Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 33

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Question 1
1 / -0

Two wires of the same material and length are stretched by the same force. Their masses are in the ratio $$3 : 2$$. Their elongations are in the ratio

A
$$3 : 2$$

B
$$9 : 4$$

C
$$2 : 3$$

D
$$4 : 9$$

Solution

HINT: Using Hooke's law we can find elongation in wire for a given load.

STEP 1: Use elongation formula

$$\Delta l = \dfrac{F\times l}{A\times Y}\longrightarrow(1)$$

where $$F =$$ force
$$A =$$ Area
$$l =$$ actual length
$$Y =$$ Young's modulus

Given same force, same length and same material means the Young's modulus is also same.

Volume = Area $$\times$$ length

Given, $$m_1 : m_2 = 3:2$$

mass = volume $$\times$$ density
= Area $$\times$$ length $$\times$$ density (length & density are same)
= $$A_1 : A_2\;=\;3:2$$

From eq.(1), we set

$$\dfrac{\Delta l_1}{\Delta l_2} = \dfrac{A_2}{A_1} = \dfrac 23$$

Thus, option C is correct.
Question 2
1 / -0

The length of a wire is increased by $$1\ mm$$ on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

A
$$0.25\ mm$$

B
$$0.5\ mm$$

C
$$2\ mm$$

D
$$4\ mm$$

Solution

Young's Modulus of elasticity =stress/strain
$$Y = \dfrac {F/a}{\triangle l/l}$$ or $$Y = \dfrac {Fl}{a\triangle l}$$
or $$\triangle l = \dfrac {Fl}{aY} = \dfrac {Fl}{\pi r^{2} Y}$$
In the given problem, $$\triangle l \propto \dfrac {l}{r^{2}}$$
When both $$l$$ and $$r$$ are doubled, $$\triangle l$$ is halved.
Question 3
1 / -0

A cube is shifted to a depth of $$100m$$ is alake. The change in volume is $$0.1$$%. The bulk modules of the material is nearly

A
$$10Pa$$

B
$${10}^{4}Pa$$

C
$${10}^{7}Pa$$

D
$${10}^{9}Pa$$

Solution

$$B=-V\dfrac{dP}{dV}$$
Now, change in pressure, $$dP=\rho gh=1000 \times 10\times 100={10}^{6}$$
Also, given $$\dfrac{dV}{V}=-0.001$$
Thus, $$B=-\dfrac{{10}^{6}}{-0.001}={10}^{9}$$
Question 4
1 / -0

Two wires of the same length and same material but radii in the ratio of $$1:2$$ are stretched by unequal forces to produce equal elongation. The ratio of the two forces is

A
$$1:1$$

B
$$1:2$$

C
$$1:3$$

D
$$1:4$$

Solution

$$Y = \dfrac {Fl}{a\triangle l}$$
In the given problem, $$Y, l$$ and $$\triangle l$$ are constants.
$$\triangle F\propto a$$
or $$F \propto \pi r^{2}$$ or $$F\propto r^{2}$$ or $$\dfrac {F_{1}}{F_{2}} = \dfrac {r_{1}^{2}}{r_{2}^{2}} = \dfrac {1}{4}$$.
Question 5
1 / -0

When the load on a wire is slowly increased from $$3$$ to $$5 kgwt$$, the elongation increases from $$0.61$$ to $$1.02 mm$$. The work done during the extension of wire is

A
$$0.16J$$

B
$$0.016J$$

C
$$1.6J$$

D
$$16J$$

Solution

Work done = Change in elastic potential energy
Which is given by $$1/2\times Stress\times Strain\times Volume$$
Thus, work $$\displaystyle =\frac{1}{2}\times \frac{3g}{A}\times \frac{0.61\times{10}^{-3}}{L}\times AL - \frac{1}{2}\times \frac{5g}{A}\times \frac{1.02\times{10}^{-3}}{L}\times AL$$

$$\displaystyle =\frac{1}{2}\times (1.83-5.10)\times{10}^{-2}$$

$$=1.635\times{10}^{-2}$$, which is nearly equal to $$0.016J$$
Question 6
1 / -0

Two wires of the same material and length but diameters in the ration $$1 : 2$$ 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

Potential energy per unit volume = $$\dfrac{1}{2}\times Stress\times Strain=\dfrac{1}{2}\times \dfrac{{(Stress)}^{2}}{Y}$$

Since same material, thus Y will be same. Thus it is linearly proportional to $${(Stress)}^{2}$$ or $${(\dfrac{F}{A})}^{2}$$

Thus it is inversely proportional to square of area or radii to power 4. Thus, ratio will be $${(1/2)}^{-4}=16:1$$
Question 7
1 / -0

What amount of work is done in increasing the length of a wire through unity?

A
$$\dfrac {YL}{2A}$$

B
$$\dfrac {YL^{2}}{2A}$$

C
$$\dfrac {YA}{2L}$$

D
$$\dfrac {YL}{A}$$

Solution

Work done=Change in potential energy=$$\dfrac{1}{2}\times Stress\times Strain\times Volume=\dfrac{1}{2}\times Y\times {(Strain)}^{2}\times AL=\dfrac{1}{2}\times Y\times AL\times {\dfrac{1}{L}}^{2}$$ (because Strain=$$\dfrac{\Delta L}{L}=\dfrac{1}{L}$$)
Thus, work=$$\dfrac{YA}{2L}$$
Question 8
1 / -0

When a certain weight is suspended from a long uniform wire, its length increases by $$1\ cm$$. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increase in length will be

A
$$0.5cm$$

B
$$2cm$$

C
$$4cm$$

D
$$8cm$$

Solution

Since force, material and length is same thus extension is dependent only on area, and inversely proportional to it. Thus for the wire with diameter half of first one, area will be one fourth and so extension will be 4 times, or 4cm
Question 9
1 / -0

A thick rope of density $$\rho$$ and length $$L$$ is hung from a rigid support. The increase in length of the rope due to its own weight is ($$Y$$ is the Young's modulus)

A
$$\dfrac {0.1}{4Y} \rho L^{2}g$$

B
$$\dfrac {1}{2Y} \rho L^{2}g$$

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

D
$$\dfrac {\rho Lg}{Y}$$

Solution

Take a small section $$\delta y$$ at distance $$y$$ from top as shown in figure. Let area of rope be A

Stress due to this section = $$ \rho A g \delta y / A = \rho d \delta y$$

Strain on section y = $$\delta l / y$$

Using Stress = Y \times Strain

=> $$\rho g \delta y = Y \times \dfrac{\delta l} {y}$$

=>$$\Delta l = \dfrac {\rho g} {y} \int_{0}^{L} y \delta y$$

=> $$ \Delta l = \dfrac {\rho g L^{2}} {2 Y}$$

Answer. $$\dfrac {\rho g L^{2}} {2 Y}$$
Question 10
1 / -0

If the work done by stretching a wire by $$1mm$$ is $$2J$$, the work necessary for stretching another wire of the same material but with half the radius of cross section and half the length by $$1mm$$ is

A
$$\cfrac{1}{4}J$$

B
$$4J$$

C
$$8J$$

D
$$16J$$

Solution

The stretching force is given by, $$F=\dfrac{YA\Delta l}{l}$$
where $$Y=$$ Young's modulus , $$A=$$ area of cross-section of the wire, $$l=$$ actual length of wire and $$\Delta l=$$ increase in length.
or $$F=\dfrac{Y(\pi r^2)\Delta l}{l}$$
As same material so $$Y$$ becomes constant.
Thus, $$\dfrac{F_1}{F_2}=\dfrac{r_1^2\Delta l_1}{l_1}\times \dfrac{l_2}{r_2^2\Delta l_2}$$
Given, $$r_2=r_1/2$$ , $$l_2=l_1/2$$ and $$\Delta l_1=\Delta l_2=1 mm$$
So, $$F_1/F_2=(4^2/2)=8$$ .....(1)
The work done in stretching wire by amount $$\Delta l$$ is $$W=(1/2)F\Delta l$$
Thus, $$W_1/W_2=F_1/F_2$$
or $$W_2=(F_2/F_1)W_1=2/8=1/4$$ as $$ (W_1=2 J)$$

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Mechanical Properties of Solids Test - 33

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Mechanical Properties of Solids Test - 33

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Question 1

$$3 : 2$$

$$9 : 4$$

$$2 : 3$$

$$4 : 9$$

Solution

Question 2

$$0.25\ mm$$

$$0.5\ mm$$

$$2\ mm$$

$$4\ mm$$

Solution

Question 3

$$10Pa$$

$${10}^{4}Pa$$

$${10}^{7}Pa$$

$${10}^{9}Pa$$

Solution

Question 4

$$1:1$$

$$1:2$$

$$1:3$$

$$1:4$$

Solution

Question 5

$$0.16J$$

$$0.016J$$

$$1.6J$$

$$16J$$

Solution

Question 6

$$16 : 1$$

$$4 : 1$$

$$2 : 1$$

$$1 : 1$$

Solution

Question 7

$$\dfrac {YL}{2A}$$

$$\dfrac {YL^{2}}{2A}$$

$$\dfrac {YA}{2L}$$

$$\dfrac {YL}{A}$$

Solution

Question 8

$$0.5cm$$

$$2cm$$

$$4cm$$

$$8cm$$

Solution

Question 9

$$\dfrac {0.1}{4Y} \rho L^{2}g$$

$$\dfrac {1}{2Y} \rho L^{2}g$$

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

$$\dfrac {\rho Lg}{Y}$$

Solution

Question 10

$$\cfrac{1}{4}J$$

$$4J$$

$$8J$$

$$16J$$

Solution

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