Mechanical Properties of Solids Test 3

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Mechanical Properties of Solids Test 3

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Weekly Quiz Competition

Question 1
1 / -0

An Indian rubber cube of side 10 cm has one side fixed while a tangential force of 1800 N is applied to the opposite side. Find the shear strain produced. Take η = 2 x 10⁶ N/ m².

A
0.9

B
0.009

C
0.09

D
9
Question 2
1 / -0

The length of the wire is increased by 1 mm on the application of a given load. In a wire of the same material but of length and radius twice that of the first, on application of the same force, extension produced is

A
0.25 mm

B
2 mm

C
4 mm

D
0.5 mm

Solution

Young's Modulus of elasticity =stress/strain
Y= [F/a/△l/l] or Y= [Fl/a△l]
or △l= [Fl/aY] = Fl/ πr²Y
In the given problem, △l∝ l/ r²
When both l and r are doubled, △l is halved.
Question 3
1 / -0

Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain. Which of the following is the unit of Young’s modulus?

A
dyne per square cm.

B
dyne cm²

C
dyne/cm

D
dyne cm

Solution

Stress = Force per area and strain is unit less.
Therefore in CGS units dyne per cm square.
Hence A is correct.
Question 4
1 / -0

The volume of a spherical body is decreased by 10^-3% when it is subjected to pressure of 40 atmospheres. Find the bulk modulus of body.
(1 atm = 1.01 x 10³ N/m²).

A
4.04 x 10¹¹ N/m²

B
4.04 x 10⁶N/m²

C
4.04 x 10³ N/m²

D
4.04 x 10² N/m²

Solution

We know that magnitude of bulk modulus
K=[P/(dV/V)]
Now, percentage change in volume is. 10^-3 %
Therefore, (dV/V)x100=10^-3
So, dV/V=10^-5
Hence, k is,
K=40atm/10^-5=40x1.01x10⁵ /10^-5 =4.04x10¹¹ N/m²
Question 5
1 / -0

The materials which have the same elastic properties in all directions are called __________.

A
Isotropic

B
Brittle

C
Homogeneous

D
Hard

Solution

Same elastic properties in all direction is called the homogenity of a material.
Question 6
1 / -0

How does the modulus of elasticity change with temperature?

A
It does not depend on temperature

B
It decreases with increase in temperature

C
It increases with increase in temperature

D
It sometimes increases, sometimes decreases with increase in temperature

Solution

As the temperature increases, the inter-atomic distance also increases. So, it results in an increase in the area (stress = force /area). As the area increases stress decreases, this results in a decrease in Young's modulus.
Question 7
1 / -0

What type of stress is produced in a body when the deforming force produces sheer strain?

A
Tangential stress

B
Longitudinal stress

C
Normal stress

D
shear stress

Solution

When the direction of the deforming force or external force is parallel to the cross-sectional area, the stress experienced by the object is called shearing stress or tangential stress. This results in the change in the shape of the body.
Question 8
1 / -0

Modulus of rigidity is defined as the

A
ratio of tangential stress to tangential strain

B
ratio of normal stress to sheer strain

C
ratio of normal stress to tangential strain

D
ratio of tangential stress to sheer strain

Solution

Modulus of rigidity is defined as the ratio of Shear stress to shear strain.
Question 9
1 / -0

Which of the following represents Hooke’s Law?

A
Stress = Estrain²

B
Stress = Estrain

C
Stress² = Estrain

D

Solution

E=stress/strain
Strain x E= Stress
Question 10
1 / -0

An iron bar of length l cm and cross section A cm² is pulled by a force F dynes from the ends so as to produce an elongation l cm. Which of the following statements is correct?

A
Elongation is directly proportional to area of cross-section A.

B
Elongation is inversely proportional to length

C
Elongation is inversely proportional to area of cross-section A.

D
Elongation is directly proportional to Young’s modulus.

Solution

F =AYΔl/L
Δl = FL/AY

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Mechanical Properties of Solids Test 3

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Mechanical Properties of Solids Test 3

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SHARING IS CARING

Question 1

0.9

0.009

0.09

9

Question 2

0.25 mm

2 mm

4 mm

0.5 mm

Solution

Question 3

dyne per square cm.

dyne cm2

dyne/cm

dyne cm

Solution

Question 4

4.04 x 1011 N/m2

4.04 x 106N/m2

4.04 x 103 N/m2

4.04 x 102 N/m2

Solution

Question 5

Isotropic

Brittle

Homogeneous

Hard

Solution

Question 6

It does not depend on temperature

It decreases with increase in temperature

It increases with increase in temperature

It sometimes increases, sometimes decreases with increase in temperature

Solution

Question 7

Tangential stress​

Longitudinal stress

Normal stress

shear stress

Solution

Question 8

ratio of tangential stress to tangential strain

ratio of normal stress to sheer strain

ratio of normal stress to tangential strain

ratio of tangential stress to sheer strain

Solution

Question 9

Stress = Estrain2

Stress = Estrain

Stress2 = Estrain

Solution

Question 10

Elongation is directly proportional to area of cross-section A.

Elongation is inversely proportional to length

Elongation is inversely proportional to area of cross-section A.

Elongation is directly proportional to Young’s modulus.

Solution

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dyne cm²

4.04 x 10¹¹ N/m²

4.04 x 10⁶N/m²

4.04 x 10³ N/m²

4.04 x 10² N/m²

Tangential stress

Stress = Estrain²

Stress² = Estrain