Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 10

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Question 1
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At $$40^oC$$, a brass wire of $$1 mm$$ radius is hung from the ceiling. A small mass, $$M$$ is hung from the free end of the wire. When the wire is cooled down from $$40^oC$$ to $$20^oC$$ it regains its original length of $$0.2 m$$. The value of $$M$$ is close to :
(Coefficient of linear expansion and Young's modulus of brass are $$10^{-5} / ^oC$$ and $$10^{11} \ N/m^2$$, respectively; $$g = 10 ms^{-2}$$)

A
$$1.5\ kg$$

B
$$9 \ kg$$

C
$$0.9 \ kg$$

D
$$0.5 \ kg$$

Solution

Load = Stress $$\times$$ Area
$$Mg = \left(\dfrac{Y\Delta \ell}{\ell}\right) A$$
$$\dfrac{\Delta \ell}{\ell} = \alpha \Delta T$$

Putting this value in above equation

$$Mg = (A Y) \alpha \Delta T = 2\pi$$

It is closest to $$9$$.
Question 2
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Temperature of a gas is $$20^{0}\mathrm{C}$$ and pressure is changed from $$1.01\times 10^{5}$$ Pa to $$1.165\times 10^5\mathrm{P}\mathrm{a}$$. If volume is decreased isothermally by 10%. The bulk modulus of gas is (in $$Pa$$):

A
$$1.55\times 10^{5}$$

B
$$0.155\times 10^{5}$$

C
$$1.4\times 10^{5}$$

D
$$1.01\times 10^{5}$$

Solution

Bulk modulus $$B = \dfrac{Pressure}{Strain}= \dfrac{-\Delta P}{\dfrac{\Delta l \times A}{L \times A}} =-\dfrac{\Delta P}{\dfrac{\Delta V}{V}} = \dfrac{(1.01 - 1.165)\times 10^5}{-0.1}= 1.55 \times 10^5\ Pa$$
Question 3
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$$\mathrm{A}$$ student performs an experiment to determine the Young's modulus of a wire, exactly 2 $$\mathrm{m}$$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $$0.8 mm$$ with an uncertainty of $$\pm 0.05$$ mm at a load of exactly $$1.0 kg$$. The student also measures the diameter of the wire to be $$0.4 mm$$ with an uncertainty of $$\pm 0.01$$ mm. Take $$\mathrm{g}=9.8\mathrm{m}/\mathrm{s}^{2}$$ (exact). The Young's modulus obtained from the reading is

A
$$(2.0\pm 0.3)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

B
$$(2.0\pm 0.2)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

C
$$(2.0\pm 0.1)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

D
$$(2.0\pm 0.05)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

Solution

Young's modulus $$= \dfrac{FL}{Al}$$

$$Y = \dfrac{mg\times 2}{\pi \times (\dfrac{d}{2})^2\times 0.8 \times 10^{-3}} = 1.96 \times 10^{11} N/m^2$$

$$\dfrac{\Delta Y}{Y} = \dfrac{2\Delta d}{d} + \dfrac{\Delta (\Delta l)}{\Delta l}$$

$$= \dfrac{2(0.01)}{0.4} + \dfrac{0.05}{0.8}$$

$$=0.1125$$

$$\Delta Y = .1125Y =0.2205 \times 10^{11} N/m^2$$

$$Y = (2 \pm 0.2 ) \times 10^{11} N/m^2$$
Question 4
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One end of a horizontal thick copper wire of length $$2L$$ and radius $$2R$$ is welded to an end of another horizontal thin copper wire of length $$L$$ and radius $$R$$. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:

A
$$0.25$$

B
$$0.50$$

C
$$2.00$$

D
$$4.00$$

Solution

We have change in length
$$\triangle l=\cfrac { F.l }{ YA } $$
Since the two rods are in series
$${ \left( { F }_{ 1 } \right) }_{ rest }={ \left( { F }_{ 2 } \right) }_{ rest }$$
$$\triangle l\propto \cfrac { l }{ { R }^{ 2 } } $$ $$ \left( \because A=\pi { R }^{ 2 } \right) $$
$$\therefore$$ $$ \cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } }$$ $$ =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \times \cfrac { { { R }_{ 2 } }^{ 2 } }{ { { R }_{ 1 } }^{ 2 } } $$
$$ \cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } }$$ $$=\cfrac { 2L }{ L } \times \cfrac { { { R }_{ 1 } }^{ 2 } }{ 4{ { R }_{ 1 } }^{ 2 } } $$
$$\cfrac { \triangle { l }_{ 1 } }{ \triangle { l }_{ 2 } } =\cfrac { 1 }{ 2 } $$
$$\therefore \cfrac { \triangle { l }_{ 2 } }{ \triangle { l }_{ 1 } } =\cfrac { 2 }{ 1 } =2$$
Question 5
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Dimensions of stress are :

A
$$[ML^2T^{-2}]$$

B
$$[ML^0T^{-2}]$$

C
$$[ML^{-1}T^{-2}]$$

D
$$[MLT^{-2}]$$

Solution

The stress is defined as:
$$stress=\dfrac{force}{Area}$$

Substitute the dimensions:
$$stress=\dfrac{[MLT^{-2}]}{[L^2]}$$

$$stress=[ML^{-1}T^{-2}]$$
Question 6
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The bulk modulus of a spherical objects is '$$B$$'. If it is subjected to uniform pressure '$$P$$', the fractional decrease in radius is:

A
$$\dfrac{P}{3B}$$

B
$$\dfrac{P}{B}$$

C
$$\dfrac{B}{3P}$$

D
$$\dfrac{3P}{B}$$

Solution

Bulk modulus $$B = -V\dfrac{\triangle P}{\triangle V}$$
$$B=\dfrac{-V(P)}{dV} r^3$$
$$B=\dfrac{-r^3P}{3r^2dr}$$
$$\dfrac{dr}{r} = \dfrac{P}{3B}$$
Question 7
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Copper of fixed volume 'V' is drawn into wire of length 'l'. When this wire is subjected to a constant force 'F', the extension produced in the wire is $$'\Delta l'$$. Which of the following graph is a straight line?

A
$$\Delta l \, vs \, \dfrac{1}{l}$$

B
$$\Delta l \, vs \, l^2$$

C
$$\Delta l \, vs \,\dfrac {1}{l^2}$$

D
$$\Delta l \, vs \, l$$

Solution

$$V=Al$$ (standard result)
$$Y=\dfrac {F/A}{\dfrac {\Delta l}{l}}$$ (standard formula)
$$\dfrac {Y\Delta l}{l}=\dfrac {F}{A}$$
$$\Delta l=\dfrac {Fl}{YA}=\dfrac {F}{Y}\cdot \dfrac {ll}{V}$$
$$\Delta l=\dfrac {F}{YV}l^2$$
$$\Delta l\propto l^2$$
Question 8
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The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

A
$$Length = 50\ cm, diameter = 0.5\ mm$$

B
$$Length = 100\ cm, diameter = 1\ mm$$

C
$$Length = 200\ cm, diameter = 2\ mm$$

D
$$Length = 300\ cm, diameter = 3\ mm$$

Solution

Using Hook'e law $$Y = \dfrac{F L}{A \Delta L}$$ where $$A = \dfrac{\pi D^2}{4}$$
$$\implies$$ $$\Delta L = K \dfrac{L}{D^2}$$ where $$K = constant$$
For wire A : $$\dfrac{L}{D^2} = \dfrac{50 }{(0.05)^2} =20000$$

For wire B : $$\dfrac{L}{D^2} = \dfrac{100 }{(0.1)^2} =10000$$

For wire C : $$\dfrac{L}{D^2} = \dfrac{200 }{(0.2)^2} =5000$$

For wire D : $$\dfrac{L}{D^2} = \dfrac{300 }{(0.3)^2} =3333.33$$
Thus wire A has the largest extension.
Question 9
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The Young's modulus of steel is twice that of brass. Two wires of sample length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:

A
$$1:1$$

B
$$1:3$$

C
$$2:1$$

D
$$4:1$$

Solution

$$\textbf{Step 1 - Find Expression of elongation in steel Refer Figure 1}$$
Let Young's modulus of steel $$= Y_{S}$$
and let length of steel and brass $$= l$$
Let Area of cross-section of steel and brass $$= A$$
Assume $$W_{1}$$ weight added to the lower end of steel wire
As we know,
$$Stress = Y\times Strain$$ $$\textbf{[here Y is Young's modulus]}$$
$$Stress = \dfrac {force}{area}$$
$$Strain = \dfrac {\text {Change in length}}{\text {Natural length}}$$
$$\dfrac {W_{1}}{A} = Y_{S} \dfrac {\triangle L}{l}$$ $$....(1)$$ $$\text{[Assume}$$ $$\triangle l$$ $$\text {is elongation in steel wire]}$$

$$\textbf{Step 2 - Expression of elongation in brass Refer Figure 2}$$
Let Young's modulus of brass $$= Y_{b}$$
Assume $$W_{2}$$ weight added to the lower end of brass wire.
Apply
$$Stress = Y_{b}\times (Strain)$$
Lower end of both wire should be at same level,
so elongation in brass $$= \triangle l$$
$$\dfrac {W_{2}}{A} = Y_{b} \dfrac {\triangle l}{l}$$ $$....(2)$$

$$\textbf{Step 3 - Ratio of weights}$$
Divide equation $$(1)$$ by equation $$(2)$$
$$\dfrac {W_{1}}{W_{2}} = \dfrac {Y_{S}}{Y_{b}}$$
$$\dfrac {W_{1}}{W_{2}} = \dfrac {2}{1}$$ $$[\because given\ Y_{S} = 2Y_{b}]$$
$$\dfrac {W_{1}}{W_{2}} = 2 : 1$$

Hence option (C) correct.
Question 10
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The length of a metal wire is $$l_1$$ when the tension in it is $$T_1$$ and is $$I_2$$ when the tension is $$T_2$$. The natural length of the wire is?

A
$$\dfrac{l_1+l_2}{2}$$

B
$$\sqrt{l_1/I_2}$$

C
$$\dfrac{I_1T_2-I_2T_1}{T_2-T_1}$$

D
$$\dfrac{I_1T_2+l_2T_1}{T_1+T_2}$$

Solution

Let $$ l$$ = original length of material – wire.
$$A$$ = original length of metal – wire.
Change in length in first case = $$l_1-l$$
change in length in second case = $$l_2-l$$

Young Modulus= $$\dfrac{\text{Normal Stress}}{\text{Longitudinal strain}}$$

$$Y= \dfrac{\dfrac T A}{\dfrac{\Delta l}l}$$

here $$Y=$$ Young Modulus
$$T=$$ Tension
$$A= $$ Area
$$\Delta l $$= Change in Length
$$l$$ Original length

In First Case
$$Y=\dfrac{T_1}{A}*\dfrac{l}{l_1-l}$$

In Second case :
$$Y=\dfrac{T_2}{A}*\dfrac{l}{l_2-l}$$

Since Young Modulus remains the same
So,
$$\dfrac{T_1}{A}*\dfrac{l}{l_1-l}=\dfrac{T_2}{A}*\dfrac{l}{l_2-l} $$

On solving we get

$$l=\dfrac{I_1T_2-I_2T_1}{T_2-T_1}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 10

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Mechanical Properties of Solids Test - 10

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SHARING IS CARING

Question 1

$$1.5\ kg$$

$$9 \ kg$$

$$0.9 \ kg$$

$$0.5 \ kg$$

Solution

Question 2

$$1.55\times 10^{5}$$

$$0.155\times 10^{5}$$

$$1.4\times 10^{5}$$

$$1.01\times 10^{5}$$

Solution

Question 3

$$(2.0\pm 0.3)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

$$(2.0\pm 0.2)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

$$(2.0\pm 0.1)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

$$(2.0\pm 0.05)\times 10^{11}\mathrm{N}/\mathrm{m}^{2}$$

Solution

Question 4

$$0.25$$

$$0.50$$

$$2.00$$

$$4.00$$

Solution

Question 5

$$[ML^2T^{-2}]$$

$$[ML^0T^{-2}]$$

$$[ML^{-1}T^{-2}]$$

$$[MLT^{-2}]$$

Solution

Question 6

$$\dfrac{P}{3B}$$

$$\dfrac{P}{B}$$

$$\dfrac{B}{3P}$$

$$\dfrac{3P}{B}$$

Solution

Question 7

$$\Delta l \, vs \, \dfrac{1}{l}$$

$$\Delta l \, vs \, l^2$$

$$\Delta l \, vs \,\dfrac {1}{l^2}$$

$$\Delta l \, vs \, l$$

Solution

Question 8

$$Length = 50\ cm, diameter = 0.5\ mm$$

$$Length = 100\ cm, diameter = 1\ mm$$

$$Length = 200\ cm, diameter = 2\ mm$$

$$Length = 300\ cm, diameter = 3\ mm$$

Solution

Question 9

$$1:1$$

$$1:3$$

$$2:1$$

$$4:1$$

Solution

Question 10

$$\dfrac{l_1+l_2}{2}$$

$$\sqrt{l_1/I_2}$$

$$\dfrac{I_1T_2-I_2T_1}{T_2-T_1}$$

$$\dfrac{I_1T_2+l_2T_1}{T_1+T_2}$$

Solution

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