Important Questions Class 11 Physics Chapter 9

Class 11 Physics Chapter 9 Important Questions 

Important Questions for CBSE Class 11 Physics Chapter 9 – Mechanical Properties of Solids

Chapter 9 of Class 11 Physics discusses various laws and concepts related to solid bodies. Because the chapter is theoretically dense, students should first understand the concepts before trying to memorise them well.

To help students gain a better understanding of the chapter, Extramarks has provided Class 11 Physics Chapter 9 Important Questions and answers so that students can go through them and score well in their exam.

CBSE Class 11 Physics Chapter 9 Important Questions

Study Important Questions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

Here are some of the Important Questions for Class 11 Physics Chapter 9, click here to get the complete set of Class 11 Physics Chapter 9 Important Questions.

CBSE Class 11 Physics Chapter 9 Important Questions 

1 Mark Answers and Questions

Q.1 The stretching of a coil spring is determined by its shear modulus. Why?

Ans: The stretching of a coil spring is determined by its shear modulus, as when a coil spring is stretched, neither its length nor its volume nor its shape change.

Q.2 What are ductile and brittle materials?

Ans: Ductile materials are those that have a large plastic range beyond the elastic limit. Copper, iron, and other ductile materials are examples.

Brittle materials include glass, ceramics, graphite, and some alloys with extremely low plasticity, in which cracks can initiate without plastic deformation and can soon evolve into brittle breakage. Apparently, the micro-cutting model is not suitable for such materials.

2 Marks Answers and Questions

Q.1 Write the characteristics of displacement.

Ans: Following are the characteristics of displacement.

(1) Displacement is a vector quantity having both magnitude and direction.

(2) The displacement of a given body can be positive, negative, or zero.

Q.2 What cause the velocity of a particle to vary?

Ans: Variation in the velocity of a particle happens when:

(1) magnitude of velocity changes

(2) direction of motion changes.

Q.3 A hollow shaft is found to be stronger than a solid shaft made of the same material. Why?

Ans. The torque required to produce a given twist in a hollow cylinder is greater than that required to produce a given twist in a solid cylinder of the same length and material through the same angle. Hence, a hollow shaft is found to be stronger than a solid shaft made of equal material.

Q.4 An elastic wire is cut to half its original length. How would it affect the maximum load that the wire can support?

Ans: We know that,

Breaking load = Breaking Stress × Area Breaking load = Breaking Stress × Area

Hence, if the cable is cut to half of its original length, there will be no change in its area of cross-section. Therefore, there is no effect on the maximum load that the wire can support.

3 Marks Questions and Answers

Q.1 A police Jeep on patrol duty on a national highway was moving at a speed of 54km/hr. in the same direction. It finds a thief rushing up in a car at a rate of 126km/hr in the same direction. A police sub inspector fired at the car of the thief with his service revolver, which had a muzzle speed of100m/s. With what speed will the bullet hit the thief’s car?

Ans: In the above question it is given that,

VPJ = 54 km/hr = 15m/s

VTC = 126 km/hr = 35m/s

vb = 100m/s

Hence,

Velocity of car w.r.t. police, VCP = 35−15 = 20m/s

Velocity of bullet w.r.t. car, VBC = 100−20 = 80  m/s

Hence, the bullet will hit the car with a velocity of 80 m/s.

Q.2 Explain the following.

1) Elastic Body

2) Plastic Body

3) Elasticity.

Ans:

1. An elastic body is defined as a body that immediately returns to its original configuration after a deforming force is removed from it. For example, quartz and phosphor bronze.
2. A plastic body is defined as a body that does not regain its original configuration after removing a deforming force, regardless of the deforming force. Paraffin wax is an example.
3. Elasticity is defined as the ability of a body to return to its original configuration after deforming forces are removed.

Q.3. Why is the force of repulsion responsible for the formation of a solid and not the forces of attraction?

Ans: It has been observed that in the motion of a large number of spheres, two hard spheres do not attract each other but rebound immediately on collision, that is, they do not come closer than their diameter “d.”The interaction potential ‘V’ for a pair of hard spheres is where,

d is the diameter, and r is the distance of interaction between two spheres.

It indicates that there is infinite repulsion when r = d  and no potential when r>d and thus repulsive forces bind them together.

4 Marks Questions and Answers

Q1. The Mariana Trench is located in the Pacific Ocean, and at one place it is nearly 11 kms beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1×108Pa. A steel ball with an initial volume 0.32m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in volume of the ball when it reaches the bottom?

Ans: In the above question, it is given that:

The water pressure at the bottom of the trench is about 1.1×108Pa.

Initial volume of the steel ball is V = 0.32m3.

The bulk modulus of steel is 1.6×1011N/m2.

The ball is said to fall at the bottom of the Pacific Ocean, which is 11 km beneath the surface.

Consider the change in the volume of the ball on reaching the bottom of the trench to be ΔV.

Bulk modulus is given by the relation,

B= pΔVV

ΔV= pVB

⇒ΔV= 1.1×108×0.321.6×1011 = 2.2×10−4m3

Hence, the change in volume of the ball on reaching the bottom of the trench is 2.2×10−4m3.

5 Marks Questions and Answers

Q1. Calculate the volume contraction of a solid copper cube with an edge of 10 cm when subjected to a hydraulic pressure of 7×106Pa.

Ans: In the above question, it is given that,

The length of an edge of the solid copper cube is l = 10 cm = 0.1m.

Hydraulic pressure is p = 7×106Pa.

Bulk modulus of copper is B = 140×109Pa.

Bulk modulus is given by the relation,

B= pΔVV

Where,

ΔVV is the volumetric strain

ΔV is the change in volume.

V is the original volume.

ΔV = pVB

The original volume of the cube is V = l3.

ΔV = pl3B

⇒ΔV = 7106 0.13140109 = 5×10−8m3 = 5×10−2cm3

Clearly, the volume contraction of the solid copper cube is 5×10−2cm3.

Q.2 A steel wire of length 4.7 m and cross-sectional area 3.0×10−5m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0×10−5m2 of under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Ans: In the above question it is given that,

Length of the steel wire is  L1 = 4.7m.

Area of a cross-section of the steel wire is A1 = 3.0×10−5m2.

Length of the copper wire is L2 = 3.5m.

Area of a cross-section of the copper wire is A2 = 4.0×10−5m2 .

Now,

The change in length is given by:

ΔL = L1−L2 = 4.7−3.5 = 1.2m

Let the force applied in both cases be F.

Therefore, Young’s modulus of the steel wire is given by,

y1= FA1 L1L = F 4.73.010-51.2 …….(1)

And Young’s modulus of the copper wire is given by,

y2= FA2 L2L = F 3.54.010-51.2 …….(2)

Therefore,

y1y2 = F×4.7×4.0×10-51.23.010-5×1.2×F×3.5 = 1.791

Hence, the ratio of Young’s modulus of steel to that of copper is 1.79:1.