Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal is α. The sphere is heated a little by a temperature ?T so that its new temperature is T+ ?T. The increase in the volume of the sphere is approximately

(a) 2παR?T

(b) πR²α?T

(c) 4πR³?T/3

(d) 4παR³ $\mathit{\alpha }$ ?T

Heat is associated with

(a) Kinetic energy of random motion of molecules

(b) Kinetic energy of orderly motion of molecules

(c) Total kinetic energy of random and orderly motion of molecules

(d) Kinetic energy of random motion in some cases and kinetic energy of orderly motion in other

As the temperature is increased, the time period of a pendulum

(a) Increases as its effective length increases even though its centre of mass still remains at the centre of the bob

(b) Decreases as its effective length increases even though its centre of mass still remains at the centre of the bob

(c) Increases as its effective length increases due to shifting of centre of mass below the centre of the bob

(d) Decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

An aluminium sphere is dipped into water. Which of the following is true?

(a) Buoyancy will be less in water at 0°C than that in water at 4°C.

(b) Buoyancy will be more in water at 0°C than that in water at 4°C.

(c) Buoyancy in water at 0°C will be same as that in water at 4°C.

(d) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.

The graph between two temperature scales A and B is shown in Fig. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

(a) $\frac{{\mathit{t}}_{\mathit{A}}-180}{100}=\frac{{\mathit{t}}_{\mathit{B}}}{150}$

(b) $\frac{{\mathit{t}}_{\mathit{A}}-30}{150}=\frac{{\mathit{t}}_{\mathit{B}}}{100}$

(c) $\frac{{\mathit{t}}_{\mathit{B}}-180}{150}=\frac{{\mathit{t}}_{\mathit{A}}}{100}$

(d) $\frac{{\mathit{t}}_B-40}{100}=\frac{{\mathit{t}}_{\mathit{A}}}{180}$

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

(a) Its speed of rotation increases

(b) Its speed of rotation decreases

(c) Its speed of rotation remains same

(d) Its speed increases because its moment of inertia increases

A bimetallic strip is made of aluminium and steel (α_Al > α_steel ) . On heating, the strip will

(a) Remain straight

(b) Get twisted

(c) Will bend with aluminium on concave side

(d) Will bend with steel on concave side

One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 minutes before he could take bath. Now he had two options:

(i) Fill the remaining bucket completely by cold water and then attend to the work,

(ii) First attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer ? Explain

100 g of water is supercooled to –10° C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze? [S_w=1cal/g/ $°\mathit{C}$  and L_fusion= 80cal/g]

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of P–T diagram of water.