## Why Air pressure in a car tyre increases during driving ?

Explain why:

(a) Air pressure in a car tyre increases during driving ?

(b)Two bodies at different temperatures T1 and T2 , if brought in thermal contact to not necessarily settle to the mean temperature (T1+T2)/2 ?

(c) The coolant in a chemical or nuclear plant (i.e. the liquid used to prevent different parts of a plant from getting too hot) should have high specific heat. Comment.

(d) The climate of a harbour town has more temperature (i.e. without extremes of heat and cold) than that of a town a desert at the same latitude, why?

Sol:

(a)Temperature of air inside the tyre increases due to motion during driving. Air pressure therefore, increase inside the tyre (by Charle’s law, P ∝ T)

(b)In thermal contact, heat from the body at higher temperature transfers to the body at lower temperature, till temperature of both becomes equal. Final temperature can be mean temperature. (T1+T2)/2 only when thermal capacities of two bodies are equal.

(c)Because heat absorbed ∝ specific heat of substance.

(d)In a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbor town is without extreme of hot and cold.

## Explain how the range of your car’s headlights limits the safe driving speed at night …

Q: Explain how the range of your car’s headlights limits the safe driving speed at night .

Sol:When a car moves , there is a rolling friction between the tyres and road . Rolling friction is smaller in comparison to sliding friction . When the brakes are applied , the wheels are clamped and hence arises enormous sliding friction . This sliding friction stops the car If S is the range of headlight , then the car must be stopped within this distance . Suppose M be the mass of the car and μs be the coefficient of sliding friction .
Then limiting frictional force = μs M g = M a ; Where ‘ a ‘ is the retardation produced .

a = μs g

But , $\displaystyle v_{safe \, speed}^2 = 2 a S$

$\displaystyle = 2 \mu_s g S$

$\displaystyle v_{safe \, speed} = \sqrt{2 \mu_s g S}$

## When A body explodes in mid-air. Why does its momentum remain conserved ?

Q : When a body explodes in mid-air . Why does its momentum remain conserved  ?

Sol: During explosion, the net force acting on the system is the weight of the system, that is mg, where m is the mass of the body. In this case we see the change in momentum in time Δt is given as

$\displaystyle \Delta P = \int_{0}^{\Delta t} F_{ext}dt$

ΔP = Fext Δt

Since Δt → 0 ⇒  ΔP→ 0

The momentum of the system first before and after the impact remains practically equal. Then the external force, that is gravitational force changes the momentum of the c.m. of the system for considerable time interval not during explosion.

## When walking on ice , why it is better to take short steps ?

Q: When walking on ice , why it is  better to take short steps ?

Sol: Ice is almost frictionless . While walking there must be sufficient friction to prevent slipping of feet . Frictional force depends on the normal reaction and normal reaction depends on the inclination of the reaction of ground to the vertical . If we take long steps the inclination of the reaction of ground to the vertical becomes large and hence frictional force is less . If we take short steps , the reaction remains close to vertical . This helps to develop more frictional force on a slippery road and makes walking possible . This is why while walking on ice it is better to take short steps .

## Why cricketer moves his hands backwards when holding a catch ?

Sol: When a cricketer holds a catch the impulse received at hand = F × t = change in linear momentum of the ball = constant. By moving the hand backward, the cricketer increases the time of impact (Since Force = Impulse/time , As time increases Force decreases )and reduces the force, so as the reaction, thereby reduces the changes of hurting severely.