6m$ and a steel wire having

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

$4 m$ each of cross section

$10^{-5}m^{2}$ are fastened end to end stretched by a tension of

$100 N$ . The elongation produced in the copper wire is :

0%
$0.2 mm$
0%
$0.4 mm$
0%
$0.6 mm$
0%
$0.8 mm$

Explanation

Elongation.

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

$\Delta l = \dfrac{100 \times 6}{10^{11} \times 10^{-5}} + \dfrac{100 \times 4}{2 \times 10^{11} \times 10^{-5}} = 0.8mm$

The bulk modulus of rubber is

$9.8\times 10^{8}N/m^{2}$ . To what depth a rubber ball be taken in a lake so that its volume is decreased by

$0.1$ % ?

0%
$1 km$
0%
$25 km$
0%
$100 km$
0%
$200 km$

Explanation

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

$P = h \rho g = h \times 10^3 \times 9.8 = 9.8h \times 10^3$

$9.8 \times 10^8 =\dfrac{ 9.8h \times 10^3}{\dfrac{0.1}{100}}$

$h = 100 \ km$

When a force is applied on a wire of uniform cross-sectional area

$3\times 10^{-6}m^{2}$ and length

$4 m$ , the increase in length is

$1\ mm$ . Energy stored in it will be (

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

0%
$6250\ J$
0%
$0.177\ J$
0%
$0.075\ J$
0%
$0.150\ J$

Explanation

$U = \dfrac{1}{2} F \Delta l = \dfrac{1}{2} \times \dfrac{Y \Delta lA}{L} \times \Delta l = \dfrac{1}{2} \times \dfrac{2 \times 10^{11} \times 3\times 10^{-6} \times (10^{-3})^2}{4} = 0.075 J$

If a rubber ball is taken at the depth of 200 m in a pool its volume decreases by 0.1% If the density of the water is

$\displaystyle 1\times 10^{3}kg/m^{3}\:$ and

$\: g=10m/s^{2}$ then the volume elasticity in

$\displaystyle N/m^{2}$ will be

0%
$\displaystyle 10^{8}$
0%
$\displaystyle 2\times 10^{8}$
0%
$\displaystyle 10^{9}$
0%
$\displaystyle 2\times 10^{9}$

Explanation

Pressure change on the rubber ball at the depth of

$200m$ ,

$P=200 \times 1 \times{10}^{3} \times 10$

$2 \times {10}^{6}N/{m}^{2}.$

$\therefore$ fractional change in volume,

$\cfrac{\triangle V}{V}=\cfrac{0.1}{100}$

$\therefore$ Volume elasticity

$(K)=\cfrac{p}{\cfrac{\triangle V}{V}}$

$=\cfrac{2 \times {10}^{6}}{\cfrac{0.1}{100}}$

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

If the potential energy of a spring is V on stretching it by 2 cm then its potential energy when it is stretched by 10 cm will be

0%
V/25
0%
5 V
0%
V/5
0%
25 V

Explanation

Potential Energy

$\propto x^{2}$

Therefore

$\dfrac{V_1}{V_2}=\dfrac{(x_1)^{2}}{(x_2)^{2}}$

Here

$V_1=V$

$x_1=4$

$x_2=10$

$\dfrac{V}{V_2}=\dfrac{(2)^{2}}{(10)^{2}}$

$\dfrac{V}{V_2}=\dfrac{4}{100}$

$\dfrac{V}{V_2}=\dfrac{1}{25}$

$V_2=25V$

Hence the correct option is (D).

If the ratio of lengths, radii and Youngs modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be:

0%
$\displaystyle \dfrac{2ac}{b^{2}}$
0%
$\displaystyle \dfrac{3a}{2b^{2}c}$
0%
$\displaystyle \dfrac{3c}{2ab^{2}}$
0%
$\displaystyle \dfrac{2a^{2}c}{b}$

Explanation

Ref image

Given ;

$l_S / l_B = a , \, \dfrac{y_S}{y_B} = c$

$\dfrac{r_S}{r_B} = b$

By definition, young's Modulus =

$Y = \dfrac{linear \, stress}{linear \, strain}$

$\Rightarrow Y = \dfrac{T/A}{\Delta l/l}$

where, T - tension in wire

A - cross sectional area of wire

$\Delta l$ - change in length

l - original length

$\Rightarrow \boxed {\Delta l = \dfrac{T.l}{Y.A}}$ __(i)

FBD : Ref. image

$T_3 = mg + T_B$ __(ii) Ref. image

$T_B = 2mg$ __(iii)

Put (iii) in (ii) :

$T_S = mg + 2mg = 3mg$

$\Rightarrow \dfrac{T_S}{T_B} = \dfrac{3mg}{2mg} = \dfrac{3}{2}$ __(iv)

From (i) :

$\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{\left(\dfrac{T_S l_S}{Y_S A_S} \right)}{\left(\dfrac{T_B l_B}{Y_B A_B} \right)} \begin{matrix} = \left(\dfrac{T_S}{T_B} \right) . \left(\dfrac{l_S}{l_B} \right) \dfrac{Y_B \, A_B}{Y_S \, A_S} & / for \, a \, wire: A = \pi r^2 \\ = \dfrac{3}{2} \times a \times \dfrac{1}{c} \times \dfrac{\pi r_B^2}{\pi r_S^2} &/ from \, given \, ratios \end{matrix}$

$\boxed {\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{3 a}{2 c.b^2}}$

No correct option given

1179383_293473_ans_7132b5ae20ea4902af03c526542c8f35.png

A force F is needed to break a copper wire having radius R The force needed to break a copper wire of radius 2R will be

[assume F is applied along the wire and the wire obeys Hooke's law until it breakes]

0%
F/2
0%
2 F
0%
4 F
0%
F/4

Explanation

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

Force

$\propto$ area

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

$F_1\propto \pi R^{2}$

$F_2\propto \pi (2R)^{2}=\pi\times 4R^{2}$

$\dfrac{F_1}{F_2}=\dfrac{ \pi \times R^{2}}{\pi\times 4R^{2}}$

$\dfrac{F_1}{F_2}=\dfrac{1}{4}$

$F_2=4F_1$

Hence the correct option is (C).

Two wires of equal length and cross section area suspended as shown in figure. Their Youngs modulus are

$\displaystyle Y_{1}\: \: and\: \: Y_{2}$ respectively The equivalent Youngs modulus will be

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

Explanation

$k=\dfrac{YA}{L}$

$k_1=\dfrac{Y_1A}{L}$

$k_2=\dfrac{Y_2A}{L}$

$k_1$ and

$k_2$ are in parallel.

$k_{eq}=k_1+k_2$

$\dfrac{Y_{eq}(2A)}{L}=(Y_1+Y_2)\dfrac{A}{L}$

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

A metal block is experiencing an atmospheric pressure of

$\displaystyle 1\times 10^{5}N/m^{2}$ when the same block is placed in a vaccum chamber the fractional change in its volume is (the bulk modulus of metal is

$\displaystyle 1.25\times 10^{11}N/m^{2}$ )

0%
$\displaystyle 4\times 10^{-7}$
0%
$\displaystyle 2\times 10^{-7}$
0%
$\displaystyle 8\times 10^{-7}$
0%
$\displaystyle 1\times 10^{-7}$

Explanation

Bulk modulus

$(K)$ =

$\cfrac{p}{\cfrac{\triangle V}{V}}$

where,

$p=$ pressure experienced

$=1 \times {10}^{5} N/{m}^{2}$

$K=1.25 \times {10}^{11} N/{m}^{2}$

$\therefore \cfrac{\triangle V}{V}=\cfrac{p}{K}$

$=\cfrac {1 \times {10}^{5}}{1.25 \times {10}^{11}}$

$\therefore \cfrac{\triangle V}{V}=8 \times {10}^{-7}$

The high domes of ancient buildings have structural value (besides beauty). It arises from pressure difference on the two faces due to curvature (as in soap bubbles). There is a dome of radius 5 m and uniform (but small) thickness. The surface tension of its masonry structure is about 500 N/m. Treated as hemispherical, the maximum load the dome can support is nearest to

0%
1500 kg wt.
0%
3000 kg wt.
0%
6000 kg wt.
0%
12000 kg wt.

Explanation

$\begin{array}{l}\quad P_{\text {ext }}=\frac{4 T}{R}\\\text { Max. Load that can be supported }\\\qquad\begin{aligned}F&=\operatorname{P_{ext}} \times \pi r^{2} \\&=\frac{4 T}{R}\times \pi R^{2}=4 T \times \pi R\end{aligned}\end{array}$

$\begin{aligned}F &=4 \times 500 \times \frac{22}{7} \times 5 \\&=2000 \times 5 \times \frac{22}{7}=31400\mathrm{~N}\\F\text { in } \mathrm{kg} \omega t=\frac{31400}{\mathrm{~g}} \mathrm{~N} \approx3000&,\therefore \text {Option (B) is correct}\end{aligned}$

The diameter of a brass rod is

$4mm$ and Youngs modulus of brass is

$\displaystyle 9\times 10^{10}N/m^{2}$ The force required to stretch it by

$0.1\%$ of its length is

0%
$\displaystyle 360\pi N$
0%
$36 N$
0%
$\displaystyle 144\pi \times10^{3} N$
0%
$\displaystyle 36\pi \times10^{5} N$

Explanation

Let

$F$ be force

$Y$ be Young's modulus

$\Delta$ be change in length

$l$ be length

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

Given:

$\dfrac{\Delta l}{l}=0.001$

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

$r=\dfrac{d}{2}=2mm$

$A=\pi r^{2}=\pi\times (2\times 10^{-3})^{2}m^{2}$

$F=\dfrac{YA\Delta l}{l}=9\times 10^{10} \times \pi\times 4\times (10)^{-6}\times 0.001=360\pi$

Hence the force is

$360\pi N$

Two wires of the same material and length but diameter in the ratio 1 : 2 are stretched by the same force. The ratio of potential energy per unit volume for the two wires when stretched will be :

0%
1 : 1
0%
2 : 1
0%
4 : 1
0%
16 : 1

Explanation

Hint:

When a wire of length $L$ , Diameter $D$ , Young's Modulus of rigidity $Y$ is stretched by a Force $F$ from both the sides, then Potential Energy per unit volume stored in the wire is,

$E = \dfrac{8F^2}{Y \pi^2 D^{4}}$

Step 1: Calculate Energy stored in both the wires given in the question.

It is given that two wires of same length and material but diameter in ratio

$1:2$ are stretched by same force. Let diameter of first Wire is

$D$ , then diameter of second wire will be

$2D$ .

Potential Energy per unit volume stored in first wire is,

$E_1 = \dfrac{8F^2}{Y \pi^(2 D)^{4}}$ . ....eq.1

Potential Energy per unit volume stored in second wire is,

$E_2 = \dfrac{8F^2}{Y \pi^2 {2D}^{4}}$

$E_2 = \dfrac{8F^2}{16Y \pi^2 {D}^{4}}$ . ....eq.2

Step 2: Calculate the ratio of Energies Stored.

By dividing eq.1 and eq.2, we get

$\dfrac{E_1}{E_2} = \dfrac{\dfrac{8F^2}{Y \pi^2 D^{4}}}{\dfrac{8F^2}{16Y \pi^2 D^4}}$

$\Rightarrow \dfrac{E_1}{E_2} = \dfrac{16}{1}$

Thus, ratio of Potential Energy per unit volume stored in the wires is $16:1$ .

Option D is correct.

One end of a uniform rope of length L and of weight w is attached rigidIy to a point in the roof and a weight w

$_1$ is suspended from its lower. If s is the area of cross-section of the wire, the stress in the wire at a height

$\displaystyle \frac{3 L}{4}$ from its lower end is:

0%
$\displaystyle \frac{w}{s}$
0%
$\displaystyle \frac{\displaystyle w_1 + \frac{w}{4}}{s}$
0%
$\displaystyle \frac{\displaystyle w_1 + \frac{2w}{4}}{s}$
0%
$\displaystyle \frac{\displaystyle w_1 + w}{s}$

Explanation

Weight of the rope =

$w$

Weight at the point A

$= \displaystyle \frac{w}{4}$

Area of cross section =

$s$

Downward force of A

$= \displaystyle \frac{w}{4} + w_1$

Stress in the wire at A

$= \displaystyle \frac{\text{force}}{\text{area of cross section}} = \frac{\displaystyle \frac{w}{4} + w_1}{s}$

With what minimum acceleration can a fireman slide down a rope whose breaking strength is 3/4 th of his weight ?

0%
1/4 g
0%
1/2 g
0%
3/4 g
0%
zero

Explanation

Here,

$\displaystyle mg-T=ma$

$\displaystyle \Rightarrow \quad mg-\frac { 3 }{ 4 } mg=ma\\$

$\displaystyle \therefore \quad a={ g }/{ 4 }$

Assertion: Stress is the internal force per unit area of a body.
Reason: Rubber is more elastic than stee

0%
If both assertion and reason are true but the reason is the correct explanation of assertion.
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both the assertion and reason are false.
0%
If reason is true but assertion is false.

If in a wire of Young's modulus Y, longitudinal strain X is produced then the potential energy stored in its unit volume will be :

0%
$\displaystyle 0.5Y{ X }^{ 2 }$
0%
$\displaystyle 0.5{ Y }^{ 2 }X$
0%
$\displaystyle 2Y{ X }^{ 2 }$
0%
$\displaystyle Y{ X }^{ 2 }$

Explanation

Potential energy stored per unit volume of a wire

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

$Y = \dfrac{stress}{strain}$

$stress = Y \times strain$

$U = \dfrac{1}{2} \times YX \times X = \dfrac{1}{2}YX^2$

$S$ is stress and

$Y$ is Young's modulus of material of wire, then energy stored in the wire per unit volume is:

0%
$\displaystyle 2{ S }^{ 2 }Y$
0%
$\displaystyle \frac { S }{ Yx }$
0%
$\displaystyle \frac { 2Y }{ { S }^{ 2 } }$
0%
$\displaystyle \frac { { S }^{ 2 } }{ 2Y }$

Explanation

Energy stored per unit volume can be given as:

$E =\dfrac{1}{2} \times stress \times strain$ -----------(1)

From Hooke's law :

Young's modulus,

$Y = \dfrac{Stress}{Strain}$

$\implies$

$Strain = \dfrac{Stress}{Y} = \dfrac{S}{Y}$ -----------(2)

From equation (1) and (2):

$\therefore$

$E = \dfrac{1}{2} \times S \times \dfrac{S}{Y}$

$\Rightarrow E= \dfrac{1}{2} \dfrac{S^2}{Y}$

Hence, the correct option is

$(D)$

Longitudinal strain is possible in

0%
Liquid
0%
Gases
0%
Solid
0%
All of these

The adjacent graph shows the extension (

$\displaystyle \Delta l$ ) of a wire of length 1 m suspended from the top of a . roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is

$\displaystyle { 10 }^{ -6 }{ m }^{ 2 }$ , calculate the Young's modulus of the material of the wire.

0%
$\displaystyle 2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$
0%
$\displaystyle 2\times { 10 }^{ -11 }{ N }/{ { m }^{ 2 } }$
0%
$\displaystyle 3\times { 10 }^{ -12 }{ N }/{ { m }^{ 2 } }$
0%
$\displaystyle 2\times { 10 }^{ -13 }{ N }/{ { m }^{ 2 } }$

Explanation

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

from the graph we can see

$\dfrac{\Delta l}{W} = \dfrac{10^{-4}}{20}$

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

$Y = \dfrac{1\times 20}{10^{-6} \times 10^{-4}}$

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

For a given material, the Young's modulus is

$2.4$ times that of rigidity modulus. Its poisson's ratio is.

0%
$2.4$
0%
$1.2$
0%
$0.4$
0%
$0.2$

Explanation

Hint: The “ratio of transverse contraction strain to longitudinal extension strain in the direction of the stretching force” is known as the poisson ratio.

Explanation:

Step 1: Formula used:
The relation of Young's modulus '

$Y$ ' and rigidity modulus '

$\eta$ ' and poisson's ratio '

$\sigma$ ' :

$Y=2\eta(1+\sigma)$ ..........(1)

Step 2: Finding the value of poisson's ratio:

Given:

$Y= 2.4 \ \eta$
Putting the value in eq. (1)

$\Rightarrow 2.4 \ \eta =2 \ \eta \ (1+\sigma)$

$\Rightarrow 1.2=1+\sigma$

$\Rightarrow \sigma=0.2$

Option D is correct.

The length of an elastic string is

$a$ metre when the longitudinal tension is

$4$ N and

$b$ metre when the longitudinal tension is

$5$ N. The length of the string in metre when longitudinal tension is

$9$ N is :

0%
$a-b$
0%
$5b-4a$
0%
$2b-\displaystyle\frac{1}{4}a$
0%
$4a-3b$

Explanation

Let L is the original length of the wire and k is force constant of wire.
Final length

$=$ initial length

$+$ elongation

$L'=L+\displaystyle\frac{F}{k}$
For first condition

$a=L+\displaystyle\frac{4}{k}$ .....

$(i)$
For second condition

$b=\displaystyle L+\frac{5}{k}$ ......

$(ii)$
By solving Eqs.

$(i)$ and

$(ii)$ , we get

$L=5a-4b$ and

$k=\displaystyle\frac{1}{b-a}$
Now, when the longitudinal tension is

$9N$ . length of the string.

$=L+\displaystyle\frac{9}{k}=5a-4b+9(b-a)$

$=5b-4a$

For a constant hydraulic stress on an object, the fractional change in the object's volume (

$\displaystyle \Delta { V }/{ V }$ ) and its bulk modulus (B) are related as:

0%
$\displaystyle \frac { \Delta V }{ V } \propto B$
0%
$\displaystyle \frac { \Delta V }{ V } \propto \frac { 1 }{ B }$
0%
$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ 2 }$
0%
$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ -2 }$

Explanation

$B = \dfrac{stress}{volume \ strain} = \dfrac{stress}{\dfrac{\Delta V}{V}}$

$\dfrac{\Delta V}{V} = \dfrac{stress}{B}$

As stress is constant

$\dfrac{\Delta V}{V} \propto \dfrac{1}{B}$

The length of a metal wire is

$L_1$ when the tension is

$T_1$ and

$L_2$ when the tension is

$T_2$ . The unstretched length of wire is :

0%
$\displaystyle \frac{L_1+L_2}{2}$
0%
$\displaystyle \sqrt{L_1L_2}$
0%
$\displaystyle \frac{T_2L_1-T_1L_2}{T_2-T_1}$
0%
$\displaystyle \frac{T_2L_1+T_1L_2}{T_2+T_1}$

Explanation

Let the initial length of the metal wire is L.
The strain at tension

$T_1$ is

$\triangle L_1=L_1-L$
The strain at tension

$T_2$ is

$\triangle L_2=L_2-L$
Suppose, the Youngs modulus of the wire is Y,

$\displaystyle \frac{\dfrac{T_1}{A}}{\dfrac{\triangle L_1}{L}}=\dfrac{\dfrac{T_2}{A}}{\dfrac{\triangle L_2}{L}}$
Where, A is an cross-section of the wire assume to be same at all the situations.

$\Rightarrow \displaystyle \frac{T_1}{A}\times \frac{L}{\triangle L_1}=\frac{T_2}{A}\times \frac{L}{\triangle L_2}$

$\Rightarrow \displaystyle \frac{T_1}{(L_1-L)}=\frac{T_2}{(L_2-L)}$

$\displaystyle T_1(L_2-L)=T_2(L_1-L); L=\frac{T_2L_1-T_1L_2}{T_2-T_1}$

A metal wire of length l, area of cross-section A and Young's modulus Y behaves as a spring of spring constant k given by.

0%
$k=\displaystyle\frac{YA}{l}$
0%
$k=\displaystyle\frac{2YA}{l}$
0%
$k=\displaystyle\frac{YA}{2l}$
0%
$k=\displaystyle\frac{Yl}{A}$

In Young's double slit experiment, the ratio of intensities of bright and dark bands is

$16$ which means

0%
The ratio of their amplitudes is $5$
0%
Intensities of individual sources are $25$ and $9$ units respectively
0%
The ratio of their amplitudes is $4$
0%
Intensities of individuals sources are $4$ and $3$ units respectively

The increase in pressure required in

$kPa$ , to decrease the

$200$ litres volume of a liquid by

$0.004$ % is (bulk modulus of the liquid

$= 2100\ MPa$ )

0%
$8.4$
0%
$84$
0%
$92.4$
0%
$168$

Explanation

Bulk modulus of elasticity is defined as the ratio of normal stress to the volumetric strain, within the elastic limit.
Thus,

$B = \dfrac {\text {normal stress}}{\text {volumetric strain}}$

$B = \dfrac {\triangle p}{-\triangle V/V}$
Here, negative sign shows that volume is decreased, when pressure is increased.
Here,

$B = 2100\times 10^{6} Pa$

$V = 200\ L$

$\triangle V = 200\times \dfrac {0.004}{100} = 0.008\ L$

$\therefore 2100\times 10^{6} = \dfrac {\triangle p}{\left (\dfrac {0.008}{200}\right )}$

$\therefore \triangle p = 84 kPa$
Compressibility (C) of a material is the reciprocal of its bulk modulus of elasticity (B) ie,

$C = \dfrac {1}{B}$

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

0%
L = 100 cm, r = 1 mm
0%
L = 200 cm, r = 3 mm
0%
L = 300 cm, r = 3 mm
0%
L = 400 cm, r = 4 mm

Explanation

$\begin{array}{l}\text { L=length of the wire } \\r=\text { radius of the wine } \\Y \text { is same for all as made up of } \\\text { same material }\end{array}$

$Y=\frac{F / A}{\frac{d l}{l}} \Rightarrow d l=\frac{F l}{A Y}=\left(\frac{f}{\gamma}\right) \times \frac{\lambda}{\pi r^{2}}$

$\begin{aligned}\therefore d l &=\frac{c l}{\pi r^{2}} \\\therefore\left(\frac{e}{r^{2}}\right) \uparrow \Rightarrow d l \uparrow & \\\therefore \text { checking options we get }(L=100 \mathrm{~cm} \& r=\operatorname{lm} m) \text { wire elongate most} \\\text { }\end{aligned}$

Hence option A is correct

1990559_461518_ans_a19f6595cfe2493ebb459ba9226e3081.PNG

A long spring is stretched by

$2 cm$ and its potential energy is

$U$ . If the spring is stretched by

$10 cm$ ; its potential energy will be (in terms of

$U$ )

0%
${U}/{5}$
0%
${U}/{25}$
0%
$5 U$
0%
$25 U$

Explanation

The potential energy of a stretched spring is

$U = \dfrac{1}{2} k{x}^{2}$
Here,

$k =$ spring constant,

$x =$ elongation in spring.
But given that, the elongation is

$2 cm$ .
So,

$U = \dfrac{1}{2} k{\left(2\right)}^{2}$

$\Rightarrow U = \dfrac{1}{2} k \times 4$ ......(i)
If elongation is

$10 cm$ then potential energy

${ U }^{ \prime }=\dfrac { 1 }{ 2 } k{ \left( 10 \right) }^{ 2 }$

${ U }^{ \prime }=\dfrac { 1 }{ 2 } k\times 100$ ......(ii)
On dividing equation (ii) by equation (i), we have

$\dfrac { { U }^{ \prime } }{ U } =\dfrac { \dfrac { 1 }{ 2 } k\times 100 }{ \dfrac { 1 }{ 2 } k\times 4 }$
or

$\dfrac { { U }^{ \prime } }{ U } =25\Rightarrow { U }^{ \prime }=25U$

Two wires A and B are of the same materials. Their lengths are in the ratio

$1:2$ and the diameters are in the ratio

$2:1$ , When stretched by force

$F_A$ and

$F_B$ respectively they get equal increase in their lengths. Then the ratio

$\dfrac{F_A}{F_B}$ should be:

0%
$1:2$
0%
$1:1$
0%
$2:1$
0%
$8:1$

Explanation

Using Hooke's law

$Y = \dfrac{F/A}{\Delta L/L}$ where

$A = \dfrac{\pi D^2}{4}$

$\implies$

$F =Y \dfrac{\pi D^2}{4} \dfrac{\Delta L}{L}$

$Y$ and

$\Delta L$ are constant, thus

$F \propto \dfrac{D^2}{L}$

$\implies$

$\dfrac{F_A}{F_B} = \dfrac{D^2_A}{D^2_B} \times \dfrac{L_B}{L_A}$

Given :

$D_A:D_B = 2:1$ and

$L_A:L_B = 1:2$

$\therefore$

$\dfrac{F_A}{F_B} = \dfrac{4}{1} \times \dfrac{2}{1} = \dfrac{8}{1}$

There is no change in volume of a wire due to change in its length of stretching. The Poisson's ratio of the material of the wire is:

0%
0.50
0%
- 0.50
0%
0.25
0%
- 0.25

Explanation

Hint: The negative of the ratio of transverse strain to lateral or axial strain is Poisson's ratio.
Explanation:
Step 1: Concept used:

Volume of a wire of radius

$r$ and length

$l$ is,

$V=\pi{r}^{2}l$ .........................(1)
Differentiate the equation from both sides-

$\therefore dV=2\pi rldr + \pi{r}^{2}dl$

$(dV=0,$ as volume is unchanged

$)$

$0=2\pi rldr + \pi{r}^{2}dl$

$\therefore 2rldr = -{r}^{2}dl$

$\cfrac{dr}{r}= -\cfrac{1}{2} \cfrac{dl}{l}$

Step 2: Calculating the Poisson's ratio:

“The ratio of transverse contraction strain to longitudinal extension strain in the direction of the stretching force,” as defined by Poisson.

$\eta = \frac{Transverse\ strain}{Longitudional\ strain}$

$\therefore \cfrac{\cfrac{dr}{r}}{\cfrac{dl}{l}}=-\cfrac{1}{2}$

$\therefore \sigma= -\cfrac{1}{2}$

As, here

$-ve$ sign implies that if, length increased, radius decreased, so, we can write

$\sigma=\cfrac{1}{2}=0.5$

Option A is correct.

The value of Poisson's ratio (theoretically) lies between

0%
$-1$ to $\dfrac { 1 }{ 2 }$
0%
$-\dfrac { 3 }{ 4 }$ to $-\dfrac { 1 }{ 2 }$
0%
$-\dfrac { 1 }{ 2 }$ to $1$
0%
$1$ to $2$

Bulk modulus of water is

$2\times 10^9N/m^2$ . The change in pressure required to increase the density of water by

$0.1\%$ is.

0%
$2\times 10^9N/m^2$
0%
$2\times 10^8N/m^2$
0%
$2\times 10^6N/m^2$
0%
$2\times 10^4N/m^2$

Explanation

$Mass=volume\times density=Vd=constant$

$V\Delta d+d\Delta V=0$

$\Rightarrow \displaystyle\frac{\Delta V}{V}=-\frac{\Delta d}{d}=-\frac{0.1}{100}=10^{-3}$

$\therefore K=-\displaystyle\frac{\Delta p}{\Delta V/V}$

$\Rightarrow \Delta p=-K\frac{\Delta V}{V}$

$\Delta p=2\times 10^9\times 10^{-3}$

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

A tension of

$22$ N is applied to a copper wire of cross-sectional area

$0.02\ cm^2$ . Young's modulus of copper is

$1.1\times 10^{11}N/m^2$ and Poisson's ratio

$0.32$ . The decrease in cross sectional area will be:

0%
$1.28\times 10^{-6}cm^2$
0%
$1.6\times 10^{-6}cm^2$
0%
$2.56\times 10^{-6}cm^2$
0%
$0.64\times 10^{-6}cm^2$

Explanation

Young's modulus,

$Y=\dfrac{Fl}{A\Delta l}$ and Poisson's ratio ,

$\sigma=\dfrac{\Delta r/ r}{\Delta l/l}$

From these we get,

$\Delta r/ r=\dfrac{F\sigma}{AY}=\dfrac{22\times0.32 }{0.02\times 10^{-4}\times 1.1\times 10^{11}}=32\times 10^{-6}$

Cross sectional area ,

$A=\pi r^2$ and

$\Delta A=2\pi r \Delta r$

Thus,

$\dfrac{\Delta A}{A}=\dfrac{2\Delta r}{r}=64\times 10^{-6}$

$\therefore \Delta A= 64\times 10^{-6}\times 0.02=1.28\times 10^{-6} cm^2$

The materials, which do not show a fixed trend of deformation vs. applied force, are called:

0%
inelastic materials
0%
plastic materials
0%
elastic materials
0%
rigid materials

A spring is stretched by applying a load to its free end. The strain produced in the spring is

0%
Volumetric
0%
Shear
0%
Longitudinal and shear
0%
Longitudinal

The average depth of Indian ocean is about

$3000\ m$ . The value of fractional compression

$\dfrac{\Delta V}{V}$ of water at the bottom of the ocean is:

[Given that the bulk modulus of water is

$2.2\times 10^9\ Nm^{-2}$ ,

$g=9.8\ ms^{-2}$ and

$\rho_{H_2O}=1000\ kg.m^{-3}$ ]

0%
$3.4\times 10^{-2}$
0%
$1.34\times 10^{-2}$
0%
$4.13\times 10^{-2}$
0%
$13.4\times 10^{-2}$

Explanation

$\dfrac{\Delta V}{V}=\dfrac{h\rho g}{k}=\dfrac{3\times 10^3\times 9.8 \times 10^3}{22\times 10^8}$

$=\dfrac{3}{22}\times 10^{-3}\times 98\\ =\dfrac{147}{11}\times 10^{-3}\\=1.34\times 10^{-2}m$

Let L be the length and d be the diameter of cross-section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

0%
$L = 200\ cm, d = 0.5\ mm$
0%
$L = 300\ cm, d = 1.0\ mm$
0%
$L = 50\ cm, d = 0.05\ mm$
0%
$L = 100\ cm, d = 0.2\ mm$

Explanation

Since material is same, Young's modulus is also same.

We have:

$Y\dfrac { \Delta l }{ l } =\dfrac { F }{ A } \Rightarrow \Delta l=\dfrac { Fl }{ YA } \propto \dfrac { l }{ { d }^{ 2 } }$

where,

$Y$ is the Young's Modulus,

$F$ is the tension which is same in all cases,

$A$ is the area of cross section of the wire and

$l$ is the length of the wire.

Calculating the value of

$\dfrac { l }{ { d }^{ 2 } }$ in all cases, we get:

$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 2000 }{ 0.5\times 0.5 } =8000$

$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 3000 }{ 1.0\times 1.0 } =3000$

$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 50 }{ 0.05\times 0.05 } =20000$

$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 100 }{ 0.2\times 0.2 } =2500$

So maximum elongation will be in case C.

The Poisson's ratio of a material is

$0.5$ . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is :

0%
1%
0%
2%
0%
2.5%
0%
4%

Explanation

Poisson ratio

$=0.5$
Since, density is constant therefore change in volume is zero, we have

$V=A\times l=$ constant

$\Rightarrow \log { V } =\log { A } +\log { l }$
or

$\dfrac { dA }{ A } +\dfrac { dl }{ l } =0$

$\Rightarrow \dfrac { dl }{ l } =-\dfrac { dA }{ A }$

$\therefore$ Percentage increase in length

$=4$ %

A metal rod is fixed rigidly at two ends . If

$L, \alpha$ and

$Y$ respectively denote the length of the rod, coefficient of linear thermal expansion and Young's modulus of its material, then for an increase in temperature of the rod by

$\triangle T$ , the longitudinal stress developed in the rod is

0%
Inversely proportional to $\alpha$
0%
Inversely proportional to $Y$
0%
Directly proportional to $\dfrac{\triangle T}{Y}$
0%
Independent of $L$

Explanation

The increase in length of the rod due to increase in temperature of the rod=

$L\alpha \Delta T$

The strain in the rod due to increase in temperature of the rod=

$\alpha \Delta T$

Hence the longitudinal stress in the rod=

$Y\times Strain=Y\alpha\Delta T$

Hence correct answer is option D.

In the Searle's method to determine the Young's modulus of a wire, a steel wire of length

$156cm$ and diameter

$0.054\ cm$ is taken as experimental wire. the average increase in length for

$1.5\ kgwt$ is found to be

$0.050cm$ . then the Ypung's modulus of the wire is

0%
$3.002\times 10^{11}N/m^{2}$
0%
$1.002\times 10^{11}N/m^{2}$
0%
$2.002\times 10^{11}N/m^{2}$
0%
$2.5\times 10^{11}N/m^{2}$

Explanation

We know, Young's modulus is given by :

$Y = \dfrac {\text {Longitudinal stress}}{\text {Longitudinal strain}}$
if the radius of the wire of

$'r'$ then

$A = \pi r^{2}$

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

The figure shows the Searle's apparatus.

The adjacent graph shows the extension (l) of a wire of length 1 m suspended from the top of a roof at one end and with load 'W' connected to the other end. If the cross-sectional area of the wire is

$10^{-6}$

$m^{-2}$ , calculate the Young modulus of the material of the wire:

0%
$2 \times 10^{-11}$
0%
$2 \times 10^{10}$
0%
$2 \times 10^{11}$
0%
$2 \times 10^{-10}$

Explanation

From the graph if we choose a point for which

$W=40N$ and correspondingly

$\triangle L=2\times {10}^{-4}m$

Also given,

$A={10}^{-6}{m}^{2}$

$l=1m$

$\therefore$ Youngs modulus,

$Y=\cfrac{\cfrac{F}{A}}{\cfrac{\triangle l}{l}}$

putting values we get,

$Y=\cfrac{F\times l}{A\times\triangle l}$

$=\cfrac{40\times 1}{{10}^{-6}\times2\times{10}^{-4}}$

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

An iron rod of length 2m and cross- sectional area of $50mm^{2}$ stretched by 0.5mm, when a mass of 250 kg is hung from its lower end. Young's modulus of iron rod is

0%
$\displaystyle19.6\times 10^{20} N/m^{2}$
0%
$\displaystyle19.6\times 10^{18} N/m^{2}$
0%
$\displaystyle19.6\times 10^{10} N/m^{2}$
0%
$\displaystyle19.6\times 10^{15} N/m^{2}$

Explanation

$\displaystyle Y=\frac{F/A}{\Delta l/l}=\dfrac{\dfrac{250\times 9.8}{50\times 10^{-6}}}{\dfrac{0.5\times 10^{-3}}{2}}$

$\displaystyle =\dfrac{250\times 9.8}{50\times 10^{-6}}\times \dfrac{2}{0.5\times 10^{-3}}\Rightarrow 19.6\times 10^{10}N/m^{2}$

A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter?

0%
Youngs modulus
0%
Shear modulus
0%
Poissons ratio
0%
both Youngs modulus and Shear modulus

Explanation

Young's modulus

$=\cfrac{\cfrac{F}{A}}{\cfrac{\triangle l}{l}}$

Shear Modulus

$=\cfrac{p}{\cfrac{\triangle V}{V}}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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