Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 40

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Weekly Quiz Competition

Question 1
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The bulk modulus of rubber is $$9.8\times 10^{8}N/m^{2}$$. To what depth a rubber ball be taken in a lake so that its volume is decreased by $$0.1$$% ?

A
$$1 km$$

B
$$25 km$$

C
$$100 km$$

D
$$200 km$$

Solution

$$B = \dfrac{P}{\dfrac{\Delta V}{V}}$$
$$P = h \rho g = h \times 10^3 \times 9.8 = 9.8h \times 10^3$$

$$9.8 \times 10^8 =\dfrac{ 9.8h \times 10^3}{\dfrac{0.1}{100}}$$
$$h = 100 \ km$$
Question 2
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When a force is applied on a wire of uniform cross-sectional area $$3\times 10^{-6}m^{2}$$ and length $$4 m$$, the increase in length is $$1\ mm$$. Energy stored in it will be ($$Y=2\times 10^{11}N/m^{2}$$):

A
$$6250\ J$$

B
$$0.177\ J$$

C
$$0.075\ J$$

D
$$0.150\ J$$

Solution

$$U = \dfrac{1}{2} F \Delta l = \dfrac{1}{2} \times \dfrac{Y \Delta lA}{L} \times \Delta l = \dfrac{1}{2} \times \dfrac{2 \times 10^{11} \times 3\times 10^{-6} \times (10^{-3})^2}{4} = 0.075 J$$
Question 3
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If the potential energy of a spring is V on stretching it by 2 cm then its potential energy when it is stretched by 10 cm will be

A
V/25

B
5 V

C
V/5

D
25 V

Solution

Potential Energy $$\propto x^{2}$$
Therefore
$$\dfrac{V_1}{V_2}=\dfrac{(x_1)^{2}}{(x_2)^{2}}$$
Here $$V_1=V$$
$$x_1=4$$
$$x_2=10$$
So $$\dfrac{V}{V_2}=\dfrac{(2)^{2}}{(10)^{2}}$$
$$\dfrac{V}{V_2}=\dfrac{4}{100}$$
$$\dfrac{V}{V_2}=\dfrac{1}{25}$$
$$V_2=25V$$
Hence the correct option is (D).
Question 4
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If the ratio of lengths, radii and Youngs modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be:

A
$$\displaystyle \dfrac{2ac}{b^{2}} $$

B
$$\displaystyle \dfrac{3a}{2b^{2}c} $$

C
$$\displaystyle \dfrac{3c}{2ab^{2}} $$

D
$$\displaystyle \dfrac{2a^{2}c}{b} $$

Solution

Ref image
Given ; $$l_S / l_B = a , \, \dfrac{y_S}{y_B} = c$$
$$\dfrac{r_S}{r_B} = b$$
By definition, young's Modulus = $$Y = \dfrac{linear \, stress}{linear \, strain}$$
$$\Rightarrow Y = \dfrac{T/A}{\Delta l/l}$$
where, T - tension in wire
A - cross sectional area of wire
$$\Delta l$$ - change in length
l - original length
$$\Rightarrow \boxed {\Delta l = \dfrac{T.l}{Y.A}}$$ __(i)
FBD : Ref. image $$T_3 = mg + T_B$$ __(ii) Ref. image $$T_B = 2mg$$ __(iii)
Put (iii) in (ii) : $$T_S = mg + 2mg = 3mg$$
$$\Rightarrow \dfrac{T_S}{T_B} = \dfrac{3mg}{2mg} = \dfrac{3}{2}$$ __(iv)
From (i) :
$$\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{\left(\dfrac{T_S l_S}{Y_S A_S} \right)}{\left(\dfrac{T_B l_B}{Y_B A_B} \right)} \begin{matrix} = \left(\dfrac{T_S}{T_B} \right) . \left(\dfrac{l_S}{l_B} \right) \dfrac{Y_B \, A_B}{Y_S \, A_S} & / for \, a \, wire: A = \pi r^2 \\ = \dfrac{3}{2} \times a \times \dfrac{1}{c} \times \dfrac{\pi r_B^2}{\pi r_S^2} &/ from \, given \, ratios \end{matrix}$$
$$\boxed {\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{3 a}{2 c.b^2}}$$
No correct option given
Question 5
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Copper wire of length $$3m$$ and area of cross-section $$1\:mm^{2}$$, passes through an arrangement of two frictionless pulleys, $$P_{1}$$ and $$P_{2}$$. One end of the wire is rigidly clamped and a mass of $$1 kg$$ is hanged from the other end. If Young's modulus for copper is $$10\times 10^{10}N/m^{2}$$, the elongation in the wire is :

A
$$0.05 mm$$

B
$$0.1 mm$$

C
$$0.2 mm$$

D
$$0.3 mm$$

Solution

$$\Delta l = \dfrac{FL}{AY}$$
$$\Delta l = \dfrac{10 \times 3}{10\times 10^{10} \times 10^{-6}} = 3\times 10^{-4} =0.3mm$$
Question 6
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The diameter of a brass rod is $$4mm$$ and Youngs modulus of brass is $$\displaystyle 9\times 10^{10}N/m^{2}$$ The force required to stretch it by $$0.1\%$$ of its length is

A
$$\displaystyle 360\pi N$$

B
$$36 N$$

C
$$\displaystyle 144\pi \times10^{3} N$$

D
$$\displaystyle 36\pi \times10^{5} N$$

Solution

Let $$F$$ be force
$$Y$$ be Young's modulus
$$\Delta$$ be change in length
$$l$$ be length
$$F=\dfrac{YA\Delta l}{l}$$

Given:
$$\dfrac{\Delta l}{l}=0.001$$
$$Y=9\times 10^{10} N/m^{2}$$
$$r=\dfrac{d}{2}=2mm$$
$$A=\pi r^{2}=\pi\times (2\times 10^{-3})^{2}m^{2}$$
$$F=\dfrac{YA\Delta l}{l}=9\times 10^{10} \times \pi\times 4\times (10)^{-6}\times 0.001=360\pi$$
Hence the force is $$360\pi N$$
Question 7
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The graph showing the extension of is wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. of the cross-sectional area of the wire is $$1{mm}^{2}$$, then the Young's modulus of the material of the wire .( In graph, X-axis 1 unit = 10mm) .

A
$$2\times {10}^{11}N{m}^{-1}$$

B
$$2\times {10}^{10}N{m}^{-2}$$

C
$$\dfrac {1}{2}\times {10}^{11}N{m}^{-2}$$

D
None of these

Solution

Let $$Y$$ denote the Young's Modulus.
Then

$$Y\dfrac{\triangle L}{L_{0}}=\dfrac{F}{A}$$

$$L=1m$$ , $$A=10^{-6}m$$

Hence

$$Y(\dfrac{(5-1)\times 10^{-3}}m{1m}=\dfrac{100-20N}{10^{-6}m^{2}}$$

$$Y(4\times 10^{-3})=80\times 10^{6} N m^{-2}$$

$$Y=20\times 10^{9} N m^{-2}$$

$$=2\times 10^{10} N m^{-2}$$
Question 8
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A metal block is experiencing an atmospheric pressure of $$\displaystyle 1\times 10^{5}N/m^{2}$$ when the same block is placed in a vaccum chamber the fractional change in its volume is (the bulk modulus of metal is $$\displaystyle 1.25\times 10^{11}N/m^{2}$$)

A
$$\displaystyle 4\times 10^{-7}$$

B
$$\displaystyle 2\times 10^{-7}$$

C
$$\displaystyle 8\times 10^{-7}$$

D
$$\displaystyle 1\times 10^{-7}$$

Solution

Bulk modulus$$(K)$$=$$\cfrac{p}{\cfrac{\triangle V}{V}}$$
where,$$p=$$pressure experienced$$=1 \times {10}^{5} N/{m}^{2}$$
$$K=1.25 \times {10}^{11} N/{m}^{2}$$
$$\therefore \cfrac{\triangle V}{V}=\cfrac{p}{K}$$
$$=\cfrac {1 \times {10}^{5}}{1.25 \times {10}^{11}}$$
$$\therefore \cfrac{\triangle V}{V}=8 \times {10}^{-7}$$
Question 9
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Two wires of equal length and cross section area suspended as shown in figure. Their Youngs modulus are $$\displaystyle Y_{1}\: \: and\: \: Y_{2}$$ respectively The equivalent Youngs modulus will be

A
$$\displaystyle Y_{1}+ Y_{2}$$

B
$$\displaystyle \frac{Y_{1}+Y_{2}}{2}$$

C
$$\displaystyle \frac{Y_{1}Y_{2}}{Y_{1}+Y_{2}}$$

D
$$\displaystyle \sqrt{Y_{1}Y_{2}}$$

Solution

$$k=\dfrac{YA}{L}$$
$$k_1=\dfrac{Y_1A}{L}$$
$$k_2=\dfrac{Y_2A}{L}$$
$$k_1$$ and $$k_2 $$ are in parallel.
$$k_{eq}=k_1+k_2$$
$$\dfrac{Y_{eq}(2A)}{L}=(Y_1+Y_2)\dfrac{A}{L}$$
$$Y_{eq}=\dfrac{Y_1+Y_2}{2}$$
Question 10
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A force F is needed to break a copper wire having radius R The force needed to break a copper wire of radius 2R will be
[assume F is applied along the wire and the wire obeys Hooke's law until it breakes]

A
F/2

B
2 F

C
4 F

D
F/4

Solution

$$F=\dfrac{YA\Delta l}{l}$$

Force $$\propto $$ area

$$F\propto \pi r^{2}$$
$$F_1\propto \pi R^{2}$$
$$F_2\propto \pi (2R)^{2}=\pi\times 4R^{2}$$
$$\dfrac{F_1}{F_2}=\dfrac{ \pi \times R^{2}}{\pi\times 4R^{2}}$$
$$\dfrac{F_1}{F_2}=\dfrac{1}{4}$$
$$F_2=4F_1$$
Hence the correct option is (C).