Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 44

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two wires of the same material and length but diameter in the ratio $$1:2$$ are stretched by the same force. The elastic potential energy per unit volume for the two wires when stretched by the same force will be in the ratio:

A
$$16:1$$

B
$$1:1$$

C
$$2:1$$

D
$$4:1$$

Solution

Elastic energy per unit volume: $$E = \dfrac{1}{2} \times Stress \times Strain$$
Using Hooke's law: $$Strain = \dfrac{Stress}{Y}$$
$$\implies$$ $$E = \dfrac{(Stress)^2}{2Y} = \dfrac{F^2}{2Y A^2}$$ where $$A = \dfrac{\pi D^2}{4}$$

$$\implies$$ $$E = \dfrac{8F^2}{\pi^2 D^4 Y}$$

$$\therefore$$ $$\dfrac{E_1}{E_2} = \bigg (\dfrac{D_2}{D_1}\bigg )^4$$ $$ (\because F $$ and $$Y $$ are same for both$$)$$
GIven : $$\dfrac{D_1}{D_2} = \dfrac{1}{2}$$

$$\dfrac{E_1}{E_2} = \bigg (\dfrac{2}{1}\bigg )^4 = \dfrac{16}{1}$$
Question 2
1 / -0

Two wires A and B are of the same materials. Their lengths are in the ratio $$1:2$$ and the diameters are in the ratio $$2:1$$, When stretched by force $$F_A$$ and $$F_B$$ respectively they get equal increase in their lengths. Then the ratio $$\dfrac{F_A}{F_B}$$ should be:

A
$$1:2$$

B
$$1:1$$

C
$$2:1$$

D
$$8:1$$

Solution

Using Hooke's law $$Y = \dfrac{F/A}{\Delta L/L}$$ where $$A = \dfrac{\pi D^2}{4}$$
$$\implies$$ $$F =Y \dfrac{\pi D^2}{4} \dfrac{\Delta L}{L}$$
As $$Y$$ and $$\Delta L$$ are constant, thus $$F \propto \dfrac{D^2}{L}$$
$$\implies$$ $$\dfrac{F_A}{F_B} = \dfrac{D^2_A}{D^2_B} \times \dfrac{L_B}{L_A}$$

Given : $$D_A:D_B = 2:1$$ and $$L_A:L_B = 1:2$$
$$\therefore$$ $$\dfrac{F_A}{F_B} = \dfrac{4}{1} \times \dfrac{2}{1} = \dfrac{8}{1}$$
Question 3
1 / -0

The average depth of Indian ocean is about $$3000\ m$$. The value of fractional compression $$ \dfrac{\Delta V}{V}$$ of water at the bottom of the ocean is:
[Given that the bulk modulus of water is $$2.2\times 10^9\ Nm^{-2}$$, $$ g=9.8\ ms^{-2}$$ and $$\rho_{H_2O}=1000\ kg.m^{-3}$$]

A
$$3.4\times 10^{-2}$$

B
$$1.34\times 10^{-2}$$

C
$$4.13\times 10^{-2}$$

D
$$13.4\times 10^{-2}$$

Solution

$$\dfrac{\Delta V}{V}=\dfrac{h\rho g}{k}=\dfrac{3\times 10^3\times 9.8 \times 10^3}{22\times 10^8} $$

$$=\dfrac{3}{22}\times 10^{-3}\times 98\\ =\dfrac{147}{11}\times 10^{-3}\\=1.34\times 10^{-2}m$$
Question 4
1 / -0

Let L be the length and d be the diameter of cross-section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

A
$$L = 200\ cm, d = 0.5\ mm$$

B
$$L = 300\ cm, d = 1.0\ mm$$

C
$$L = 50\ cm, d = 0.05\ mm$$

D
$$L = 100\ cm, d = 0.2\ mm$$

Solution

Since material is same, Young's modulus is also same.
We have: $$Y\dfrac { \Delta l }{ l } =\dfrac { F }{ A } \Rightarrow \Delta l=\dfrac { Fl }{ YA } \propto \dfrac { l }{ { d }^{ 2 } } $$
where, $$Y$$ is the Young's Modulus, $$F$$ is the tension which is same in all cases, $$A$$ is the area of cross section of the wire and $$l$$ is the length of the wire.

Calculating the value of $$ \dfrac { l }{ { d }^{ 2 } } $$ in all cases, we get:
A: $$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 2000 }{ 0.5\times 0.5 } =8000$$

B: $$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 3000 }{ 1.0\times 1.0 } =3000$$

C: $$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 50 }{ 0.05\times 0.05 } =20000$$

D: $$\dfrac { l }{ { d }^{ 2 } } =\dfrac { 100 }{ 0.2\times 0.2 } =2500$$

So maximum elongation will be in case C.
Question 5
1 / -0

The Poisson's ratio of a material is $$0.5$$. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is :

A
1%

B
2%

C
2.5%

D
4%

Solution

Poisson ratio $$=0.5$$
Since, density is constant therefore change in volume is zero, we have
$$V=A\times l=$$ constant
$$\Rightarrow \log { V } =\log { A } +\log { l } $$
or $$\dfrac { dA }{ A } +\dfrac { dl }{ l } =0$$
$$\Rightarrow \dfrac { dl }{ l } =-\dfrac { dA }{ A } $$
$$\therefore $$ Percentage increase in length $$=4$$%
Question 6
1 / -0

A uniform aluminium wire of length $$3$$m and area of cross-section $$2$$ $$mm^2$$ is extended through $$12\ mm$$. The energy stored in the wire is ___________. ($$Y_{Al}=7\times 10^{10}N/m^2$$)

A
$$336\ J$$

B
$$33.6\ J$$

C
$$3.36\ J$$

D
$$0.336\ J$$

Solution

$$A= $$ area of cross section
$$ L=$$ length of the wire
$$ K=$$ spring constant

$$K=\dfrac { A{ Y }_{ al } }{ L } \\ E=\dfrac { 1 }{ 2 } K{ x }^{ 2 }=\dfrac { A{ Y }_{ al }{ x }^{ 2 } }{ 2L } =\dfrac { 2\times { { 10 }^{ -6 }\times 7\times { 10 }^{ 10 }\times }{ (12\times { 10 }^{ -3 }) }^{ 2 } }{ 2\times 3 } =3.36J$$
Question 7
1 / -0

The ability of the material to deform without breaking is called :

A
Elasticity

B
Plasticity

C
Creep

D
None of these

Solution

The quality of being easily shaped or moulded is called Plasticity
So Plasticity is the ability of the material to deform without breaking
Therefore option $$B$$ is correct
Question 8
1 / -0

A ball is falling in a lake of depth $$200 m$$ creates a decrease of $$0.1$$% in its volume at the bottom. The bulk modulus of the material of the ball will be:

A
$$19.6 \times 10^{-16}$$ $$N\ m^{-2}$$

B
$$19.6 \times 10^{8}$$ $$N\ m^{-2}$$

C
$$19.6 \times 10^{-8}$$ $$N\ m^{-2}$$

D
$$19.6 \times 10^{16}$$ $$N\ m^{-2}$$

Solution

The density of water is $$d=10^{3}Kg/m^{3}$$
The depth of lake is $$h=200m$$
The decrease in volume is $$0.1$$%
The bulk modulus will be $$B=V\frac{P}{dV}=\frac{dgh}{\frac{dV}{V}}=\frac{10^{3} \times 9.8 \times 200}{10^{-3}}=19.6 \times 10^{8}$$
Question 9
1 / -0

A tension of $$22$$ N is applied to a copper wire of cross-sectional area $$0.02\ cm^2$$. Young's modulus of copper is $$1.1\times 10^{11}N/m^2$$ and Poisson's ratio $$0.32$$. The decrease in cross sectional area will be:

A
$$1.28\times 10^{-6}cm^2$$

B
$$1.6\times 10^{-6}cm^2$$

C
$$2.56\times 10^{-6}cm^2$$

D
$$0.64\times 10^{-6}cm^2$$

Solution

Young's modulus, $$Y=\dfrac{Fl}{A\Delta l} $$ and Poisson's ratio , $$\sigma=\dfrac{\Delta r/ r}{\Delta l/l}$$

From these we get, $$\Delta r/ r=\dfrac{F\sigma}{AY}=\dfrac{22\times0.32 }{0.02\times 10^{-4}\times 1.1\times 10^{11}}=32\times 10^{-6}$$

Cross sectional area , $$A=\pi r^2 $$ and $$\Delta A=2\pi r \Delta r$$

Thus, $$\dfrac{\Delta A}{A}=\dfrac{2\Delta r}{r}=64\times 10^{-6} $$

$$\therefore \Delta A= 64\times 10^{-6}\times 0.02=1.28\times 10^{-6} cm^2$$
Question 10
1 / -0

A steel wire if $$1\ m$$ long and $$1\ mm^2$$ cross section area is hanged from rigid end when weight of $$1\ kg$$ is hang from it, then change in length will be
(Young's coefficient for wire $$Y=2\times 10^{11}N/m^2)$$

A
$$0.5\ mm$$

B
$$0.25\ mm$$

C
$$0.05\ mm$$

D
$$5\ mm$$

Solution

Young's coefficient $$Y$$ of a wire of length $$l$$ and cross section $$A$$ , having weight $$mg$$ is given by ,
$$Y=\dfrac{mgl}{A\Delta l}$$
or $$\Delta l=\dfrac{mgl}{AY}$$
where $$\Delta l=$$ change in length
given - $$Y=2\times10^{11}N/m^{2} , l=1m , A=1mm^{2}=10^{-6}m^{2} , m=1kg$$
Hence , $$\Delta l=\dfrac{1\times10\times1}{10^{-6}\times2\times10^{11}}=5\times10^{-5}m=0.05mm$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 44

Result

Mechanical Properties of Solids Test - 44

Score

-

Rank

-

SHARING IS CARING

Question 1

$$16:1$$

$$1:1$$

$$2:1$$

$$4:1$$

Solution

Question 2

$$1:2$$

$$1:1$$

$$2:1$$

$$8:1$$

Solution

Question 3

$$3.4\times 10^{-2}$$

$$1.34\times 10^{-2}$$

$$4.13\times 10^{-2}$$

$$13.4\times 10^{-2}$$

Solution

Question 4

$$L = 200\ cm, d = 0.5\ mm$$

$$L = 300\ cm, d = 1.0\ mm$$

$$L = 50\ cm, d = 0.05\ mm$$

$$L = 100\ cm, d = 0.2\ mm$$

Solution

Question 5

1%

2%

2.5%

4%

Solution

Question 6

$$336\ J$$

$$33.6\ J$$

$$3.36\ J$$

$$0.336\ J$$

Solution

Question 7

Elasticity

Plasticity

Creep

None of these

Solution

Question 8

$$19.6 \times 10^{-16}$$ $$N\ m^{-2}$$

$$19.6 \times 10^{8}$$ $$N\ m^{-2}$$

$$19.6 \times 10^{-8}$$ $$N\ m^{-2}$$

$$19.6 \times 10^{16}$$ $$N\ m^{-2}$$

Solution

Question 9

$$1.28\times 10^{-6}cm^2$$

$$1.6\times 10^{-6}cm^2$$

$$2.56\times 10^{-6}cm^2$$

$$0.64\times 10^{-6}cm^2$$

Solution

Question 10

$$0.5\ mm$$

$$0.25\ mm$$

$$0.05\ mm$$

$$5\ mm$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.