Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 39

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Question 1
1 / -0

The dimensions of two wires $$A$$ and $$B$$ are the same. But their materials are different, Their load- extension graphs are shown. If $$Y_{A}$$ and $$Y_{B}$$ are the values of Young's modulus of elasticity of $$A$$ and $$B$$ respectively then :

A
$$Y_{A} >Y_{B}$$

B
$$Y_{A}< Y_{B}$$

C
$$Y_{A} =Y_{B}$$

D
$$Y_{B} =2Y_{A}$$

Solution

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

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

For same load elongation in A is greater than in B therefore

$$\Delta l_A > \Delta l_B$$

$$\dfrac{FL}{Y_A A} > \dfrac{FL}{Y_B A}$$

$$Y_B >Y_A$$
Question 2
1 / -0

The diameter of a brass rod is 4 mm and Young's modulus of brass is $$9\times 10^{10}N/m^{2}$$. The force required to stretch by 0.1% of its length is :-

A
$$360\pi \:N$$

B
36 N

C
$$144\pi \times 10^{3}\:N$$

D
$$36\pi \times 10^{5}\:N$$

Solution

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

$$F = \dfrac{Y \Delta l A}{L} = \dfrac{YA\times 0.1L}{L} = 0.1YA = 0.1 \times 9 \times 10^{10}\times \pi \times (\dfrac{4\times 10^{-3}}{2})^2 = 360\pi$$
Question 3
1 / -0

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 :-

A
$$16 :1$$

B
$$4:1$$

C
$$2:1$$

D
$$1:1$$

Solution

$$U = \dfrac{1}{2} F\Delta l = \dfrac{1}{2} F \dfrac{FL}{YA} = \dfrac{1}{2} \dfrac{F^2L}{YA}$$
same material means they have same bulk modulus and length are also same.
therefore
$$d_1:d_2 = {r_1}:{r_2} =\dfrac{1}{2}$$
$$\dfrac{U_1}{U_2} = \dfrac{\dfrac{1}{2}\dfrac{F^2L}{Y\pi r_1^4}}{\dfrac{1}{2}\dfrac{F^2L}{Y\pi(2r_1)^4}} = \dfrac{16}{1}$$
Question 4
1 / -0

The Young's modulus of a rubber string $$8\ cm$$ long and density $$1.5\ kg/m^ {3}$$ is $$5\times 10^{8}N/m^{2},$$ is suspending on the ceiling in a room. The increase in the length due to its own weight will be:

A
$$9.6\times 10^{-5}\ m$$

B
$$9.6\times 10^{-11}\ m$$

C
$$9.6\times 10^{-3}\ m$$

D
$$9.6\ m$$

Solution

$$\displaystyle Y=\dfrac{F/A}{\dfrac{\Delta l}{l}}$$ $$\displaystyle F=\dfrac{Mg}{2}$$

$$\displaystyle \Delta l=\dfrac{FA}{AY}$$ $$\displaystyle F=\frac{Al\rho g}{2}$$

$$\displaystyle \Delta l=\dfrac{Al\rho gl}{2AY}=\frac{\rho l^{2}g}{2y}$$

$$\displaystyle \Delta l=\dfrac{1.5\times 64\times 10^{-14}\times 10}{2\times 5\times 10^{8}}$$

$$ \Delta l =9.6 \times 10^{-11}m $$
Question 5
1 / -0

On increasing the length by $$0.5 mm$$ in a steel wire of length $$2 m$$ and area of cross-section $$2\:cm^{2}$$, the force required is [$$Y$$ for steel $$=2.2\times 10^{11}N/m^{2}$$]:

A
$$1.1\times 10^{5}\:N$$

B
$$1.1\times 10^{4}\:N$$

C
$$1.1\times 10^{3}\:N$$

D
$$1.1\times 10^{2}\:N$$

Solution

Force, $$F = \dfrac{YA \Delta l}{L} = \dfrac{2.2 \times 10^{11} \times 2 \times 10^{-4} \times 0.5 \times 10^{-3}}{2} = 1.1 \times 10^3 N$$
Question 6
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A brass rod of cross-sectional area $$1\:cm^{2}$$ and length $$0.2 m$$ is compressed lengthwise by a weight of $$5 $$kg. If Young's modulus of elasticity of brass is $$1\times 10^{11}N/m^{2}$$ and $$g=10\:m/s^{2}$$, then increase in the energy of the rod will be :

A
$$10^{-5}$$ joule

B
$$2.5\times 10^{-5}$$ joule

C
$$5\times 10^{-5}$$ joule

D
$$2.5\times 10^{-4}$$ joule

Solution

$$U = \dfrac{1}{2} F \Delta l $$

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

$$U = \dfrac{1}{2} \dfrac{F^2L}{YA} = \dfrac{1}{2} \times \dfrac{50^2 \times 0.2}{10^{11} \times 10^{-4}} = 2.5 \times 10^{-5} J$$
Question 7
1 / -0

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 $$=2100\:MPa$$):

A
$$188\ kPa$$

B
$$8.4\ kPa$$

C
$$18.8 \ kPa$$

D
$$84 \ kPa$$

Solution

Bulk modulus $$\displaystyle B=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$$
$$\displaystyle B=2100\times 10^{6} \ Pa$$
$$\Delta P=\dfrac{B\Delta V}{V}$$

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

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

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

$$\Delta P=84000\:Pa$$

$$\Delta P=84\:kPa$$
Question 8
1 / -0

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:-

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

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

C
$$19.6\times 10^{10}N/m^{2}$$

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

Solution

$$B = \dfrac{P}{\dfrac{\Delta V}{V}}$$
$$P = h \rho g = 200 \times 10^3 \times 9.8 = 19.6 \times 10^5$$

$$B =\dfrac{ 19.6 \times 10^5}{\dfrac{0.1}{100}} = 19.6 \times 10^8 N/m^2$$
Question 9
1 / -0

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

A
$$1.77\times 10^{-3}m$$

B
$$7.17\times 10^{-3}m$$

C
$$3.17\times 10^{-7}m$$

D
$$1.37\times 10^{-7}m$$

Solution

Forces at the lowest point are gravitational force and centrifugal force

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

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

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

$$\Delta l = \dfrac{(30+ 3 \times 16 \pi^2\times 1.5)\times 1.5}{2 \times 10^{11} \times \pi \times (10^{-3})^2} = 1.77 \times 10^{-3} m$$
Question 10
1 / -0

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 :

A
$$+0.50$$

B
$$-0.50$$

C
$$0.25$$

D
$$-0.25$$

Solution

Let the material of length $$l$$ and side $$s$$
If a material maintains constant volume during stretching
$$V = l \times s^2$$
Differentiate wrt $$dl$$
$$dV = s^2.dl+ l .2s.ds$$
$$dl .s = 2l .ds$$
$$\dfrac{ds}{dl} = -\dfrac{1}{2}\dfrac{s}{l}$$
$$\eta = -\dfrac{\dfrac{ds}{s}}{\dfrac{dl}{l}} = \dfrac{1}{2}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 39

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Mechanical Properties of Solids Test - 39

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SHARING IS CARING

Question 1

$$Y_{A} >Y_{B}$$

$$Y_{A}< Y_{B}$$

$$Y_{A} =Y_{B}$$

$$Y_{B} =2Y_{A}$$

Solution

Question 2

$$360\pi \:N$$

36 N

$$144\pi \times 10^{3}\:N$$

$$36\pi \times 10^{5}\:N$$

Solution

Question 3

$$16 :1$$

$$4:1$$

$$2:1$$

$$1:1$$

Solution

Question 4

$$9.6\times 10^{-5}\ m$$

$$9.6\times 10^{-11}\ m$$

$$9.6\times 10^{-3}\ m$$

$$9.6\ m$$

Solution

Question 5

$$1.1\times 10^{5}\:N$$

$$1.1\times 10^{4}\:N$$

$$1.1\times 10^{3}\:N$$

$$1.1\times 10^{2}\:N$$

Solution

Question 6

$$10^{-5}$$ joule

$$2.5\times 10^{-5}$$ joule

$$5\times 10^{-5}$$ joule

$$2.5\times 10^{-4}$$ joule

Solution

Question 7

$$188\ kPa$$

$$8.4\ kPa$$

$$18.8 \ kPa$$

$$84 \ kPa$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$1.77\times 10^{-3}m$$

$$7.17\times 10^{-3}m$$

$$3.17\times 10^{-7}m$$

$$1.37\times 10^{-7}m$$

Solution

Question 10

$$+0.50$$

$$-0.50$$

$$0.25$$

$$-0.25$$

Solution

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