Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 41

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Question 1
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A copper wire having $$Y=1\times 10^{11}N/m^{2}$$ with length $$6m$$ and a steel wire having $$Y=2\times 10^{11}N/m^{2}$$ with length $$4 m$$ each of cross section $$10^{-5}m^{2}$$ are fastened end to end stretched by a tension of $$100 N$$. The elongation produced in the copper wire is :

A
$$0.2 mm$$

B
$$0.4 mm$$

C
$$0.6 mm$$

D
$$0.8 mm$$

Solution

Elongation.$$\Delta l = \dfrac{FL}{YA}$$

$$\Delta l = \dfrac{100 \times 6}{10^{11} \times 10^{-5}} + \dfrac{100 \times 4}{2 \times 10^{11} \times 10^{-5}} = 0.8mm$$
Question 2
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If a rubber ball is taken at the depth of 200 m in a pool its volume decreases by 0.1% If the density of the water is $$\displaystyle 1\times 10^{3}kg/m^{3}\: $$and$$\: g=10m/s^{2}$$ then the volume elasticity in $$\displaystyle N/m^{2}$$ will be

A
$$\displaystyle 10^{8}$$

B
$$\displaystyle 2\times 10^{8}$$

C
$$\displaystyle 10^{9}$$

D
$$\displaystyle 2\times 10^{9}$$

Solution

Pressure change on the rubber ball at the depth of $$200m$$,
$$P=200 \times 1 \times{10}^{3} \times 10$$
$$2 \times {10}^{6}N/{m}^{2}.$$
$$\therefore$$ fractional change in volume,
$$\cfrac{\triangle V}{V}=\cfrac{0.1}{100}$$
$$\therefore$$ Volume elasticity$$(K)=\cfrac{p}{\cfrac{\triangle V}{V}}$$
$$=\cfrac{2 \times {10}^{6}}{\cfrac{0.1}{100}}$$
$$=2 \times {10}^{9} N/{m}^{2}$$.
Question 3
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One end of a uniform rope of length L and of weight w is attached rigidIy to a point in the roof and a weight w$$_1$$ is suspended from its lower. If s is the area of cross-section of the wire, the stress in the wire at a height $$\displaystyle \frac{3 L}{4}$$ from its lower end is:

A
$$\displaystyle \frac{w}{s}$$

B
$$\displaystyle \frac{\displaystyle w_1 + \frac{w}{4}}{s}$$

C
$$\displaystyle \frac{\displaystyle w_1 + \frac{2w}{4}}{s}$$

D
$$\displaystyle \frac{\displaystyle w_1 + w}{s}$$

Solution

Weight of the rope = $$w$$

Weight at the point A $$= \displaystyle \frac{w}{4}$$

Area of cross section = $$s$$

Downward force of A $$= \displaystyle \frac{w}{4} + w_1$$

Stress in the wire at A $$= \displaystyle \frac{\text{force}}{\text{area of cross section}} = \frac{\displaystyle \frac{w}{4} + w_1}{s}$$
Question 4
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The high domes of ancient buildings have structural value (besides beauty). It arises from pressure difference on the two faces due to curvature (as in soap bubbles). There is a dome of radius 5 m and uniform (but small) thickness. The surface tension of its masonry structure is about 500 N/m. Treated as hemispherical, the maximum load the dome can support is nearest to

A
1500 kg wt.

B
3000 kg wt.

C
6000 kg wt.

D
12000 kg wt.

Solution

$$\begin{array}{l}\quad P_{\text {ext }}=\frac{4 T}{R}\\\text { Max. Load that can be supported }\\\qquad\begin{aligned}F&=\operatorname{P_{ext}} \times \pi r^{2} \\&=\frac{4 T}{R}\times \pi R^{2}=4 T \times \pi R\end{aligned}\end{array}$$

$$\begin{aligned}F &=4 \times 500 \times \frac{22}{7} \times 5 \\&=2000 \times 5 \times \frac{22}{7}=31400\mathrm{~N}\\F\text { in } \mathrm{kg} \omega t=\frac{31400}{\mathrm{~g}} \mathrm{~N} \approx3000&,\therefore \text {Option (B) is correct}\end{aligned}$$
Question 5
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A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the Bulk modulus of the ball material? (Take density of water = 1000 kg/m$$^3$$):

A
$$19.6 \times 10^8 N/m^2$$

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

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

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

Solution

$$\displaystyle \frac{\Delta V}{V} = \frac{0.1}{100},$$

$$p = h \rho g = 200 \times 1000 \times 9.8$$

$$= 19.6 \times 10^5$$

$$K = \displaystyle \frac{P}{\displaystyle \frac{\Delta V}{V}} = \frac{19.6 \times 10^5}{\displaystyle \frac{0.1}{100}} = 19.6 \times 10^5 \times 10^3$$

$$= 19.6 \times 10^8 N/m^2$$
Question 6
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Find out Bulk stress on the spherical object of radius $$\displaystyle \frac{10}{\pi }$$ cm if area and mass of piston is 50 $$\displaystyle cm^{2}$$ and 50 kg respectively for a cylinder filled with gas:

A
$$4\times 10^{5}N/m^{2}$$

B
$$6\times 10^{5}N/m^{2}$$

C
$$2\times 10^{5}N/m^{2}$$

D
$$5\times 10^{5}N/m^{2}$$

Solution

$$\displaystyle P_{gas}=\dfrac{mg}{A}+p_{a}=\dfrac{50\times 10}{50\times 10^{-4}}+1\times 10^{5}=2\times 10^{5}=2\times 10^{5}N/m^{2}$$
Bulk stress = $$\displaystyle P_{gas}=2\times 10^{5}N/m^{2}$$
Question 7
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With what minimum acceleration can a fireman slide down a rope whose breaking strength is 3/4 th of his weight ?

A
1/4 g

B
1/2 g

C
3/4 g

D
zero

Solution

Here, $$\displaystyle mg-T=ma$$
$$\displaystyle \Rightarrow \quad mg-\frac { 3 }{ 4 } mg=ma\\ $$
$$\displaystyle \therefore \quad a={ g }/{ 4 }$$
Question 8
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Two wires of the same material and length but diameter in the ratio 1 : 2 are stretched by the same force. The ratio of potential energy per unit volume for the two wires when stretched will be :

A
1 : 1

B
2 : 1

C
4 : 1

D
16 : 1

Solution

Hint:
When a wire of length$$L$$, Diameter$$D$$, Young's Modulus of rigidity$$Y$$ is stretched by a Force $$F$$ from both the sides, then Potential Energy per unit volume stored in the wire is,
$$E = \dfrac{8F^2}{Y \pi^2 D^{4}}$$

Step 1: Calculate Energy stored in both the wires given in the question.
It is given that two wires of same length and material but diameter in ratio $$1:2$$ are stretched by same force. Let diameter of first Wire is $$D$$, then diameter of second wire will be $$2D$$.
Potential Energy per unit volume stored in first wire is,
$$E_1 = \dfrac{8F^2}{Y \pi^(2 D)^{4}}$$. ....eq.1

Potential Energy per unit volume stored in second wire is,
$$E_2 = \dfrac{8F^2}{Y \pi^2 {2D}^{4}}$$

$$E_2 = \dfrac{8F^2}{16Y \pi^2 {D}^{4}}$$. ....eq.2

Step 2: Calculate the ratio of Energies Stored.
By dividing eq.1 and eq.2, we get

$$\dfrac{E_1}{E_2} = \dfrac{\dfrac{8F^2}{Y \pi^2 D^{4}}}{\dfrac{8F^2}{16Y \pi^2 D^4}}$$

$$\Rightarrow \dfrac{E_1}{E_2} = \dfrac{16}{1}$$

Thus, ratio of Potential Energy per unit volume stored in the wires is $$16:1$$.
Option D is correct.
Question 9
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Directions For Questions

These questions consist of two statements, each printed as assertion and reason. While answering these question you are required to choose anyone of the following five responses.

...view full instructions

Assertion: Stress is the internal force per unit area of a body.
Reason: Rubber is more elastic than stee

A
If both assertion and reason are true but the reason is the correct explanation of assertion.

B
If both assertion and reason are true but the reason is not the correct explanation of assertion.

C
If assertion is true but reason is false.

D
If both the assertion and reason are false.

E
If reason is true but assertion is false.

Solution

Stress is internal force (restoring force) developed with the body of the object. Since it is easier to stretch rubber so it is less stressful and therefore less elastic.
Question 10
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If in a wire of Young's modulus Y, longitudinal strain X is produced then the potential energy stored in its unit volume will be :

A
$$\displaystyle 0.5Y{ X }^{ 2 }$$

B
$$\displaystyle 0.5{ Y }^{ 2 }X$$

C
$$\displaystyle 2Y{ X }^{ 2 }$$

D
$$\displaystyle Y{ X }^{ 2 }$$

Solution

Potential energy stored per unit volume of a wire
$$\dfrac{1}{2}stress \times strain$$
$$Y = \dfrac{stress}{strain}$$
$$stress = Y \times strain $$
$$U = \dfrac{1}{2} \times YX \times X = \dfrac{1}{2}YX^2$$

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Mechanical Properties of Solids Test - 41

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Mechanical Properties of Solids Test - 41

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Question 1

$$0.2 mm$$

$$0.4 mm$$

$$0.6 mm$$

$$0.8 mm$$

Solution

Question 2

$$\displaystyle 10^{8}$$

$$\displaystyle 2\times 10^{8}$$

$$\displaystyle 10^{9}$$

$$\displaystyle 2\times 10^{9}$$

Solution

Question 3

$$\displaystyle \frac{w}{s}$$

$$\displaystyle \frac{\displaystyle w_1 + \frac{w}{4}}{s}$$

$$\displaystyle \frac{\displaystyle w_1 + \frac{2w}{4}}{s}$$

$$\displaystyle \frac{\displaystyle w_1 + w}{s}$$

Solution

Question 4

1500 kg wt.

3000 kg wt.

6000 kg wt.

12000 kg wt.

Solution

Question 5

$$19.6 \times 10^8 N/m^2$$

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

$$19.6 \times 10^{10} N/m^2$$

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

Solution

Question 6

$$4\times 10^{5}N/m^{2}$$

$$6\times 10^{5}N/m^{2}$$

$$2\times 10^{5}N/m^{2}$$

$$5\times 10^{5}N/m^{2}$$

Solution

Question 7

1/4 g

1/2 g

3/4 g

zero

Solution

Question 8

1 : 1

2 : 1

4 : 1

16 : 1

Solution

Question 9

If both assertion and reason are true but the reason is the correct explanation of assertion.

If both assertion and reason are true but the reason is not the correct explanation of assertion.

If assertion is true but reason is false.

If both the assertion and reason are false.

If reason is true but assertion is false.

Solution

Question 10

$$\displaystyle 0.5Y{ X }^{ 2 }$$

$$\displaystyle 0.5{ Y }^{ 2 }X$$

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

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

Solution

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