Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

The Young's modulus of the material of a rod is $$20 \times 10^{10}$$ pascal. When the longitudinal strain is 0.04%, The energy stored per unit volume is

A
$$4 \times 10^{3} \ J/m^3$$

B
$$8 \times 10^{-3} \ J/m^3$$

C
$$16 \times 10^{-3} \ J/m^3$$

D
$$16 \times 10^{3} \ J/m^3$$

Solution

$$\begin{array}{l}\text { Energy stored per unit volume }=\frac{1}{2} \times \text { stress}\times\text{strain. } \\\qquad \begin{array}{c}\text { Giuen } y=20 \times 10^{10}\mathrm{Pa}\text { and } \text { strain }=0.04\% \\y=\frac{\text { Stress }}{\text { strain. }} \Rightarrow \text { stress }=y(\text { strain })\end{array}\end{array}$$

$$\begin{aligned}\Rightarrow \text { Energy stored }&\text { Per unit volume }=\frac{1}{2}\times Y(\operatorname{strain} )^{2} \\&=\frac{1}{2} \times 20 \times 10^{10} \times\left(4 \times 10^{-4}\right)^{2} \\&=10 \times 10^{10} \times 16\times 10^{-8} \\&=16 \times 10^{3} \mathrm{~J} /\mathrm{m}^{3}\end{aligned}$$
Question 2
1 / -0

A woman with a mass of $$65\ kg$$ puts all her weight on one heel of her high-heel shoe. The cross-sectional area of the heel is $$1\ cm^{2}$$. According to the table, if she is standing on a pane of glass that is flat against the ground, does the glass break :

A
No, because the stress is less than the ratio for the ultimate strength to Young's modulus for glass

B
No, because the stress is less than the ultimate strength of glass

C
Yes, because the stress is greater than the ratio for the ultimate strength to Young's modulus for glass

D
No, because the stress is greater than the ultimate strength of glass

Solution

$$(B)$$ No, because the stress is less than the ultimate strength of glass.
$$m=65kg$$
$$A=1{ cm }^{ 2 }$$
$$=1\times { \left( { 10 }^{ -2 } \right) }^{ 2 }$$
$$=1\times { 10 }^{ -4 }{ m }^{ 2 }$$
Question 3
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A metal wire of length $$1\ m$$ and cross-section area $$2\ mm^{2}$$ and Young's modulus of elasticity $$Y=4\times 10^{11}\ N/m^{2}$$ is stretched by $$2\ mm$$. Then

A
the restoring force developed in the wire is $$1600\ N$$

B
the energy density in the wire is $$4\times 10^{5}\ J/m^{3}$$

C
the restoring force developed in the wire is $$400\ N$$

D
the total elastic energy stored in the wire is $$1.6\ J$$

Solution

$$L=1m$$
$$\alpha=2{ mm }^{ 2 }=\left( \dfrac { 2 }{ { \left( 1000 \right) }^{ 2 } } \right) =2\times { 10 }^{ -6 }m$$
$$Y=4\times { 10 }^{ 11 }N/{ m }^{ 2 }$$
$$\Delta l=2mm=\dfrac { 2 }{ 10.00 } =0.002m$$
$$F=\dfrac { Y\alpha l }{ L } $$
$$=\dfrac { 4\times { 10 }^{ 11 }\times 2\times { 10 }^{ -6 }\times 2\times { 10 }^{ -3 } }{ 1m } $$
  $$=16\times { 10 }^{ 11-6-3 }$$
  $$=16\times { 10 }^{ 11-9 }$$
  $$=16\times { 10 }^{ 2 }$$
$$=1600N$$
Where $$\alpha =$$ cross sectional area $$=2\times { 10 }^{ -6 }m$$
  $$l=2mm=2\times { 10 }^{ -3 }m$$
$$L=1m$$
Question 4
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A force F doubles the length of wire of cross-section a. The Young modulus of wire is

A
$$\dfrac{F}{a}$$

B
$$\dfrac{F}{3a}$$

C
$$\dfrac{F}{2a}$$

D
$$\dfrac{F}{4a}$$

Solution

As youngsters modulus is given as$$,$$
$$Y=f/a.$$ $$L/\Delta L$$
From above we can conclude that
$$Y' = 2Y$$
That is youngs modulus becomes $$2$$ times to its original value$$.$$
Hence,
option $$(C)$$ is correct answer.
Question 5
1 / -0

An elastic metal rod will change its length when it

A
falls vertically under its weight

B
is pulled along its lengthy by a force acting at one end

C
both

D
none

Solution

$$\begin{array}{l}\text { An elastic metal rod will change it's length when } \\\text { it is pulled along it's length by a force acting at one } \\\text { end. }\end{array}$$
Question 6
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If the speed of longitudinal wave equals $$10$$ times the speed of the transverse waves in a stretched wire of material which has modulus of elasticity E, then the stress in the wire is

A
$$100E$$

B
$$\dfrac {E}{ 100 } $$

C
$$\dfrac {E}{ 10 } $$

D
$$10E$$

Solution

$${v}_{l}=\sqrt{\dfrac{Y}{\rho}}$$ and $${v}_{t}=\sqrt{\dfrac{Y\dfrac{\Delta l}{l}}{\rho}}={v}_{l}\sqrt{\dfrac{\Delta l}{l}}$$
$$\Rightarrow \sqrt{\dfrac{l}{\Delta l}}=\dfrac{{v}_{l}}{{v}_{t}}=10$$
$$\therefore \dfrac{\Delta l}{l}=\dfrac{1}{100}$$
Stress$$=Y\dfrac{\Delta l}{l}=E\dfrac{\Delta l}{l}=\dfrac{E}{100}$$
Question 7
1 / -0

Two wires of the same material have masses in the ratio 3:4. The ratio of their extensions under the same load if their lengths are in the ratio 9:10 is

A
5 : 3

B
27 : 40

C
6 : 5

D
27 : 25

Solution

$$Y=\dfrac { FL }{ \left( Ae \right) } $$
For same material $$Y$$ is constant and $$A$$ is constant for two similar wires.
$$\dfrac { FL }{ e } =$$ constant
here, $$F=mg$$
$$\dfrac { { e }_{ 1 } }{ { e }_{ 2 } } =\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } \times \dfrac { { L }_{ 1 } }{ { L }_{ 2 } } $$
$$=\dfrac { 3 }{ 4 } \times \dfrac { 9 }{ 10 } $$
$$=\dfrac { 27 }{ 40 } $$
Question 8
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In case of bending of a beam, depression $$\delta$$ depends on Young modulus of elasticity $$Y$$ as

A
$$\propto Y$$

B
$$\propto Y^{2}$$

C
$$\propto Y^{-1}$$

D
$$\propto Y^{-2}$$

Solution

In case of bending of a beam, depression
$$\delta$$ depends on Young modulus of elasticity $$Y$$ as
$$ \delta = \dfrac{PL^3}{48YI}$$
$$\therefore \delta \propto Y^{-1}$$
Question 9
1 / -0

A steel bar of cross-section $$500$$ mm and length $$1$$ m acted upon by two forces as shown If Young modulus of elasticity of steel is $$200 \times 10^9$$ $$N/m^2$$ then elongation of the rod is

A
$$0.05$$ mm

B
$$0.5$$ mm

C
$$0.1$$ mm

D
$$0.2$$ mm

Solution

$$\begin{array}{l}\text { Tension in } b a r=50\times10^{3}N\\\text { length }=12 m \\\text { Area of cross section =500mm }^{2} \\\qquad Y=200\times10^{9}\mathrm{~N} / \mathrm{m}^{2}\end{array}$$

$$\begin{aligned}\Delta l &=\frac{F L}{A Y}\\&=\frac{50_{}\times 10^{3} \times 1}{\operatorname{500}\times10^{6}\times200\times10^{9}}\\&=\frac{1}{2} \times 10^{-3} \\\Delta l&=0.5\mathrm{~mm}\end{aligned}$$
Question 10
1 / -0

The extension produced in a wire by the application of a load is $$3.0$$ mm. The extension produced in a wire of the same material and length but half the radius by the same load is:

A
$$5$$ mm

B
$$9$$ mm

C
$$12$$ mm

D
$$13$$ mm

Solution

Formulae,

$$\dfrac{\Delta l_2}{\Delta l_1}=\dfrac{a_1}{a_2}$$

$$=\dfrac{a_1}{\dfrac{a_1}{4}}$$

$$\Rightarrow \Delta l_2=4\Delta l_1$$

$$\therefore \Delta l_2=4\times 3=12mm$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 65

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Mechanical Properties of Solids Test - 65

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Question 1

$$4 \times 10^{3} \ J/m^3$$

$$8 \times 10^{-3} \ J/m^3$$

$$16 \times 10^{-3} \ J/m^3$$

$$16 \times 10^{3} \ J/m^3$$

Solution

Question 2

No, because the stress is less than the ratio for the ultimate strength to Young's modulus for glass

No, because the stress is less than the ultimate strength of glass

Yes, because the stress is greater than the ratio for the ultimate strength to Young's modulus for glass

No, because the stress is greater than the ultimate strength of glass

Solution

Question 3

the restoring force developed in the wire is $$1600\ N$$

the energy density in the wire is $$4\times 10^{5}\ J/m^{3}$$

the restoring force developed in the wire is $$400\ N$$

the total elastic energy stored in the wire is $$1.6\ J$$

Solution

Question 4

$$\dfrac{F}{a}$$

$$\dfrac{F}{3a}$$

$$\dfrac{F}{2a}$$

$$\dfrac{F}{4a}$$

Solution

Question 5

falls vertically under its weight

is pulled along its lengthy by a force acting at one end

both

none

Solution

Question 6

$$100E$$

$$\dfrac {E}{ 100 } $$

$$\dfrac {E}{ 10 } $$

$$10E$$

Solution

Question 7

5 : 3

27 : 40

6 : 5

27 : 25

Solution

Question 8

$$\propto Y$$

$$\propto Y^{2}$$

$$\propto Y^{-1}$$

$$\propto Y^{-2}$$

Solution

Question 9

$$0.05$$ mm

$$0.5$$ mm

$$0.1$$ mm

$$0.2$$ mm

Solution

Question 10

$$5$$ mm

$$9$$ mm

$$12$$ mm

$$13$$ mm

Solution

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