Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 75

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Question 1
1 / -0

Two wires having same length and material are stretched by same force. Their diameters are in the ratio 1 : 3. The ratio of strain energy per unit volume for these two wires (smaller to larger diameter) when stretched is

A
3 : 1

B
9 : 1

C
27 : 1

D
81 : 1

Solution

Strain energy per unit volume $$U = \dfrac{1}{2}\times Stress \times Strain$$
From Hooke's law, $$Stress = Y \times Strain$$
$$\implies$$ $$U = \dfrac{1}{2Y} \times (Stress)^2$$
where $$Stress = \dfrac{F}{A} = \dfrac{F}{\pi d^2/4}$$
$$\implies$$ $$U \propto \dfrac{1}{d^4}$$
Given : $$\dfrac{d_s}{d_l} = \dfrac{1}{3}$$
Thus ratio of strain energy per unit volume $$\dfrac{U_s}{U_l} = \bigg(\dfrac{d_l}{d_s}\bigg)^4$$
$$\implies$$ $$\dfrac{U_s}{U_l} = \bigg(\dfrac{3}{1}\bigg)^4 = 81$$
Question 2
1 / -0

A steel wire of length $$4.5m$$ and cross-sectional area $$3\times { 10 }^{ -5 }{ m }^{ 2 }$$ stretches by the same amount as a copper wire of length $$3.5m$$ and cross-sectional area of $$4\times { 10 }^{ -5 }{ m }^{ 2 }$$ under a given load. The ratio of the Young's modulus of steel to that of copper is:

A
$$1.3$$

B
$$1.5$$

C
$$1.7$$

D
$$1.9$$

Solution

For copper wire, $${ L }_{ C }=3.5m,{ A }_{ C }=4\times { 10 }^{ -5 }{ m }^{ 2 }\quad $$
for steel wire, $${ L }_{ S }=4.5m,{ A }_{ S }=3\times { 10 }^{ -5 }{ m }^{ 2 }$$

As Young's modulus, $$\quad Y=\cfrac { (F/A) }{ (\Delta L/L) } $$

As applied force $$F$$ and extension $$\Delta L$$ are same for steel and copper wire
$$\therefore \cfrac { F }{ \Delta L } =\cfrac { { Y }_{ S }{ A }_{ S } }{ { L }_{ S } } =\cfrac { { Y }_{ C }{ A }_{ C } }{ { L }_{ C } } $$

where the subscripts $$C$$ and $$S$$ refers to copper and steel respectively
$$\therefore \cfrac { { Y }_{ S } }{ { Y }_{ C } } =\cfrac { { L }_{ S } }{ { L }_{ C } } \times \cfrac { { A }_{ C } }{ { A }_{ S } } =\cfrac { \left( 4.5m \right) \left( 4\times { 10 }^{ -5 }{ m }^{ 2 } \right) }{ \left( 3.5m \right) \left( 3\times { 10 }^{ -5 }{ m }^{ 2 } \right) } =1.7$$
Question 3
1 / -0

A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is?

A
$$\dfrac{2}{3}P$$

B
P

C
$$\dfrac{3}{2}P$$

D
$$2P$$

Solution

$$\begin{array}{l}\text { Given, an ideal gas at a pressure } \\\text { and atsolite temparature T. } \\\text { Volume Stress }=-\beta \text { ( Volume strain) }\end{array}$$\

$$\begin{array}{l}\frac{F}{A}=-\beta\left(\frac{\Delta V}{V}\right) \\\therefore \quad \beta=-\frac{\Delta P}{\Delta V / V} \\\text { Now, in isethermal conditions } \\P V=n R T=\text { constant }\end{array}$$

$$\begin{array}{l}\text { Taking log on beth sides, } \log (P V)=\log (n R T) \\\qquad \begin{array}{c}\ln P+\ln V=\ln (n R T) \\\text { Differmtiating, } \frac{1}{P} d P+\frac{1}{V} d V=0 \\\therefore \quad \frac{d V}{V}=-\frac{d P}{P}\end{array}\end{array}$$

$$\begin{array}{l}\therefore \text { Bule Medulus ( } \beta \text { ) in iscthermal }\\\text { conditions is, P. } \\\beta=\frac{\Delta P}{-\Delta P / P}=P \text { ( using equation 1) }\end{array}$$
Question 4
1 / -0

A brass rod of length $$1 \ m$$ is fixed to a vertical wall at one end, with the other end keeping free to expand. When the temperature of the rod is increased by $$120^{\circ}C$$ , the length increases by $$3 \ cm$$. What is the strain?

A
$$0$$

B
$$10$$

C
$$20$$

D
$$30$$

Solution

$$\triangle L=\alpha \triangle T$$
Strain$$=\cfrac { \triangle L }{ L } =\cfrac { 0.03m }{ 1m } \approx 0$$
Question 5
1 / -0

An amusement park ride consists of airplane shaped cars attached to steel rods. Each rod has a length of 20.0 m and a cross-sectional area of 8.00 $$cm^2$$. Young's modulus for steel is $$2 \, \times \, 10^{11} \, N/m^2.$$
b. When operating, the ride has a maximum angular speed of $$\sqrt{1}$$9/5 rad/s. How much is the rod stretched (in mm) then?

A
0.38 mm

B
0.55 mm

C
0.45 mm

D
0.34 mm

Solution

0.38 mm
Sln :
The force that each car exerts on the cable is F = $$m\omega^2l_0 \, = \, \dfrac{W}{g} \omega^2 l_0.$$
$$\Delta l \, = \, \dfrac{Fl_0}{YA} \, = \, \dfrac{W\omega^2l_0^2}{gYA}$$
Question 6
1 / -0

Two metal wire 'P' and 'Q' of same length and material are stretched by same load. Their masses are in the ratio $$m_1 : m_2$$. The ratio of elongations of wire 'P' to that of 'Q' is

A
$$m_{1}^{2} : m_{2}^{2}$$

B
$$m_{2}^{2} : m_{2}^{1}$$

C
$$m_2 : m_1$$

D
$$m_1 : m_2$$

Solution

We know that:-

$$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$$ $$Y$$=Young's Modulus

$$\implies \Delta l=\dfrac{Fl}{AY}$$

$$\implies \Delta l\propto \dfrac{1}{A}$$

Now, $$m=Al\rho\implies m\propto A$$

Hence, $$\dfrac{\Delta l_P}{\Delta l_Q}=\dfrac{A_Q}{A_P}$$

$$\implies \dfrac{\Delta l_P}{\Delta l_Q}=\dfrac{m_2}{m_1}$$

Hence, answer is option-(C).
Question 7
1 / -0

A solid sphere of radius $$R$$, made up of a material of bulk modulus $$K$$ is surrounded by a liquid in a cylindrical container. A massless piston of area $$A$$ floats on the surface of the liquid. When a mass $$M$$ is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is

A
$$\dfrac {Mg}{2AK}$$

B
$$\dfrac {Mg}{3AK}$$

C
$$\dfrac {Mg}{AK}$$

D
$$\dfrac {2Mg}{3AK}$$

Solution

Step 1: Find change in pressure when mass $$M$$ is placed on massless piston.
It is given that a sphere of radius $$R$$ made of material of bulk modulus$$K$$ is kept in a cylindrical container filled with liquid.
A massless piston of area $$A$$ is floating on the surface of liquid. Since the piston is massless, initial force by piston on liquid is zero, $$F_i = 0$$
When mass $$M$$ is kept on piston force exerted by piston on liquid becomes, $$F_f = Mg$$
Thus, change in pressure on liquid is given as,
$$dP = \dfrac{dF}{A}$$

$$\Rightarrow dP = \dfrac{Mg - 0}{A}$$
$$\Rightarrow dP = \dfrac{Mg}{A}$$

Step 2: Use formula for Bulk Modulus for Sphere to find fractional change in radius of sphere.
As radius of sphere is $$R$$, its volume is given as,
$$V = \dfrac{4}{3} \pi R^3$$

Let after putting mass on piston small chabge in volume of sphere is $$dV$$, it is viven as,
$$dV = \dfrac{4}{3} \pi 3R^2dR = 4\pi R^2dR$$

Now by using formula for Bulk Modulus of sphere, we have
$$K = \dfrac{dP}{\dfrac{dV}{V}}$$

$$\Rightarrow K = \dfrac{dP}{\dfrac{dV}{V}}$$

$$\Rightarrow K = \dfrac{\dfrac{Mg}{A}}{\dfrac{4\pi R^2dR}{\dfrac{4}{3} \pi R^3}}$$

$$\Rightarrow \dfrac{dR}{R} K = \dfrac{Mg}{3A}$$

$$\Rightarrow \dfrac{dR}{R} = \dfrac{Mg}{3KA}$$

Thus, fractional change in radius is given as, $$\dfrac{dR}{R} = \dfrac{Mg}{3KA}$$.
Option B is correct.
Question 8
1 / -0

If the ratio of diameters, lengths and Young's moduli of steel and brass wires shown in the figure are $$p,q$$ and $$r$$ respectively. Then the corresponding ratio of increase in their lengths would be:

A
$$\cfrac { 3q }{ 5{ p }^{ 2 }r } $$

B
$$\cfrac { 5q }{ 3{ p }^{ 2 }r } $$

C
$$\cfrac { 3q }{ 5{ p }^{ }r } $$

D
$$\cfrac { 5q }{ { p }^{ }r } $$

Solution

Given:
$$\cfrac { { F }_{ S } }{ { F }_{ B } }=\dfrac{5}{3}$$ $$\cfrac { { D }_{ B } }{ { D }_{ S }}=p $$ $$\cfrac { { L }_{ S } }{ { L }_{ B } } =q$$ $$\cfrac { { Y }_{ S } }{ { Y }_{ B } }=r$$

As Young's modulus $$Y=\cfrac { FL }{ A\Delta L } =\cfrac { 4FL }{ \pi { D }^{ 2 }\Delta L } $$
where the symbols have their usual meanings
$$\therefore \Delta L=\cfrac { 4FL }{ \pi { D }^{ 2 }Y } $$
$$\therefore \cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ B } } =\cfrac { { F }_{ S } }{ { F }_{ B } } \cfrac { { L }_{ S } }{ { L }_{ B } } \cfrac { { D }_{ B }^{ 2 } }{ { D }_{ S }^{ 2 } } \cfrac { { Y }_{ B } }{ { Y }_{ S } } $$

$$\quad \therefore \cfrac { \Delta { L }_{ S } }{ \Delta { L }_{ B } } =\left( \cfrac { 5}{ 3 } \right) \left( q \right) { \left( \cfrac { 1 }{ p } \right) }^{ 2 }\left( \cfrac { 1 }{ r } \right) =\cfrac { 5q }{ 3{ p }^{ 2 }r } $$
Question 9
1 / -0

A student performs an experiment to determine the Young's modulus of a wire, exactly $$2m$$ 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 $$0.05 mm$$ at a load of exactly $$1.0$$kg. The student also measures the diameter of the wire to be $$0.4mm$$ with an uncertainty of $$0.01mm$$. Take $$g = 9.8 m s^{-2}$$ (exact). The Young's modulus obtained from the reading is:

A
$$(2.0 \ \bar+ \ 0.3) \times 10^{11} N m^{-2}$$

B
$$(2.0 \ \bar+ \ 0.2) \times 10^{11} N m^{-2}$$

C
$$(2.0 \ \bar+ \ 0.1) \times 10^{11} N m^{-2}$$

D
$$(2.0 \ \bar+ \ 0.05) \times 10^{11} N m^{-2}$$

Solution

Question 10

1 / -0

A load is supported using three wires of same cross section area as shown. Then for :

	List - I		List - II
p	Equal tensile stress in all wires	1	$$Y_{2}$$ = 2$$Y_{1}$$
q	Equal tension in all wires	2	$$Y_{2}$$ = 4$$Y_{1}$$
r	Equal elastic potential energy in all wires	3	$$Y_{2}$$ = 8$$Y_{1}$$
s	Equal elastic potential energy per unit volume in all wires	4	$$Y_{2}$$ = 16$$Y_{1}$$

P- 2, Q- 3, R- 1, S- 4

P- 1, Q- 4, R- 2, S- 3

P- 3, Q- 2, R- 4, S- 1

P- 2, Q- 2, R- 3, S- 4

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 75

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Mechanical Properties of Solids Test - 75

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SHARING IS CARING

Question 1

3 : 1

9 : 1

27 : 1

81 : 1

Solution

Question 2

$$1.3$$

$$1.5$$

$$1.7$$

$$1.9$$

Solution

Question 3

$$\dfrac{2}{3}P$$

P

$$\dfrac{3}{2}P$$

$$2P$$

Solution

Question 4

$$0$$

$$10$$

$$20$$

$$30$$

Solution

Question 5

0.38 mm

0.55 mm

0.45 mm

0.34 mm

Solution

Question 6

$$m_{1}^{2} : m_{2}^{2}$$

$$m_{2}^{2} : m_{2}^{1}$$

$$m_2 : m_1$$

$$m_1 : m_2$$

Solution

Question 7

$$\dfrac {Mg}{2AK}$$

$$\dfrac {Mg}{3AK}$$

$$\dfrac {Mg}{AK}$$

$$\dfrac {2Mg}{3AK}$$

Solution

Question 8

$$\cfrac { 3q }{ 5{ p }^{ 2 }r } $$

$$\cfrac { 5q }{ 3{ p }^{ 2 }r } $$

$$\cfrac { 3q }{ 5{ p }^{ }r } $$

$$\cfrac { 5q }{ { p }^{ }r } $$

Solution

Question 9

$$(2.0 \ \bar+ \ 0.3) \times 10^{11} N m^{-2}$$

$$(2.0 \ \bar+ \ 0.2) \times 10^{11} N m^{-2}$$

$$(2.0 \ \bar+ \ 0.1) \times 10^{11} N m^{-2}$$

$$(2.0 \ \bar+ \ 0.05) \times 10^{11} N m^{-2}$$

Solution

Question 10

P- 2, Q- 3, R- 1, S- 4

P- 1, Q- 4, R- 2, S- 3

P- 3, Q- 2, R- 4, S- 1

P- 2, Q- 2, R- 3, S- 4

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