Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 74

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Weekly Quiz Competition

Question 1
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An Aluminium and Copper wire of same cross sectional area but having lengths in the ratio $$2 : 3$$ are joined end to end. This composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is $$2.1 \ mm$$, the increase in lengths of Aluminium and Copper wires are : [$$\displaystyle { Y }_{ Al }=20\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$$ and $$\displaystyle { Y }_{ Cu }=12\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$$]

A
$$0.7 \ mm; 1.4\ mm$$

B
$$0.9\ mm; 1.2\ mm$$

C
$$1.0\ mm; 1.1\ mm$$

D
$$0.6 \ mm; 1.5\ mm$$

Solution

Since the tension and cross section area of both the wires are same, stress in both the wires is same.

Since, $$\dfrac { T }{ A } =Y\times \dfrac { \Delta l }{ l } $$

Hence, $$\Delta l=k\dfrac { l }{ Y } $$

$$\dfrac { \Delta { l }_{ 1 } }{ \Delta { l }_{ 2 } } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } }\times \dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } } =\dfrac { 2 }{ 3 }\times \dfrac { 12 }{ 20 } =\dfrac { 2 }{ 5 } $$

Hence, $$2x+5x=2.1\ mm$$ (Given)

$$x=0.3\ mm$$

Hence the lengths are $$2x=0.6\ mm$$ and $$5x=1.5\ mm$$
Question 2
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A copper wire and a steel wire of the same length and same cross section are joined end to end to form a composite wire. The composite wire is hung from a rigid support and a load is suspended from the other end. If the increase in length of the composite wire is $$2.4\ mm$$, then the increase in lengths of steel and copper wires are:
$$(Y_{cu} = \, 10 \times \, 10^{10} \, N/m^2, \, Y_{steel} = \, 2 \times \, 10^{11} \, N /m^2)$$

A
$$1.2\ mm$$, $$1.2\ mm$$

B
$$0.6\ mm$$, $$1.8\ mm$$

C
$$0.8\ mm$$, $$1.6\ mm$$

D
$$0.4\ mm$$, $$2.0\ mm$$

Solution

Given : $$L_{cu} = L_{steel} = L$$ $$A_{cu} = A_{steel} = A$$
Let the increase in the lengths of individual wires be $$\Delta l_{cu}$$ and $$\Delta l_{steel}$$ .
Total increase in the length of composite wire $$\Delta l_{total} = 2.4\ mm$$
$$\therefore$$ $$\Delta l_{cu} + \Delta l_{steel} = 2.4$$ ...........(1)

Also $$Y =\dfrac{F L}{A \Delta l}$$ $$\implies \Delta l = \dfrac{FL}{AY}$$

$$\therefore$$ $$\Delta l_{cu} = \dfrac{F L_{cu}}{A_{cu} Y_{cu}} = \dfrac{FL}{A Y_{cu}}$$

Similarly $$\Delta l_{steel} = \dfrac{F L_{steel}}{A_{steel} Y_{steel}} = \dfrac{FL}{A Y_{steel}}$$

Dividing these two equations we get $$\dfrac{\Delta l_{cu}}{\Delta l_{steel}} = \dfrac{Y_{steel}}{Y_{cu}}$$
$$\therefore$$ $$\dfrac{\Delta l_{cu}}{\Delta l_{steel}} = \dfrac{2\times 10^{11}}{10 \times 10^{10} } = 2$$ $$\implies $$ $$\Delta l_{cu} = 2\Delta l_{steel}$$ ............(2)

From (1) and (2), $$ 2\Delta l_{steel} + \Delta l_{steel} = 2.4$$ $$\implies$$ $$\Delta l_{steel} =0.8\ mm$$
$$\therefore$$ $$\Delta l_{cu } = 2 \times 0.8 =1.6\ mm$$
Question 3
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A solid sphere of radius 20cm is subjected to a uniform pressure of $$ 10^{6}$$ $$N m^{-2}$$. If the bulk modulus is $$1.7 \times 10^{11}$$ $$N m^{-2}$$, the decrease in the volume of the solid is approximately equal to:

A
0.2 $$cm^3$$

B
0.3 $$cm^3$$

C
0.4 $$cm^3$$

D
0.5 $$cm^3$$

Solution

The volume of the sphere is $$\dfrac{4}{3}\pi r^{3}=\dfrac{4 \times 22 \times 0.008}{3 \times 7}=0.0335$$

The pressure applied is $$10^{6}\ Nm^{-2}$$

The Bulk modulus is defined as $$B=\dfrac{PV}{X}$$ , where $$X$$ is change in volume.

$$\Rightarrow X=\dfrac{PV}{B}=\dfrac{10^{6} \times 0.0335}{1.7 \times 10^{11}}=0.2 \ cm^{3}$$

Therefore option $$A$$ is correct.
Question 4
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When a metallic wire is stretched with a tension $$\displaystyle { T }_{ 1 }$$ its length is $$\displaystyle { l }_{ 1 }$$ and with a tension $$\displaystyle { T }_{ 2 }$$ its length is $$\displaystyle { l }_{ 2 }$$. The original length of the wire is:

A
$$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }-{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$

B
$$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }+{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } } $$

C
$$\displaystyle \sqrt { { l }_{ 1 }{ l }_{ 2 } } $$

D
$$\displaystyle \frac { { l }_{ 1 }{ l }_{ 2 } }{ 2 } $$

Solution

From the relation between stress and strain, we get,

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

Thus $$\dfrac { { T }_{ 1 } }{ A } =Y\dfrac { { l }_{ 1 }-l }{ l } $$ and $$\dfrac { { T }_{ 2 } }{ A } =Y\dfrac { { l }_{ 2 }-l }{ l } $$

Dividing the two equations, we get,

$$\dfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\dfrac { l-{ l }_{ 1 } }{ l-{ l }_{ 2 } } $$

$$l=\dfrac { { l }_{ 1 }{ T }_{ 2 }-{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$
Question 5
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A thick uniform rubber rope of density $$1.5\ g\ cm^{-3}$$ and Young's modulus $$5 \times 10^{6}$$ $$N m^{-2}$$ has a length of $$8 m$$. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

A
$$9.6 \times 10^{-2}$$

B
$$19.2 \times 10^{-2}$$

C
$$9.6 \times 10^{-3}$$

D
$$9.6 $$

Solution

The youngs modulus $$Y$$ is defined as $$Y=\frac{FL}{Ax}$$ , where $$F$$ is force apllied , $$L$$ is original length $$x$$ is change in length and $$\sigma=$$density of the wire.
$$\Rightarrow x=\frac{\sigma L^{2}g}{AY} (\therefore F=mg=\sigma \times A\times L \times g) $$
$$\Rightarrow x=\frac{\sigma L^{2}g}{Y}=\frac{1.5 \times 10^{-3} \times 64 \times 10}{5 \times 10^{6}}=19.2 \times 10^{-2}$$
Therefore option $$B$$ is correct
Question 6
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When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is

A
$$\cfrac{YS}{2}$$

B
$$\cfrac{S^2Y}{2}$$

C
$$\cfrac{S^2}{2Y}$$

D
$$\cfrac{S}{2Y}$$

Solution

Given,
Young's modulus $$=Y,$$
Streaching stress$$=S,$$
Energy stored per unit volume is, $$=\cfrac{1}{2} \times $$ stress $$\times $$ strian
We know that, stress $$=$$ strain $$\times Y$$
Therefore energy per unit volume will be $$E = \cfrac{1}{2} \times S \times \cfrac{S}{Y}=\cfrac{S^{2}}{2Y}$$
Therefore option $$C$$ is correct.
Question 7
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The length of a metal wire is $$l_{1}$$ when the tension in it is $$T_{1}$$ and is $$l_{2}$$ when the tension is $$T_{2}$$. The natural length of wire is

A
$$\dfrac {l_{1} + l_{2}}{2}$$

B
$$\sqrt {l_{1}l_{2}}$$

C
$$\dfrac {l_{1}T_{2} - l_{2}T_{1}}{T_{2} - T_{1}}$$

D
$$\dfrac {l_{1}T_{2} + l_{2}T_{1}}{T_{2} + T_{1}}$$

Solution

Let the natural length of wire be $$l_o$$.
Using Hooke's law, $$Y = \dfrac{Tl_o}{A\Delta l}$$
where $$\Delta l = l-l_o$$
We get $$l-l_o = \dfrac{Tl_o}{AY}$$
Case 1 : Tension is $$T_1$$ and length of wire $$l=l_1$$
$$\therefore$$ $$l_1-l_o = \dfrac{T_1l_o}{AY}$$ .....(1)
Case 2 : Tension is $$T_2$$ and length of wire $$l=l_2$$
$$\therefore$$ $$l_2-l_o = \dfrac{T_2l_o}{AY}$$ .....(2)
Dividing both equations $$\dfrac{l_1 - l_o}{l_2 - l_o} = \dfrac{T_1 }{T_2}$$
$$\implies$$ $$l_o = \dfrac{l_1T_2 - l_2T_1}{T_2-T_1}$$
Question 8
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The rubber cord of a catapult has cross-section area $$2m{m}^{2}$$ and a total unstretched length $$15cm$$. It is stretched to $$18cm$$ and then released to project a particle of mass $$3g$$. Calculate the velocity of projection if $$Y$$ for rubber is $$8\times { 10 }^{ 8 }N/{ m }^{ 2 }$$.

A
$$56.5{ms}^{-1}$$

B
$$10{ms}^{-1}$$

C
$$20{ms}^{-1}$$

D
$$40{ms}^{-1}$$

Solution

Given,  $$Y=8\times { 10 }^{ 8 }N/{ m }^{ 2 }$$
  $$A=2{ mm }^{ 2 }=\dfrac { 2 }{ { \left( 1000 \right) }^{ 2 } } { m }^{ 2 }$$
$$L=15cm=\dfrac { 15 }{ 100 } m$$
$$l=\left( 18-15 \right) =3cm=\dfrac { 3 }{ 100 } m$$
Let, velocity will be V m/s.
$$\therefore $$ Stored potential energy = Kinetic energy
or,  $$\dfrac { YA{ l }^{ 2 } }{ 2L } =\dfrac { 1 }{ 2 } m{ v }^{ 2 }$$
or,  $${ v }^{ 2 }=\dfrac { YA{ l }^{ 2 } }{ mL } $$
or,  $${ v }^{ 2 }=\dfrac { \left( 8\times { 10 }^{ 8 } \right) \times 2\times { 3 }^{ 2 }\times 100\times 1000 }{ { \left( 1000 \right) }^{ 2 }\times { \left( 100 \right) }^{ 2 }\times \left( 3\times 10 \right) \times 15 } $$
$$\therefore \quad v=56.5m/s$$
Question 9
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The Young's modulus of a material is $$2\times { 10 }^{ 11 }N/{ m }^{ 2 }$$ and its elastic limit is $$1.8\times { 10 }^{ 8 }N/{ m }^{ 2 }$$. For a wire of $$1m$$ length of this material, the maximum elongation achievable is

A
$$0.2mm$$

B
$$0.3mm$$

C
$$0.4mm$$

D
$$0.5mm$$

Solution

From Hooke's law $$ strain=\cfrac { stress }{ Y } $$
where $$strain = \dfrac{e}{l}$$
Given : $$stress = 1.8\times 10^8 \ N/m^2$$ $$Y = 2\times 10^{11} N/m^2$$ $$l=1 \ m$$
$$\therefore$$ $$e=\cfrac { stress }{ Y } l=\cfrac { { 10 }^{ 8 } }{ 2\times { 10 }^{ 11 } } \times 1=5\times { 10 }^{ -4 }m=0.5mm$$
Question 10
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A rigid bar of mass $$15 kg$$ is supported symmetrically by three wires each $$2 m$$ long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the tension? (Given E for copper = $$110\times 10^{9} N/m^{2}$$ and E for iron = $$190\times 10^{9} N/m^{2}$$).

A
$$12.6 : 2$$

B
$$1.31 : 1$$

C
$$4.65 : 3$$

D
$$2.69 : 4$$

Solution

Let $${ A }_{ 1 }$$ and $${ A }_{ 2 }$$ be the areas of cross-section of copper and iron wires respectively. If $${ d }_{ 1 }$$ and $${ d }_{ 2 }$$ be their respective diameters, then
$${ A }_{ 1 }=\dfrac { \pi { d }_{ 1 }^{ 2 } }{ 4 }$$
and $$ { A }_{ 2 }=\dfrac { \pi { d }_{ 2 }^{ 2 } }{ 4 }$$
$$ \therefore \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { { d }_{ 1 }^{ 2 } }{ { d }_{ 2 }^{ 2 } } ={ \left( \dfrac { { d }_{ 1 } }{ { d }_{ 2 } } \right) }^{ 2 }$$
$$L=2m$$
Let $$\Delta l$$ be the extension produced in each wire.
Let $$F=$$ tension produced in each wire.
$$\therefore $$ From the relation, $$Y=\dfrac { stress }{ strain }$$, we get
Strain for copper wire $$ =\dfrac { { F }/{ { A }_{ 1 } } }{ { Y }_{ 1 } }$$ and
Strain for iron wire $$ =\dfrac { { F }/{ { A }_{ 2 } } }{ { Y }_{ 2 } }$$
As the bar is supported symmetrically, the two strains are equal
$$ \dfrac { F }{ { A }_{ 1 }{ Y }_{ 1 } } =\dfrac { F }{ { A }_{ 2 }{ Y }_{ 2 } }$$
$$ \Rightarrow { A }_{ 1 }{ Y }_{ 1 }={ A }_{ 2 }{ Y }_{ 2 }\Rightarrow \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } } $$
$$\Rightarrow \dfrac { { \pi { d }_{ 1 }^{ 2 } }/{ 4 } }{ { \pi { d }_{ 2 }^{ 2 } }/{ 4 } } =\dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } }$$
$$ \Rightarrow \dfrac { { d }_{ 1 } }{ { d }_{ 2 } } =\sqrt { \dfrac { { Y }_{ 2 } }{ { Y }_{ 1 } } } =\sqrt { \dfrac { 190\times { 10 }^{ 9 } }{ 110\times { 10 }^{ 9 } } }$$
$$ \dfrac { { d }_{ 1 } }{ { d }_{ 2 } } =1.31$$
$$\Rightarrow { d }_{ 1 }:{ d }_{ 2 }=1.31:1$$

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Mechanical Properties of Solids Test - 74

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Mechanical Properties of Solids Test - 74

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SHARING IS CARING

Question 1

$$0.7 \ mm; 1.4\ mm$$

$$0.9\ mm; 1.2\ mm$$

$$1.0\ mm; 1.1\ mm$$

$$0.6 \ mm; 1.5\ mm$$

Solution

Question 2

$$1.2\ mm$$, $$1.2\ mm$$

$$0.6\ mm$$, $$1.8\ mm$$

$$0.8\ mm$$, $$1.6\ mm$$

$$0.4\ mm$$, $$2.0\ mm$$

Solution

Question 3

0.2 $$cm^3$$

0.3 $$cm^3$$

0.4 $$cm^3$$

0.5 $$cm^3$$

Solution

Question 4

$$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }-{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$

$$\displaystyle \frac { { l }_{ 1 }{ T }_{ 2 }+{ l }_{ 2 }{ T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } } $$

$$\displaystyle \sqrt { { l }_{ 1 }{ l }_{ 2 } } $$

$$\displaystyle \frac { { l }_{ 1 }{ l }_{ 2 } }{ 2 } $$

Solution

Question 5

$$9.6 \times 10^{-2}$$

$$19.2 \times 10^{-2}$$

$$9.6 \times 10^{-3}$$

$$9.6 $$

Solution

Question 6

$$\cfrac{YS}{2}$$

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

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

$$\cfrac{S}{2Y}$$

Solution

Question 7

$$\dfrac {l_{1} + l_{2}}{2}$$

$$\sqrt {l_{1}l_{2}}$$

$$\dfrac {l_{1}T_{2} - l_{2}T_{1}}{T_{2} - T_{1}}$$

$$\dfrac {l_{1}T_{2} + l_{2}T_{1}}{T_{2} + T_{1}}$$

Solution

Question 8

$$56.5{ms}^{-1}$$

$$10{ms}^{-1}$$

$$20{ms}^{-1}$$

$$40{ms}^{-1}$$

Solution

Question 9

$$0.2mm$$

$$0.3mm$$

$$0.4mm$$

$$0.5mm$$

Solution

Question 10

$$12.6 : 2$$

$$1.31 : 1$$

$$4.65 : 3$$

$$2.69 : 4$$

Solution

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