Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 19

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Question 1
1 / -0

The length of a wire under stress changes by 0.01%. The strain produced is

A
$$10^{-4}$$

B
$$0.01$$

C
$$1$$

D
$$10^4$$

Solution

Strain $$=\dfrac{0.0001\Delta L}{L} = 0.01$$% $$ of L=\dfrac{.01\times L}{100}=10^{-4}L$$
$$=10^{-4}$$
Question 2
1 / -0

Four wires made of same materials are stretched by the same load. Their dimensions are given below. The one which elongates more is ?

A
Wire of length 1 m and diameter 1 mm

B
Length 2m, diameter 2 mm

C
Length 3m, diameter 3 mm

D
Length 0.5m, diameter 0.5mm

Solution

$$\Delta l=\dfrac{FL}{Ay}$$
$$A\rightarrow \Delta l=\dfrac{F\times 1}{y\pi (\dfrac{1}{2})^{2}}\times 10^{6} = \dfrac{yF}{\pi y}\times 10^{6}$$
$$B\rightarrow \Delta l=\dfrac{F\times 2}{y\pi 1^{2}}\times 10^{6} = \dfrac{2F}{\pi y}\times 10^{6}$$
$$C\rightarrow \Delta l=\dfrac{F\times 3\times 4}{y\pi 1}\times 10^{6} = \dfrac{4}{3}\dfrac{F}{\pi y}\times 10^{6}$$
$$D\rightarrow \Delta l=\dfrac{F\times \dfrac{1}{2}}{y\pi (\dfrac{1}{4})^{2}}\times 10^{6} = \dfrac{8F}{Y\pi }\times 10^{6}$$
$$\Delta l$$ of wire D is longest.
Question 3
1 / -0

A wire of length '$$l$$' and radius '$$r$$' is clamped rigidly at one end. When the other end of the wire is pulled by a force '$$F$$', its length increases by $$'x'$$. Another wire of same material of length '$$2l$$' and radius $$'2r$$' is pulled by a force '$$2F'$$, the increase in its length will be :

A
$$x$$

B
$$2x$$

C
$$x/2$$

D
$$4x$$

Solution

In the first case we have
$$x=\dfrac{Fl}{\pi r^2Y}$$
and in the for the second wire we have the change in length as
$$\dfrac{2F\times 2l}{\pi(2r)^2Y}=\dfrac{Fl}{\pi r^2Y}=x$$
Question 4
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The length of a wire is $$4m$$. Its length is increased by $$2mm$$ when a force acts on it. The strain is:

A
0.5 x 10$$^{-3}$$

B
5 x 10$$^{-3}$$

C
2 x 10$$^{-3}$$

D
0.05

Solution

$${L}'=4.002 m$$

$$Strain =\dfrac{0.002}{4}$$$$=5\times 10^{-4}$$ (using standard result)
Question 5
1 / -0

A steel wire of $$2mm$$ in diameter is stretched by applying a force of $$72N$$. Stress in the wire is

A
$$2.29\times 10^7 \ N/m^2$$

B

$$1.17\times 10^7 \ N/m^2$$

C

$$3.6\times 10^7 \ N/m^2$$

D

$$0.8\times 10^7 \ N/m^2$$

Solution

Diameter=$$2\times 10^{-3}$$m
Radius=$$10^{-3}$$m
area of cross section=$$\pi r^{2}$$
$$A=\dfrac { 22 }{ 7 } \times { 10 }^{ -6 }㎡\\ F=72N\\ stress=\dfrac { F }{ A } \\ =\dfrac { 72 }{ \dfrac { 22 }{ 7 } \times { 10 }^{ -6 } } =\dfrac { 72 }{ 3.14 } \times { 10 }^{ -6 }$$
So, stress=$$22.92\times 10^{6}M/m^{2}$$
or stress=$$2.29\times10^{7}N/m^{2}$$
Question 6
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The elongation produced in a copper wire of length 2m and diameter 3mm, when a force of 30N is applied is [Y$$=$$1x10$$^{11}$$N.m$$^{-2}$$]

A
$$8.5mm$$

B
$$0.85mm$$

C
$$0.085mm$$

D
$$85mm$$

Solution

$$L=2m,$$
$$d=3mm, A=\dfrac{9\pi }{4}\times 10^{-6} m^{2}$$
$$\Delta L=\dfrac{30\times 2}{\dfrac{9\pi }{4}\times 10^{-6}\times 10^{11}}$$$$=8.48\times 10^{-5} m$$$$=0.085 mm$$
Question 7
1 / -0

The lengths of two wires of the same material and diameter are $$100 cm$$ and $$125 cm$$. If same force is applied on them, the elongation in the first wire is $$4 mm$$. The elongation in the second wire (in mm) is

A
$$4$$

B
$$5$$

C
$$0.8$$

D
$$1.25$$

Solution

$$L_{1}=100 cm ,L_{2}=125 cm$$
$$\Delta L_{1}=4 cm , \Delta L_{2}=?$$
$$A_{1}=A_{2} Y_{1}=Y_{2} , F_{1}=F_{2}$$
$$\dfrac{\Delta L_{1}}{L_{1}}=\dfrac{\Delta L_{2}}{L_{2}}$$
$$\dfrac{1250\times 4}{1000}=\Delta L_{2}$$
$$=5 mm$$
Question 8
1 / -0

The force required to double the length of a steel wire of area of cross-section $$5 \times 10^{-5} m^{2}$$ (in N) is :
($$Y=20 \times 10^{10} Pa$$)

A
$$10^{7}$$

B
$$10^{6}$$

C
$$10^{-7}$$

D
$$10^{5}$$

Solution

$$L=\Delta L$$
$$F=\dfrac{\Delta L}{L} AY=AY$$
$$=5\times 10^{-5}\times 20\times 10^{10}$$
$$=10^{7} N$$
Question 9
1 / -0

The length of two wires are in the ratio $$3 : 4$$.Ratio of the diameters is $$1:2$$; young's modulus of the wires are in the ratio $$3:2$$; If they are subjected to same tensile force, the ratio of the elongation produced is

A
1 :1

B
1 :2

C
2 : 3

D
2 : 1

Solution

$$\displaystyle \frac{L_{1}}{L_{2}}=\frac{3}{4}, \quad \frac{D_{1}}{D_{2}}=\frac{1}{2} , \quad \frac{A_{1}}{A_{2}}=\frac{1}{4}$$

$$\dfrac{y_{1}}{y_{2}}=\dfrac{3}{2}$$ same force

$$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{F.L_{1}}{A_{1}y_{1}}\times \dfrac{A_{2}y_{2}}{FL_{2}}$$$$=\dfrac{3}{4}\times \dfrac{4}{1}\times \dfrac{02}{3}$$

$$\dfrac{\Delta L_{1}}{\Delta L_{2}}=\dfrac{2}{1}$$
Question 10
1 / -0

A solid sphere hung at the lower end of a wire is suspended from a fixed point so as to give an elongation of $$0.4mm$$. When the first solid sphere is replaced by another one made of same material but twice the radius, the new elongation is

A
$$0.8mm$$

B
$$1.6mm$$

C
$$3.2mm$$

D
$$1.2mm$$

Solution

As radius is doubled volume becomes 8 times and, hence, the wt. of sphere.
The force becomes 8 times.
$$F \propto \Delta L$$
$$\therefore \Delta L$$ also becomes 8 times
$$\Delta {L}'=8\times 0.4 mm$$$$=3.2 mm$$

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Mechanical Properties of Solids Test - 19

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Mechanical Properties of Solids Test - 19

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Question 1

$$10^{-4}$$

$$0.01$$

$$1$$

$$10^4$$

Solution

Question 2

Wire of length 1 m and diameter 1 mm

Length 2m, diameter 2 mm

Length 3m, diameter 3 mm

Length 0.5m, diameter 0.5mm

Solution

Question 3

$$x$$

$$2x$$

$$x/2$$

$$4x$$

Solution

Question 4

0.5 x 10$$^{-3}$$

5 x 10$$^{-3}$$

2 x 10$$^{-3}$$

0.05

Solution

Question 5

$$2.29\times 10^7 \ N/m^2$$

$$1.17\times 10^7 \ N/m^2$$

$$3.6\times 10^7 \ N/m^2$$

$$0.8\times 10^7 \ N/m^2$$

Solution

Question 6

$$8.5mm$$

$$0.85mm$$

$$0.085mm$$

$$85mm$$

Solution

Question 7

$$4$$

$$5$$

$$0.8$$

$$1.25$$

Solution

Question 8

$$10^{7}$$

$$10^{6}$$

$$10^{-7}$$

$$10^{5}$$

Solution

Question 9

1 :1

1 :2

2 : 3

2 : 1

Solution

Question 10

$$0.8mm$$

$$1.6mm$$

$$3.2mm$$

$$1.2mm$$

Solution

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