Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

The radius of a copper wire is 4 mm. What force is required to stretch the wire by 20% of its length, assuming that the elastic limit is not exceeded (Y=$$12 \times 10^{10} N / m^2$$

A
$$7.23 \times 10^5 N$$

B
$$7.23 \times 10^7 N$$

C
$$7.23 \times 10^6 N$$

D
$$7.23 \times 10^8 N$$

Solution

We know that $$Y=\dfrac{F/A}{\delta L/L}=\dfrac{FL}{A \delta L}$$

Extension = L+0.2L=1.2 L. Substituting, we get,

$$F= 1.2YA=1.2 \times 12 \times 10^{10} \times (\pi 16 \times 10^{-6})=723.16 \times 10^4 N$$

The correct option is (c)
Question 2
1 / -0

A wire of length L can support a load W. If the wire is broken in to two equal parts , then how much load can be suspended by one of those cut wires?

A
Half

B
Same

C
Double

D
One fourth

Solution

Breaking force area of cross-section of wire i.e. load hold by the wire does not depend upon the length of the wire.
The correct option is (b)
Question 3
1 / -0

A rubber cord 10 m long is suspended vertically. How much does it stretch under its own weight. ( [Density of rubber is $$1500 (kg / m^3), Y = 5 \times 10^8 N/m^2)$$

A
0.3 mm

B
0.15 mm

C
0.015 mm

D
0.03 mm

Solution

$$\begin{array}{l}\text { According to the question, } \\\text { length of cord }=10 \mathrm{~m} \\\text { Density of rubber }=1500\mathrm{~kg}/ \mathrm{m}^{3} \\\text { young's Modulus }=5 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\end{array}$$

$$Stress = Weight/area = mg/A = (\rho A g)/A=\rho g$$ and
$$strain=\delta L/L \implies $$ change in length = $$\rho g L/Y$$
Also the length to be considered here is from the centre of gravity to the point of suspension, since the entire mass is assumed to be concentrated at the centre

Substituting the values, we get, extension = $$(1500 \times 10 \times 5)/(5 \times 10^8) =0.15 mm$$

The correct option is (b)
Question 4
1 / -0

Poisson' ratio is defined as the ratio of

A
longitudinal stress and longitudinal strain

B
longitudinal stress and lateral stress

C
lateral stress and longitudinal stress

D
lateral stress and lateral strain

Solution

Poisson' ratio is defined as the ratio of lateral stress and longitudinal stress

The correct option is (c)
Question 5
1 / -0

A student measures the poisson's ratio to be greater than 1 in an experiment. The meaning of this statement would be

A
An increase in length would also result in decrease in area of cross section of the wire

B
An increase in length would also result in increase in area of cross section of the wire

C
An decrease in length would also result in decrease in area of cross section of the wire

D
An increase in length will not change the area of cross section of the wire

Solution

Poisson's ratio = change in area / change in length. If poisson's ratio >1, then change in area > change in length. Thus area expands when length increases

The option (b) is the correct option
Question 6
1 / -0

When the temperature of a gas is $$20^0C$$ and pressure is changed from $$P_1=1.01\times 10^{5}\, Pa$$ to $$P_2=1.165\times 10^5\,Pa$$, then the volume changes by $$10$$%. The bulk modulus is

A
$$1.55\times10^5 \,Pa$$

B
$$1.05\times10^5 \,Pa$$

C
$$1.4\times10^5 \,Pa$$

D
$$0.115\times10^5 \,Pa$$

Solution

Bulk Modulus, $$B=\dfrac{\triangle P}{\triangle V/ V}$$
$$\dfrac{\triangle V}{V}\times 100$$% $$=10$$%
$$\dfrac{\triangle V}{V}=\dfrac{10}{100}=\dfrac{1}{10}$$
$$\triangle P=(1.165-1.01)\times10^5\,Pa$$
$$=0.155\times10^5 \,Pa$$
$$B=\dfrac{0.155\times 10^5}{1/10}\,Pa$$
$$=1.55\times 10^5\,Pa$$
Question 7
1 / -0

The theoretical limits of poisson's ratio lies between -1 to 0.5 because

A
Shear modulus and bulk's modulus should be positive

B
Bulk's modulus is negative during compression

C
Shear modulus is negative during compression

D
Young's modulus should be always positive

Solution

Let Y, K, n and $$\sigma$$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively.
Y = 3K (1 - 2$$\sigma$$) [Standard formula]
Y = 2n (1 + $$\sigma$$) [Standard formula]
Hence, 3K (1 - 2$$\sigma$$) = 2n (1 + $$\sigma$$)
Now K and n are always positive, so
i) If $$\sigma$$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$$\sigma$$ < 1 or $$\sigma$$ <1/2
ii) If $$\sigma$$ be -ve, then LHS will always be +ve. Therefore, 1+$$\sigma$$ > 0 or $$\sigma$$ > -1
Thus the limiting values of Poisson's ratio are -1 < $$\sigma$$ < 1/2

The correct option is (a)
Question 8
1 / -0

A $$1$$m long metal wire of cross sectional area $$10^{-6}m^2$$ is fixed at one end from a rigid support and a weight W is hanging at its other end. The graph shows the observed extension of length $$\Delta l$$ of the wire as a function of W. Young's modulus of material of the wire in SI units is?

A
$$5\times 10^4$$

B
$$2\times 10^5$$

C
$$2\times 10^{11}$$

D
$$5\times 10^{11}$$
Question 9
1 / -0

A uniform wire (Young's modulus $$2\times 10^{11}Nm^{-2}$$ ) is subjected to longitudinal tensile stress of $$5\times 10^7\ Nm^{-2}$$. If the overall volume change in the wire is $$0.02\%$$, the frictional decrease in the radius of the wire is close to

A
$$1.0 \times 10^{-4}$$

B
$$1.5 \times 10^{-4}$$

C
$$0.25 \times 10^{-4}$$

D
$$5 \times 10^{-4}$$

Solution

$$\dfrac{\triangle V}{V}=\dfrac{\triangle l}{l}+ \dfrac{2\triangle r}{r}, \dfrac{\triangle l}{l}= \dfrac{stress}{Y} = 2.5 \times 10^{-4}$$
Question 10
1 / -0

The ratio of the coefficient of volume expansion of glass container to that of a viscous liquid kept inside the container is $$1:4$$. What fraction of the inner volume of the container should the liquid occupy so that the volume of the remaining vacant space will be same at all the temperature?

A
$$2/5$$

B
$$1/4$$

C
$$1/64$$

D
$$1/8$$

Solution

We know that,

When there is no change in liquid level in vessel then,
$$\gamma _{real}^{'}=\gamma _{vessel}^{'}$$

Change in volume in liquid relative to vessel

$$ \Delta {{V}_{app}}=V\gamma _{app}^{'}\Delta \theta $$

$$ \Delta {{V}_{app}}=V\left( \gamma _{real}^{'}-\gamma _{vessel}^{'} \right) $$

Hence, the fraction is $$1:4$$

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Mechanical Properties of Solids Test - 54

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Mechanical Properties of Solids Test - 54

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Question 1

$$7.23 \times 10^5 N$$

$$7.23 \times 10^7 N$$

$$7.23 \times 10^6 N$$

$$7.23 \times 10^8 N$$

Solution

Question 2

Half

Same

Double

One fourth

Solution

Question 3

0.3 mm

0.15 mm

0.015 mm

0.03 mm

Solution

Question 4

longitudinal stress and longitudinal strain

longitudinal stress and lateral stress

lateral stress and longitudinal stress

lateral stress and lateral strain

Solution

Question 5

An increase in length would also result in decrease in area of cross section of the wire

An increase in length would also result in increase in area of cross section of the wire

An decrease in length would also result in decrease in area of cross section of the wire

An increase in length will not change the area of cross section of the wire

Solution

Question 6

$$1.55\times10^5 \,Pa$$

$$1.05\times10^5 \,Pa$$

$$1.4\times10^5 \,Pa$$

$$0.115\times10^5 \,Pa$$

Solution

Question 7

Shear modulus and bulk's modulus should be positive

Bulk's modulus is negative during compression

Shear modulus is negative during compression

Young's modulus should be always positive

Solution

Question 8

$$5\times 10^4$$

$$2\times 10^5$$

$$2\times 10^{11}$$

$$5\times 10^{11}$$

Question 9

$$1.0 \times 10^{-4}$$

$$1.5 \times 10^{-4}$$

$$0.25 \times 10^{-4}$$

$$5 \times 10^{-4}$$

Solution

Question 10

$$2/5$$

$$1/4$$

$$1/64$$

$$1/8$$

Solution

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