Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$S$$ is stress and $$Y$$ is Young's modulus of material of wire, then energy stored in the wire per unit volume is:

A
$$\displaystyle 2{ S }^{ 2 }Y$$

B
$$\displaystyle \frac { S }{ Yx } $$

C
$$\displaystyle \frac { 2Y }{ { S }^{ 2 } } $$

D
$$\displaystyle \frac { { S }^{ 2 } }{ 2Y } $$

Solution

Energy stored per unit volume can be given as:
$$E =\dfrac{1}{2} \times stress \times strain $$ -----------(1)
From Hooke's law :
Young's modulus, $$Y = \dfrac{Stress}{Strain}$$

$$\implies$$ $$Strain = \dfrac{Stress}{Y} = \dfrac{S}{Y}$$ -----------(2)

From equation (1) and (2):
$$\therefore$$ $$E = \dfrac{1}{2} \times S \times \dfrac{S}{Y} $$

$$\Rightarrow E= \dfrac{1}{2} \dfrac{S^2}{Y}$$
Hence, the correct option is $$(D)$$
Question 2
1 / -0

A long spring is stretched by $$2 cm$$ and its potential energy is $$U$$. If the spring is stretched by $$10 cm$$; its potential energy will be (in terms of $$U$$)

A
$${U}/{5}$$

B
$${U}/{25}$$

C
$$5 U$$

D
$$25 U$$

Solution

The potential energy of a stretched spring is
$$U = \dfrac{1}{2} k{x}^{2}$$
Here, $$k =$$ spring constant,
$$x =$$ elongation in spring.
But given that, the elongation is $$2 cm$$.
So, $$U = \dfrac{1}{2} k{\left(2\right)}^{2}$$
$$\Rightarrow U = \dfrac{1}{2} k \times 4$$ ......(i)
If elongation is $$10 cm$$ then potential energy
$${ U }^{ \prime }=\dfrac { 1 }{ 2 } k{ \left( 10 \right) }^{ 2 }$$
$${ U }^{ \prime }=\dfrac { 1 }{ 2 } k\times 100$$ ......(ii)
On dividing equation (ii) by equation (i), we have
$$\dfrac { { U }^{ \prime } }{ U } =\dfrac { \dfrac { 1 }{ 2 } k\times 100 }{ \dfrac { 1 }{ 2 } k\times 4 } $$
or $$\dfrac { { U }^{ \prime } }{ U } =25\Rightarrow { U }^{ \prime }=25U$$
Question 3
1 / -0

Longitudinal strain is possible in

A
Liquid

B
Gases

C
Solid

D
All of these

Solution

Longitudinal strain is possible-only in solids because only solids can have length which can be stretched by applying force. while molecules and liquid are loosely packed together and hence strain cannot occur in these mediums.
Question 4
1 / -0

The adjacent graph shows the extension ($$\displaystyle \Delta l$$) of a 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. If the cross-sectional area of the wire is $$\displaystyle { 10 }^{ -6 }{ m }^{ 2 }$$, calculate the Young's modulus of the material of the wire.

A
$$\displaystyle 2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$$

B
$$\displaystyle 2\times { 10 }^{ -11 }{ N }/{ { m }^{ 2 } }$$

C
$$\displaystyle 3\times { 10 }^{ -12 }{ N }/{ { m }^{ 2 } }$$

D
$$\displaystyle 2\times { 10 }^{ -13 }{ N }/{ { m }^{ 2 } }$$

Solution

$$Y = \dfrac{Fl}{A\Delta l}$$
from the graph we can see $$\dfrac{\Delta l}{W} = \dfrac{10^{-4}}{20}$$
$$Y = \dfrac{l}{A\dfrac{\Delta l}{W}}$$
$$ Y = \dfrac{1\times 20}{10^{-6} \times 10^{-4}}$$
$$Y = 2 \times 10^{11} N/m^2$$
Question 5
1 / -0

For a given material, the Young's modulus is $$2.4$$ times that of rigidity modulus. Its poisson's ratio is.

A
$$2.4$$

B
$$1.2$$

C
$$0.4$$

D
$$0.2$$

Solution

Hint: The “ratio of transverse contraction strain to longitudinal extension strain in the direction of the stretching force” is known as the poisson ratio.

Explanation:

Step 1: Formula used:
The relation of Young's modulus '$$Y$$' and rigidity modulus '$$ \eta $$' and poisson's ratio '$$ \sigma $$' :
$$Y=2\eta(1+\sigma)$$ ..........(1)

Step 2: Finding the value of poisson's ratio:

Given: $$ Y= 2.4 \ \eta $$
Putting the value in eq. (1)

$$\Rightarrow 2.4 \ \eta =2 \ \eta \ (1+\sigma)$$
$$\Rightarrow 1.2=1+\sigma$$
$$\Rightarrow \sigma=0.2$$

Option D is correct.
Question 6
1 / -0

For a constant hydraulic stress on an object, the fractional change in the object's volume ($$\displaystyle \Delta { V }/{ V }$$) and its bulk modulus (B) are related as:

A
$$\displaystyle \frac { \Delta V }{ V } \propto B$$

B
$$\displaystyle \frac { \Delta V }{ V } \propto \frac { 1 }{ B } $$

C
$$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ 2 }$$

D
$$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ -2 }$$

Solution

$$B = \dfrac{stress}{volume \ strain} = \dfrac{stress}{\dfrac{\Delta V}{V}}$$
$$\dfrac{\Delta V}{V} = \dfrac{stress}{B}$$

As stress is constant
$$\dfrac{\Delta V}{V} \propto \dfrac{1}{B}$$
Question 7
1 / -0

The length of an elastic string is $$a$$ metre when the longitudinal tension is $$4$$N and $$b$$ metre when the longitudinal tension is $$5$$N. The length of the string in metre when longitudinal tension is $$ 9 $$N is :

A
$$a-b$$

B
$$5b-4a$$

C
$$2b-\displaystyle\frac{1}{4}a$$

D
$$4a-3b$$

Solution

Let L is the original length of the wire and k is force constant of wire.
Final length $$=$$ initial length $$+$$ elongation
$$L'=L+\displaystyle\frac{F}{k}$$
For first condition $$a=L+\displaystyle\frac{4}{k}$$ .....$$(i)$$
For second condition $$b=\displaystyle L+\frac{5}{k}$$ ......$$(ii)$$
By solving Eqs. $$(i)$$ and $$(ii)$$, we get
$$L=5a-4b$$ and $$k=\displaystyle\frac{1}{b-a}$$
Now, when the longitudinal tension is $$9N$$. length of the string.
$$=L+\displaystyle\frac{9}{k}=5a-4b+9(b-a)$$
$$=5b-4a$$
Question 8
1 / -0

A $$5$$m long aluminium wire $$(Y=7\times 10^{10}Nm^{-2})$$ of diameter $$3$$mm supports a $$40$$kg mass. In order to have the same elongation in the copper wire $$(Y=12\times 10^{10}Nm^{-2})$$ of the same length under the same weight, the diameter should now be (in mm).

A
$$1.75$$

B
$$1.5$$

C
$$2.5$$

D
$$5.0$$

Solution

$$l=\displaystyle\frac{FL}{\pi r^2Y}$$
$$\Rightarrow r^2\propto \displaystyle\frac{1}{Y}(F, L$$ and $$l$$ are constant$$)$$
$$\displaystyle\frac{r_2}{r_1}=\left[\displaystyle\frac{Y_1}{Y_2}\right]^{1/2}=\left[\displaystyle\frac{7\times 10^{10}}{12\times 10^{10}}\right]^{1/2}$$
$$\Rightarrow r_2=1.5\times \displaystyle\left(\frac{7}{12}\right)^{1/2}=1.145$$mm
$$\therefore$$ diameter$$=2.29$$mm
Question 9
1 / -0

If a wire having initial diameter of $$2 mm$$ produced the longitudinal strain of $$0.1$$%, then the final diameter of wire is $$\left( \sigma = 0.5 \right)$$

A
$$2.002 mm$$

B
$$1.999 mm$$

C
$$1.998 mm$$

D
$$2.001 mm$$

Solution

When a deforming force is applied at the free end of a suspend wire of length $$l$$ and radius $$R$$, then its length increases by $$dl$$, but its radius decreases by $$dR$$. Now two types of strains are produced by a single force
(i) Longitudinal strain $$= \dfrac{\Delta l}{l}$$
(ii) Lateral strain $$= -\dfrac{\Delta R}{R}$$
Then the Poisson's ratio
$$\sigma = \dfrac{lateral strain}{longitudinal strain}$$
$$=-\dfrac { { \Delta R }/{ R } }{ { \Delta l }/{ l } } $$
$$\sigma =-\dfrac { \Delta R }{ R } \times \dfrac { l }{ \Delta l } $$
or $$\left| \sigma \right| =\dfrac { \Delta R }{ R } \times \dfrac { l }{ \Delta l } $$
$$\therefore \dfrac { \Delta R }{ R } =\left| \sigma \right| \left( \dfrac { \Delta l }{ l } \right) $$
or $$\Delta R=0.5\times \dfrac { 0.1 }{ 100 } \times \left( \dfrac { 2\times { 10 }^{ -3 } }{ 2 } \right) $$
$$= 0.0005 mm$$
$$\therefore $$ Final radius $$R - \Delta R = 1 mm - 0.0005 mm$$
$$= 0.9995 mm$$
$$\therefore $$ The final diameter $$=2\times 0.9995 mm$$
$$= 1.9990 mm$$
Question 10
1 / -0

The length of a metal wire is $$L_1$$ when the tension is $$T_1$$ and $$L_2$$ when the tension is $$T_2$$. The unstretched length of wire is :

A
$$\displaystyle \frac{L_1+L_2}{2}$$

B
$$\displaystyle \sqrt{L_1L_2}$$

C
$$\displaystyle \frac{T_2L_1-T_1L_2}{T_2-T_1}$$

D
$$\displaystyle \frac{T_2L_1+T_1L_2}{T_2+T_1}$$

Solution

Let the initial length of the metal wire is L.
The strain at tension $$T_1$$ is $$\triangle L_1=L_1-L$$
The strain at tension $$T_2$$ is $$\triangle L_2=L_2-L$$
Suppose, the Youngs modulus of the wire is Y,
$$\displaystyle \frac{\dfrac{T_1}{A}}{\dfrac{\triangle L_1}{L}}=\dfrac{\dfrac{T_2}{A}}{\dfrac{\triangle L_2}{L}}$$
Where, A is an cross-section of the wire assume to be same at all the situations.
$$\Rightarrow \displaystyle \frac{T_1}{A}\times \frac{L}{\triangle L_1}=\frac{T_2}{A}\times \frac{L}{\triangle L_2}$$
$$\Rightarrow \displaystyle \frac{T_1}{(L_1-L)}=\frac{T_2}{(L_2-L)}$$
$$\displaystyle T_1(L_2-L)=T_2(L_1-L); L=\frac{T_2L_1-T_1L_2}{T_2-T_1}$$

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 42

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Mechanical Properties of Solids Test - 42

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SHARING IS CARING

Question 1

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

$$\displaystyle \frac { S }{ Yx } $$

$$\displaystyle \frac { 2Y }{ { S }^{ 2 } } $$

$$\displaystyle \frac { { S }^{ 2 } }{ 2Y } $$

Solution

Question 2

$${U}/{5}$$

$${U}/{25}$$

$$5 U$$

$$25 U$$

Solution

Question 3

Liquid

Gases

Solid

All of these

Solution

Question 4

$$\displaystyle 2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$$

$$\displaystyle 2\times { 10 }^{ -11 }{ N }/{ { m }^{ 2 } }$$

$$\displaystyle 3\times { 10 }^{ -12 }{ N }/{ { m }^{ 2 } }$$

$$\displaystyle 2\times { 10 }^{ -13 }{ N }/{ { m }^{ 2 } }$$

Solution

Question 5

$$2.4$$

$$1.2$$

$$0.4$$

$$0.2$$

Solution

Question 6

$$\displaystyle \frac { \Delta V }{ V } \propto B$$

$$\displaystyle \frac { \Delta V }{ V } \propto \frac { 1 }{ B } $$

$$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ 2 }$$

$$\displaystyle \frac { \Delta V }{ V } \propto { B }^{ -2 }$$

Solution

Question 7

$$a-b$$

$$5b-4a$$

$$2b-\displaystyle\frac{1}{4}a$$

$$4a-3b$$

Solution

Question 8

$$1.75$$

$$1.5$$

$$2.5$$

$$5.0$$

Solution

Question 9

$$2.002 mm$$

$$1.999 mm$$

$$1.998 mm$$

$$2.001 mm$$

Solution

Question 10

$$\displaystyle \frac{L_1+L_2}{2}$$

$$\displaystyle \sqrt{L_1L_2}$$

$$\displaystyle \frac{T_2L_1-T_1L_2}{T_2-T_1}$$

$$\displaystyle \frac{T_2L_1+T_1L_2}{T_2+T_1}$$

Solution

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