Mechanical Properties of Solids Test

Result

Mechanical Properties of Solids Test - 24

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

A ball falling in a lake to a depth $$200m$$ shows a decrease of $$0.1$$% in its volume at the bottom. the bulk modulus of the ball is

A
19.6 x 10$$^{8}$$ N/m$$^{2}$$

B
19.6 x 10$$^{-10}$$ N/m$$^{2}$$

C
19.6 x 10$$^{10}$$ N/m$$^{2}$$

D
19.6 x 10$$^{-8}$$ N/m$$^{2}$$

Solution

$$B=\dfrac{\rho gh}{\Delta v/v}$$ (standard result)

$$=\dfrac{1000\times 200\times 9.8}{\dfrac{0.1}{100}}$$

$$=19.6\times 10^{8}N/m^{2}$$
Question 2
1 / -0

The Poisson's ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is :

A
$$3$$%

B
$$2.5$$%

C
$$1$$%

D
$$0.5$$%

Solution

Poisson's ratio, $$0.4=\cfrac{\Delta d/d}{\Delta l/l}\Rightarrow\cfrac{\Delta l}{l}=\cfrac{\Delta d/d}{0.4}$$-------------(1)
$$A=\pi r^2=\cfrac{\pi d^2}{4}$$
Taking log both sides:
$$ \log A=\log {(\frac{\pi}{4})}+2\log {d}$$
Taking relative change both the sides:
$$ \cfrac{\Delta A}{A}=2\cfrac{\Delta d}{d}\\ \cfrac{\Delta d}{d}=\cfrac{1}{2}\cfrac{\Delta A}{A}=\dfrac{1}{2}\times 2\%=1\%$$------------------------------(1)
From equation (1) and (2), the percentage increase length can be given as:
$$ \cfrac{\Delta l}{l}=\cfrac{1\%}{0.4}=2.5\%$$
Option B is correct
Question 3
1 / -0

A stress $$10^{7}$$ Pa produces a strain of 4 x 10$$^{-3}$$. The energy stored per unit volume of the body (in J.m$$^{-3}$$) is

A
2 x 10$$^{3}$$

B
2 x 10$$^{4}$$

C
2.5 x 10$$^{10}$$

D
0.8 x 10$$^{4}$$

Solution

$$energy=\dfrac{1}{2}\times stress\times strain\times volume$$

$$\dfrac{energy}{volume}=\dfrac{1}{2}\times stress\times strain$$

$$=\dfrac{1}{2}\times 10^{7}\times 4\times 10^{-3}$$

$$=2\times 10^{4} J/m^{3}$$
Question 4
1 / -0

On taking a solid rubber ball from the surface to the bottom of a lake 200m deep, the reduction in volume is found to be 0.5%. If the density of water is 10$$^{3}$$ kgm$$^{-3}$$ and g$$=$$10 ms$$^{-2}$$, find the bulk modulus of rubber.

A
2 x10$$^{8}$$Pa

B
4 x10$$^{8}$$Pa

C
6 x10$$^{8}$$Pa

D
8 x10$$^{8}$$Pa

Solution

$$B=\dfrac{\rho g\Delta h}{\Delta v/v}$$

$$=\dfrac{1000\times 10\times 200}{0.5/100}$$

$$=4\times 10^{8}pa$$
Question 5
1 / -0

A rubber cord of length $$40cm$$ and area of cross section 4 x 10$$^{-6}$$ m$$^{2}$$ is extended by 10cm. If the energy gained is $$20 joule$$, young's modulus of rubber is

A
$$10^{8} Nm^{-2}$$

B
$$2\times10^{8} Nm^{-2}$$

C
$$4\times10^{8} Nm^{-2}$$

D
$$1.5\times10^{6} Nm^{-2}$$

Solution

$$Energy =\dfrac{AY}{2L}l^{2}$$

$$20=\dfrac{4\times 10^{-6}\times Y\times \left ( \dfrac{10}{100} \right )^{2}}{2\times \dfrac{40}{100}}$$

$$Y=\dfrac{20\times 2\times 40}{4\times 10^{-6}\times \left ( \dfrac{10}{100} \right )^{2}\times 100}$$

$$=4\times 10^{8}N/m^{2}$$
Question 6
1 / -0

A steel wire of mass $$3.16 Kg$$ is stretched to a tensile strain of 1 x 10$$^{-3}$$. What is the elastic deformation energy if density $$\rho =7.9 g/cc$$ and Y$$=$$2x10$$^{11}$$ N/m$$^{2}$$?

A
$$4 KJ$$

B
$$0.4 KJ$$

C
$$0.04KJ$$

D
$$4J$$

Solution

$$Volume=\dfrac{mass}{\rho }$$

$$Y=\dfrac{stress}{strain}$$

$$stress=Y\times strain$$

$$energy=\dfrac{1}{2}\times stress\times strain\times volume$$

$$=\dfrac{1}{2}\times Y\times strain\times strain\times \dfrac{mass}{\rho }$$

$$=\dfrac{1}{2}\times 2\times 10^{11}\times 1\times 10^{-3}\times 1\times 10^{-3}\times \dfrac{3.16}{7.9\times 1000}$$

$$=0.04KJ.$$
Question 7
1 / -0

If the work done in stretching a wire by 1mm is 2 J, the work necessary for stretching another wire of same material but with double radius of cross section and half the length by 1mm is : (in joules)

A
16

B
8

C
4

D
1/4

Solution

Work done $$=\dfrac{AY}{2L}l^{2}$$$$=\dfrac{\pi r^{2}Yl^{2}}{2L}$$

Work done $$\propto \dfrac{r^{2}}{L}$$

$$\dfrac{w_{1}}{w_{2}}=\dfrac{r{_{1}}^{2}}{L_{1}}\times \dfrac{1/2L_{1}}{(2r_{1})^{2}}$$

$$\dfrac{w_{1}}{w_{2}}=\dfrac{1}{8}$$

$$w_{2}=8w_{1}$$
$$=8(2)$$
$$=16J$$
Question 8
1 / -0

When a wire of length 10m is subjected to a force of $$100 N$$ along its length , the lateral strain produced is $$0.01$$ x $$10$$$$^{-3}$$. The Poisson's ratio was found to be $$0.4$$. If the area of cross-section of wire is $$0.025m$$$$^{2}$$ , its Young's modulus is :

A
$$1.6 \times 10^{8} N /m^{2}$$

B
$$2.5 \times 10^{10} N /m^{2}$$

C
$$12.5 \times 10^{11} N /m^{2}$$

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

Solution

$$\text{Poisson's ratio} =\dfrac{\Delta d/d}{\Delta d_0/d_0}$$

$$\dfrac{\Delta l}{l}=\dfrac{\Delta d/d}{0.4}$$

$$=\dfrac{0.01\times 10^{3}}{0.4}$$

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

$$=\dfrac{100/0.025}{0.01\times 10^{3}/0.4}$$

$$=\dfrac{100\times 0.4}{0.025\times 0.01\times 10^{3}}$$

$$=1.6\times 10^{8}N/m^{2}$$
Question 9
1 / -0

The increase in length of a wire on stretching is $$0.025$$%. If its Poisson's ratio is $$0.4$$, then the percentage decrease in the diameter is :

A
$$0.01$$

B
$$0.02$$

C
$$0.03$$

D
$$0.04$$

Solution

$$\text{Poisson's ratio} =\dfrac{\Delta d/d}{\Delta l/l}$$

$$0.4=\dfrac{\Delta d/d}{0.025/100}$$

$$\dfrac{0.4\times 0.025}{100}=\Delta d/d$$

$$\Delta d/d=0.01$$ %
Question 10
1 / -0

When a wire is subjected to a force along its length, its length increases by $$0.4$$% and its radius decreases by $$0.2$$ %. Then the Poisons ratio of the material of the wire is

A
$$0.8$$

B
$$0.5$$

C
$$0.2$$

D
$$0.1$$

Solution

Poison's ratio $$=\dfrac{\Delta d/d}{\Delta r/r}$$

$$=\dfrac{0.2/100}{0.4/100}$$

$$=\dfrac{1}{2}=0.5$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mechanical Properties of Solids Test - 24

Result

Mechanical Properties of Solids Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

19.6 x 10$$^{8}$$ N/m$$^{2}$$

19.6 x 10$$^{-10}$$ N/m$$^{2}$$

19.6 x 10$$^{10}$$ N/m$$^{2}$$

19.6 x 10$$^{-8}$$ N/m$$^{2}$$

Solution

Question 2

$$3$$%

$$2.5$$%

$$1$$%

$$0.5$$%

Solution

Question 3

2 x 10$$^{3}$$

2 x 10$$^{4}$$

2.5 x 10$$^{10}$$

0.8 x 10$$^{4}$$

Solution

Question 4

2 x10$$^{8}$$Pa

4 x10$$^{8}$$Pa

6 x10$$^{8}$$Pa

8 x10$$^{8}$$Pa

Solution

Question 5

$$10^{8} Nm^{-2}$$

$$2\times10^{8} Nm^{-2}$$

$$4\times10^{8} Nm^{-2}$$

$$1.5\times10^{6} Nm^{-2}$$

Solution

Question 6

$$4 KJ$$

$$0.4 KJ$$

$$0.04KJ$$

$$4J$$

Solution

Question 7

16

8

4

1/4

Solution

Question 8

$$1.6 \times 10^{8} N /m^{2}$$

$$2.5 \times 10^{10} N /m^{2}$$

$$12.5 \times 10^{11} N /m^{2}$$

$$16 \times 10^{10} N /m^{2}$$

Solution

Question 9

$$0.01$$

$$0.02$$

$$0.03$$

$$0.04$$

Solution

Question 10

$$0.8$$

$$0.5$$

$$0.2$$

$$0.1$$

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.