Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 45

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Question 1
1 / -0

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:

A
$$2 \times 10^{-11}$$

B
$$2 \times 10^{10}$$

C
$$2 \times 10^{11}$$

D
$$ 2 \times 10^{-10}$$

Solution

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,
$$Y=\cfrac{F\times l}{A\times\triangle l}$$
$$=\cfrac{40\times 1}{{10}^{-6}\times2\times{10}^{-4}}$$
$$=2\times{10}^{11}N/{m}^{2}$$
Question 2
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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?

A
Youngs modulus

B
Shear modulus

C
Poissons ratio

D
both Youngs modulus and Shear modulus

Solution

Young's modulus$$=\cfrac{\cfrac{F}{A}}{\cfrac{\triangle l}{l}}$$
Shear Modulus$$=\cfrac{p}{\cfrac{\triangle V}{V}}$$
Question 3
1 / -0

A wire of initial length $$L$$ and radius $$r$$ is stretched by a length $$l$$. Another wire of same material but with initial length $$2L$$ and radius $$2r$$ is stretched by a length $$2l$$. The ratio of the stored elastic energy per unit volume in the first and second wire is,

A
1 : 4

B
1 : 2

C
2 : 1

D
1 : 1

Solution

Elastic energy stored per unit volume $$E = \dfrac{1}{2}\times Stress \times Strain$$
Using Hooke's law, $$Stress = Y\times Strain$$
We get $$E = \dfrac{1}{2}Y\times (Strain)^2$$ where Young's modulus $$Y$$ is a material property
$$\implies$$ $$E\propto (Strain)^2$$
Strain in first wire, $$Strain_1 = \dfrac{l}{L}$$
Strain in second wire, $$Strain_2 = \dfrac{2l}{2L} = \dfrac{l}{L}$$
$$\therefore$$ Ratio of energy stored per unit volume $$\dfrac{E_1}{E_2} = \dfrac{(l/L)^2}{(l/L)^2} =1$$
$$\implies$$ $$E_1 : E_2 = 1:1$$
Question 4
1 / -0

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

A
inelastic materials

B
plastic materials

C
elastic materials

D
rigid materials

Solution

Elastic materials are those that follow the Hooke's law, which is that the deformation produced in a material is directly proportional to the stress applied to it, and the material is recoverable after the deformation force is removed.
Inelastic materials are those that do not follow this relationship. They do not show a fixed trend of deformation vs applied force; in fact, they might not deform at all (rigid materials) or the deformation observed is not completely recoverable.
Question 5
1 / -0

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

A
Inversely proportional to $$\alpha$$

B
Inversely proportional to $$Y$$

C
Directly proportional to $$\dfrac{\triangle T}{Y}$$

D
Independent of $$L$$

Solution

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.
Question 6
1 / -0

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

A
$$3.002\times 10^{11}N/m^{2}$$

B
$$1.002\times 10^{11}N/m^{2}$$

C
$$2.002\times 10^{11}N/m^{2}$$

D
$$2.5\times 10^{11}N/m^{2}$$

Solution

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.
Question 7
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When temperature of gas is $$20^{0}C$$ and pressure is changed from $$P_1$$=1.01$$\times 10^{5}$$Pa to $$P_2$$=1.165$$\times 10^{5}$$Pa then the volume changed by 10%. the bulk modulus is:

A
$$1.55 \times 10^{5} Pa$$

B
$$0.155 \times 10^{5} Pa$$

C
$$15.5 \times 10^{5} Pa$$

D
$$155 \times 10^{5} Pa$$

Solution

Change in pressure,
$$\triangle P={P}_{2}-{P}_{1}$$
$$=0.155\times{10}^{5}{P}_{a}$$
and,
$$\cfrac{\triangle V}{V}=\cfrac{10}{100}$$
$$\therefore $$ Bulk Modulus$$=\cfrac{\triangle P}{\cfrac{\triangle V}{V}}$$
$$=\cfrac{0.155\times{10}^{5}}{\cfrac{10}{100}}$$
$$=1.55\times{10}^{5} {P}^{a}$$
Question 8
1 / -0

Stretching of a rubber band results in ___________.

A
No change in potential energy

B
Zero value of potential energy

C
Increase in potential energy

D
Decrease in potential energy

Solution

When a rubber band is stretched, work is done on it , in course of which some amount of energy in the form of elastic potential energy (strain energy) is input into it. This increases the potential energy of the system. When the rubber band is released, the potential energy is quickly converted to kinetic energy.
Option C.
Question 9
1 / -0

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

A
$$\displaystyle19.6\times 10^{20} N/m^{2}$$

B
$$\displaystyle19.6\times 10^{18} N/m^{2}$$

C
$$\displaystyle19.6\times 10^{10} N/m^{2}$$

D
$$\displaystyle19.6\times 10^{15} N/m^{2}$$

Solution

$$\displaystyle Y=\frac{F/A}{\Delta l/l}=\dfrac{\dfrac{250\times 9.8}{50\times 10^{-6}}}{\dfrac{0.5\times 10^{-3}}{2}}$$
$$\displaystyle =\dfrac{250\times 9.8}{50\times 10^{-6}}\times \dfrac{2}{0.5\times 10^{-3}}\Rightarrow 19.6\times 10^{10}N/m^{2}$$
Question 10
1 / -0

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm are connected end to end. When stretched by a force, the elonation in length 0.50 mm is produced in the copper wire. The stretching force is
$$(Y_{cu}=1.1\times 10^{11}N/m^2, Y_{steel}=2.0\times 10^{11}N/m^2)$$

A
$$5.4\times 10^2N$$

B
$$3.6\times 10^2N$$

C
$$2.4\times 10^2N$$

D
$$1.8\times 10^2N$$

Solution

Given that,
Length of the copper wire$$ l_1=2.2m$$
Length of the steel wire $$l_2=1.6m$$
Elongation in length$$\triangle{l}=0.5.mm$$
$$=0.5\times10^{-3}m$$
Radius of the$$Cu$$ wire $$r_1=1.5\times 10^{-3}m$$
$$Y_1=1.1\times 10^{11} N/m^2$$
$$Y_2=2.0\times 10^{11}N/m^2$$
Let $$F$$ be the stretching force in both the wires then,
for $$ Cu$$ wire, $$Y_1=\dfrac{F}{\pi r^2_1}\times \dfrac{l_1}{\triangle l_1}$$
$$\Rightarrow F=\dfrac{Y_1\pi r_1^2 \times \triangle l_1}{l_1}$$
$$=\dfrac{1.1\times 10^{-11}}{2.2}\times \dfrac{22}{7}\times (1.5\times 10^{-3})^2\times 0.5\times 10^3$$
$$=1.8\times 10^2N$$
So the streching force in the copper wire be $$1.8\times10^2N$$

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Mechanical Properties of Solids Test - 45

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Mechanical Properties of Solids Test - 45

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Question 1

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

$$2 \times 10^{10}$$

$$2 \times 10^{11}$$

$$ 2 \times 10^{-10}$$

Solution

Question 2

Youngs modulus

Shear modulus

Poissons ratio

both Youngs modulus and Shear modulus

Solution

Question 3

1 : 4

1 : 2

2 : 1

1 : 1

Solution

Question 4

inelastic materials

plastic materials

elastic materials

rigid materials

Solution

Question 5

Inversely proportional to $$\alpha$$

Inversely proportional to $$Y$$

Directly proportional to $$\dfrac{\triangle T}{Y}$$

Independent of $$L$$

Solution

Question 6

$$3.002\times 10^{11}N/m^{2}$$

$$1.002\times 10^{11}N/m^{2}$$

$$2.002\times 10^{11}N/m^{2}$$

$$2.5\times 10^{11}N/m^{2}$$

Solution

Question 7

$$1.55 \times 10^{5} Pa$$

$$0.155 \times 10^{5} Pa$$

$$15.5 \times 10^{5} Pa$$

$$155 \times 10^{5} Pa$$

Solution

Question 8

No change in potential energy

Zero value of potential energy

Increase in potential energy

Decrease in potential energy

Solution

Question 9

$$\displaystyle19.6\times 10^{20} N/m^{2}$$

$$\displaystyle19.6\times 10^{18} N/m^{2}$$

$$\displaystyle19.6\times 10^{10} N/m^{2}$$

$$\displaystyle19.6\times 10^{15} N/m^{2}$$

Solution

Question 10

$$5.4\times 10^2N$$

$$3.6\times 10^2N$$

$$2.4\times 10^2N$$

$$1.8\times 10^2N$$

Solution

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