Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

When the tension in a metal wire is $${T}_{1}$$, its length is $${l}_{1}$$. When the tension is $${T}_{2}$$, its length is $${l}_{2}$$. The natural length of wire is

A
$$\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } ({ l }_{ 1 }+{ l }_{ 2 })$$

B
$${ T }_{ 1 }{ l }_{ 1 }+{ i }_{ 2 }{ l }_{ 2 }$$

C
$$\cfrac { { { l }_{ 1 }T }_{ 2 }-{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$

D
$$\cfrac { { { l }_{ 1 }T }_{ 2 }+{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } } $$

Solution

$$\dfrac { \Delta { l } }{ { l } } =\dfrac { F }{ AY } \\ Thus,\quad \dfrac { { l }_{ 1 }-l }{ l } =\dfrac { { T }_{ 1 } }{ AY } \quad and\quad \dfrac { { l }_{ 2 }-l }{ l } =\dfrac { { T }_{ 2 } }{ AY } $$
Solve the 2 equations to get,
$$AY=\dfrac { { T }_{ 1 }l }{ { l }_{ 1 }-l } ,\quad giving\quad { T }_{ 2 }l({ l }_{ 1 }-l)={ T }_{ 1 }l({ l }_{ 2 }-l)\\ or\quad l=\dfrac { { T }_{ 2 }{ l }_{ 1 }-{ T }_{ 1 }{ l }_{ 2 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$
Question 2
1 / -0

A wire stretched $$1mm$$ by a force of $$1kN$$. How far would a wire of the same material and length but of four times that diameter be stretched by the same force?

A
$$\dfrac {1}{2} mm$$

B
$$\dfrac {1}{4} mm$$

C
$$\dfrac {1}{8} mm$$

D
$$\dfrac {1}{16} mm$$

Solution

$$\Delta l=\dfrac { FL }{ AY } $$
Since $$F,L$$ and $$Y$$ are same in both cases, extension depends on Area, and is inversely proportional. $$4$$ times diameter means $${4}^{2}$$ times or 16 times Area. Thus, extension$$ = 1/16$$ times $$1 mm = 1/16 mm$$
Question 3
1 / -0

Bulk modulus of water is $$2\times 10^{9}N/m^{2}$$. The change in pressure required to increase the density of water by $$0.1\%$$ is

A
$$2\times 10^{9}N/{m^{2}}$$

B
$$2\times 10^{8}N/{m^{2}}$$

C
$$2\times 10^{6}N/{m^{2}}$$

D
$$2\times 10^{4}N/{m^{2}}$$

Solution

$$B=-V\dfrac{dP}{dV}$$
Thus, $$\Delta V=-\dfrac{\Delta pV}{B}$$, where p is change in pressure.
Thus, $$V'-V=-\dfrac{\Delta pV}{B}$$ or, $$V'=V(1-\dfrac{\Delta p}{B})$$
Now, $$\rho '=\dfrac{m}{V'}=\dfrac{m}{V(1-\dfrac{\Delta p}{B})}=\dfrac{\rho}{(1-\dfrac{\Delta p}{B})}$$
Given $$\rho ' =1.001\rho$$, or $$\dfrac{1}{1.001}= 1-\dfrac{p}{2\times{10}^{9}}$$
or, $$\Delta p=\dfrac{1}{1001}\times 2\times {10}^{9}$$, which is nearly $$2\times{10}^{6}$$
Question 4
1 / -0

Two identical wires of iron and copper with their Young's modulus in the ratio $$3:1$$ are suspended at same level. They are to be loaded so as to have the same extension and hence level. Ratio of the weight is

A
$$1:3$$

B
$$2:1$$

C
$$3:1$$

D
$$4:1$$

Solution

$$\delta l\quad =\quad \dfrac { FL }{ AE } \\ \Rightarrow \quad \dfrac { { F }_{ c }.L }{ { A }.{ E }_{ c } } \quad =\quad \dfrac { { F }_{ i }.L }{ { A }.{ 3E }_{ c } } \quad or\quad { 3F }_{ c }\quad =\quad { F }_{ i }\\ Thus,\quad 3\quad { W }_{ c }\quad =\quad { W }_{ i }$$
Question 5
1 / -0

A ball falling in a lake of depth $$200\ m$$ shows a decrease of $$0.1\%$$ in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take $$g = 10\ m/s^{2})$$.

A
$$10^{9}N/m^{2}$$

B
$$2\times 10^{9}N/m^{2}$$

C
$$3\times 10^{9}N/m^{2}$$

D
$$4\times 10^{9}N/m^{2}$$

Solution

Bulk Modulus is given by;
$$B=-V\dfrac{dP}{dV}$$
Now, change in pressure, $$dP=\rho gh=1000 \times 10\times 200=2\times {10}^{6}$$
Also, given $$\dfrac{dV}{V}=-0.001$$
Thus, $$B=-\dfrac{2\times{10}^{6}}{-0.001}=2\times{10}^{9}$$
Question 6
1 / -0

A long wire hangs vertically with its upper end clamped. A torque of $$8\ Nm$$ applied to the free end twists it through $$45^{\circ}$$. The potential energy of the twisted wire is

A
$$\pi J$$

B
$$\dfrac {\pi}{2} J$$

C
$$\dfrac {\pi}{4} J$$

D
$$\dfrac {\pi}{8} J$$

Solution

the elastic potential energy twisted by a torque T by an angle $$\theta$$

$$U = \dfrac{T\theta}{2}$$
$$U = \dfrac{8\times \dfrac{\pi}{4}}{2} = \pi J$$
Question 7
1 / -0

Young's modulis of brass and steel are $$10\times 10^{10} N/m$$ and $$2\times 10^{11} N/m^{2}$$, respectively. A brass wire and a steel wire of the same length are extended by $$1\ mm$$ under the same force. The radii of the brass and steel wires are $$R_{B}$ and $$R_{S}$$, respectively. Then

A
$$R_{S} = \sqrt {2} R_{B}$$

B
$$R_{S} = \dfrac {R_{B}}{\sqrt {2}}$$

C
$$R_{S} = 4R_{B}$$

D
$$R_{S} = \dfrac {R_{B}}{4}$$

Solution

$$\delta l\quad =\quad \dfrac { FL }{ AE } \\ \Rightarrow \quad 1\quad =\quad \dfrac { { F }.L }{ { { A }_{ B } }.{ E }_{ B } } \quad =\quad \dfrac { { F }.L }{ { { A }_{ S } }.{ E }_{ S } } \quad or\quad { A }_{ B }.10\times { 10 }^{ 10 }\quad =\quad { A }_{ S }.2\times { 10 }^{ 11 }\\ Thus,\quad { A }_{ B }\quad =\quad 2A_{ S }\quad or\quad { R }_{ B }\quad =\quad { R }_{ S }\sqrt { 2 } $$
Question 8
1 / -0

A long elastic spring is stretched by $$2\ cm$$ and its potential energy is $$U$$. If the spring is stretched by $$10\ cm$$, the P.E., will be

A
$$5U$$

B
$$25U$$

C
$$U/5$$

D
$$U/20$$

Solution

Potential energy of spring = $$\dfrac{1}{2}kx^2$$
Potential energy when spring is $$2cm$$ stretched = $$\dfrac{1}{2}k(2)^2 = 2k=U$$
Potential energy when stretched by $$10cm, $$U' = $$\dfrac{1}{2}k(10)^2= 50k= 25U$$
Question 9
1 / -0

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

A
$$3:2$$

B
$$2:3$$

C
$$4:9$$

D
$$9:4$$

Solution

Let the lengths of the two wires be $$2l$$ and $$3l$$.
Two materials have same mass.
Thus we get $$\rho(2l)A_1=\rho(3l)A_2$$
Thus $$2A_1=3A_2$$
Now their elongations are given as $$\Delta l_1=\dfrac{F2l}{A_1Y}$$ and $$\Delta l_2=\dfrac{F3l}{A_2Y}$$
Thus we get $$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{4}{9}$$
Question 10
1 / -0

Each of the pictures shows four objects tied together with rubber bands being pulled to the right across a horizontal frictionless surface by a horizontal force $$F$$. All the objects have the same mass; all the rubber bands obey Hooke's law and have the same equilibrium length and the same force constant. Which of these pictures is drawn most correctly?

A

B

C

D

Solution

The block to the front will have maximum acceleration and hence will be comparitively much ahead of the rest. Similarly, the second block will have slightly lower acceleration and will be behind the first. But the distance between second and third will be lesser than first and second because first starts moving fastest, then second follows, then it takes time for next and so on. Thus B is the correct figure.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 34

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Mechanical Properties of Solids Test - 34

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SHARING IS CARING

Question 1

$$\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } ({ l }_{ 1 }+{ l }_{ 2 })$$

$${ T }_{ 1 }{ l }_{ 1 }+{ i }_{ 2 }{ l }_{ 2 }$$

$$\cfrac { { { l }_{ 1 }T }_{ 2 }-{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } $$

$$\cfrac { { { l }_{ 1 }T }_{ 2 }+{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } } $$

Solution

Question 2

$$\dfrac {1}{2} mm$$

$$\dfrac {1}{4} mm$$

$$\dfrac {1}{8} mm$$

$$\dfrac {1}{16} mm$$

Solution

Question 3

$$2\times 10^{9}N/{m^{2}}$$

$$2\times 10^{8}N/{m^{2}}$$

$$2\times 10^{6}N/{m^{2}}$$

$$2\times 10^{4}N/{m^{2}}$$

Solution

Question 4

$$1:3$$

$$2:1$$

$$3:1$$

$$4:1$$

Solution

Question 5

$$10^{9}N/m^{2}$$

$$2\times 10^{9}N/m^{2}$$

$$3\times 10^{9}N/m^{2}$$

$$4\times 10^{9}N/m^{2}$$

Solution

Question 6

$$\pi J$$

$$\dfrac {\pi}{2} J$$

$$\dfrac {\pi}{4} J$$

$$\dfrac {\pi}{8} J$$

Solution

Question 7

$$R_{S} = \sqrt {2} R_{B}$$

$$R_{S} = \dfrac {R_{B}}{\sqrt {2}}$$

$$R_{S} = 4R_{B}$$

$$R_{S} = \dfrac {R_{B}}{4}$$

Solution

Question 8

$$5U$$

$$25U$$

$$U/5$$

$$U/20$$

Solution

Question 9

$$3:2$$

$$2:3$$

$$4:9$$

$$9:4$$

Solution

Question 10

Solution

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