Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 18

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Question 1
1 / -0

A wire extends by 'l' on the application of load 'mg'. Then, the energy stored in it is :

A
$$mgl$$

B
$$\dfrac{mgl}{2}$$

C
$$\dfrac{mg}{l}$$

D
$$mgl^{2}$$

Solution

Energy stored per unit volume is:
$$\dfrac{U}{V} = \dfrac{1}{2} \times \text{Stress} \times \text{Strain}$$

Total energy stored is:
$$U= \cfrac{1}{2}\times (\dfrac{F}{A})\times (\dfrac{\Delta L} {L}) \times (AL)$$
$$=\cfrac{F}{2}\Delta L$$
$$=\dfrac{mg l}{2}$$
Question 2
1 / -0

The graph shows the behaviour of a steel wire in the region for which the wire obeys Hooke's law.The graph is a part of a parabola. The variables x and y might represent.

A
x $$=$$ stress ; y $$=$$ strain

B
x $$=$$ strain ; y $$=$$ stress

C
x $$=$$ strain ; y $$=$$ elastic energy

D
x $$=$$ elastic energy ; y $$=$$ strain

Solution

Elastic energy $$= \dfrac{1}{2} \times stress \times strain$$
whereas strain deals with change in length / original length
hence, "x' axis depends on depended variable but "y" axis depends on independant variable. So, option "D" is correct.
Question 3
1 / -0

Assertion (A) : The elastic potential energy of a spring increases when it is elongated and decreases when it is compressed.
Reason (R) : Work done on spring is stored in it as elastic potential energy.

A
Both assertion and reason are true and the reason is correct explanation of the assertion

B
Both assertion and reason are true, but reason is not correct explanation of the assertion

C
Assertion is true, but the reason is false

D
Assertion is false, but the reason is true

Solution

Elastic potential energy increases whether it is compressed or elongated. Because in both cases work done by external force is opposed by spring.
Work is done against spring. Therefore, it's PE is increased.
Question 4
1 / -0

The stress required to double the length of a wire of Young's modulus $$E$$ is :

A
$$2E$$

B
$$E$$

C
$$E/2$$

D
$$3E$$

Solution

Young's modulus, $$E$$ $$ = \dfrac{stress}{strain}$$

Where, $$ strain = \dfrac{change \ in \ length}{original\ length}$$
Let the initial length of the wire be $$x $$ m.
Thus change in length =$$ 2x - x = x$$
Original length =$$ x$$
So, strain = 1
So, we get $$E$$= stress/1
Question 5
1 / -0

A metallic rod of length 'L' and cross-section 'A' has Young's modulus 'Y' and coefficient of linear expansion $$\alpha$$. If the rod is heated to a temperature 'T', then the energy stored per unit volume is:

A
$$\dfrac{1}{2}Y\alpha ^{2}T^{2}$$

B
$$\dfrac{1}{2}YA\alpha ^{2}T^{2}$$

C
$$\dfrac{1}{2}YA\alpha T$$

D
$$\dfrac{1}{2}YA^{2}\alpha ^{2}T^{2}$$

Solution

We know that the energy stored per unit volume $$=\dfrac{1}{2} (stress)(strain)$$ = $$ \dfrac{1}{2} (Y)(strain)^{2}$$
Now strain = fractional change in length = $$ \alpha T$$ (using thermal expansion formula)
So energy stored per unit volume = $$\dfrac{1}{2}Y \alpha ^{2} T^{2}$$
Question 6
1 / -0

A force of $$30N$$ acts on a rod of area of cross section 5 x 10$$^{-6}$$ $$m^{2}$$. The stress produced in dyne/cm$$^{2}$$ is :

A
$$6 \times 10$$$$^{6}$$

B
$$6 \times 10$$$$^{7}$$

C
$$6 \times 10$$$$^{5}$$

D
$$0.16 \times 10$$$$^{-6}$$

Solution

$$1N=10^{5} dyne$$

$$stress=\dfrac{30\times 10^{5} dyne}{5\times 10^{-6}\times 10^{4} cm^{2}}$$

$$=6\times 10^{7} dyne/cm^{2}$$
Question 7
1 / -0

Three wires A,B, C made of the same material and radius have different lengths. The graphs in the figure shows the elongation-load variation. The longest wire is:

A
A

B
B

C
C

D
All

Solution

The slope C > slope B > slope A.
$$\Delta L=\dfrac{FL}{AY}$$

Slope$$=\dfrac{\Delta L}{F}=\dfrac{L}{AY}$$

$$\therefore \dfrac{\Delta L}{F} \propto L$$ and , thus, $$C$$ is the longest wire.
Question 8
1 / -0

A uniform heavy rod of length $$L$$ and area of cross-section area $$A$$ is hanging from a fixed support. If Young's modulus of the material of the rod is $$Y$$, then the increase in the length of the rod is ($$\rho$$ is a density of the material of the rod) :

A
$$\dfrac{L^{2}Y}{2\rho g}$$

B
$$\dfrac{L^{2}\rho g}{2Y}$$

C
$$\dfrac{L^{2} g}{2Y\rho}$$

D
$$\dfrac{L^{2} g}{3Y\rho}$$

Solution

Consider an element dx at a distance x from the top.
Let the wt. of rod is W.
The force acting on this because of a part which is below dx is given as $$\dfrac{(L-x)W}{L}$$

The elongation in element $$dx$$ is $$=(\dfrac{L-x}{L})\dfrac{Wdx}{AY}$$$$=\dfrac{(L-x)Wdx}{LAY}$$

Total elongation, $$\Delta L=\int_{O}^{L}\dfrac{(L-x)Wdx}{LAY}$$$$=\dfrac{WL}{2AY}$$

$$W=AL\rho g$$

$$\Delta L=\dfrac{AL\rho gL}{2AY}$$$$=\dfrac{L^{2}\rho g}{2Y}$$
Question 9
1 / -0

A $$20 kg$$ load is suspended by a wire of cross section $$0.4 mm$$$$^{2}$$. The stress produced in N/m$$^{2}$$ is :

A
4.9 x 10$$^{-6}$$

B
4.9 x 10$$^{8}$$

C
49 x 10$$^{8}$$

D
2.45 x 10$$^{-6}$$

Solution

$$stress=\dfrac{Force\, applied }{Area} = \dfrac{20\times 9.81}{0.4\times 10^{-6}}$$

$$=4.9\times 10^{8} N/m^{2}$$
Question 10
1 / -0

When an elastic material with Young's modulus $$'Y'$$ is subjected to a stretching stress $$'S'$$, then the elastic energy stored per unit volume is:

A
$$\dfrac{S^{2}}{2Y}$$

B
$$\dfrac{YS^{2}}{2}$$

C
$$\dfrac{S}{2Y}$$

D
$$\dfrac{YS}{2}$$

Solution

Elastic energy stored per unit volume $$=\dfrac{1}{2}\times stress\times strain$$
$$=\dfrac{1}{2}S\times (\dfrac{S}{Y})$$
$$=\dfrac{S^{2}}{2Y}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 18

Result

Mechanical Properties of Solids Test - 18

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SHARING IS CARING

Question 1

$$mgl$$

$$\dfrac{mgl}{2}$$

$$\dfrac{mg}{l}$$

$$mgl^{2}$$

Solution

Question 2

x $$=$$ stress ; y $$=$$ strain

x $$=$$ strain ; y $$=$$ stress

x $$=$$ strain ; y $$=$$ elastic energy

x $$=$$ elastic energy ; y $$=$$ strain

Solution

Question 3

Both assertion and reason are true and the reason is correct explanation of the assertion

Both assertion and reason are true, but reason is not correct explanation of the assertion

Assertion is true, but the reason is false

Assertion is false, but the reason is true

Solution

Question 4

$$2E$$

$$E$$

$$E/2$$

$$3E$$

Solution

Question 5

$$\dfrac{1}{2}Y\alpha ^{2}T^{2}$$

$$\dfrac{1}{2}YA\alpha ^{2}T^{2}$$

$$\dfrac{1}{2}YA\alpha T$$

$$\dfrac{1}{2}YA^{2}\alpha ^{2}T^{2}$$

Solution

Question 6

$$6 \times 10$$$$^{6}$$

$$6 \times 10$$$$^{7}$$

$$6 \times 10$$$$^{5}$$

$$0.16 \times 10$$$$^{-6}$$

Solution

Question 7

A

B

C

All

Solution

Question 8

$$\dfrac{L^{2}Y}{2\rho g}$$

$$\dfrac{L^{2}\rho g}{2Y}$$

$$\dfrac{L^{2} g}{2Y\rho}$$

$$\dfrac{L^{2} g}{3Y\rho}$$

Solution

Question 9

4.9 x 10$$^{-6}$$

4.9 x 10$$^{8}$$

49 x 10$$^{8}$$

2.45 x 10$$^{-6}$$

Solution

Question 10

$$\dfrac{S^{2}}{2Y}$$

$$\dfrac{YS^{2}}{2}$$

$$\dfrac{S}{2Y}$$

$$\dfrac{YS}{2}$$

Solution

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