Question 553

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is

$a=\frac{g}{\mathrm{cos}ec\theta}$

$a=\frac{g}{\mathrm{sin}\theta}$

a = g tan θ

a = g cos θ

Solution

C.

a = g tan θ

In non-inertial frame,

N sin θ = ma ...(i)

N cos θ = mg... (ii)

tan θ = a/g

a = g tan θ

Question 554

A cylindrical tube of uniform cross-sectional area A is fitted with two air tight frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the pressure of the gas is p0 and temperature is T0, atmospheric pressure is also p_{0}. Now, the temperature of the gas is increased to 2T_{0}, the tension of the wire will be

2p

_{0}Ap

_{0}Ap

_{0}A/24p

_{0}A

Solution

B.

p_{0}A

The volume of the gas is constant i.e., V = constant

∴ p ∝ T i.e pressure will be doubled if the temperature is doubled.

Let F be the tension in the wire. Then, the equilibrium of anyone pipes gives.

$\mathrm{F}=(\mathrm{p}-{\mathrm{p}}_{0})\mathrm{A}=2({\mathrm{p}}_{0}-{\mathrm{p}}_{0})={\mathrm{p}}_{0}\mathrm{A}$

Question 555

A skier starts from rest at point A and slides down the hill without turning or breaking. The friction coefficient is μ. When he stops at point B, his horizontal displacement is S. what is the height difference between points A and B?

(The velocity of the skier is small so that the additional pressure on the snow due to the curvature can be neglected. Neglect also the friction of air and the dependence of μ on the velocity of the skier.)

h = μS

h = μ/S

h = 2μS

h = μS

^{2}

Solution

A.

h = μS

According to the question, the condition is shown in the figure

For a sufficiently safe-horizontal displacement △S can be considered straight. If the corresponding length of path element is △L, the friction force is given by μ_{mg(}△S/△L ).△L = μ_{mg}△S

Adding up, we find that along the whole path the total work done by the friction force is μ_{mg}s. By energy conservation, this must equal the decrease mgh in potential energy of skier.

Hence, h = μS

Question 556

A bicycle wheel rolls without slipping on the horizontal floor. Which one of the following is true about the motion of points on the rim of the wheel, relative to the axis at the wheel's centre?

Points near the top move faster than points near the bottom

Points near the bottom move faster than points near the top

All points on the rim move with the same speed

All points have the velocity vectors that are pointing in the radial direction towards the centre of the wheel

Solution

A.

Points near the top move faster than points near the bottom

We have, VA = 2v sinθ/2

Hence, the velocity of a point on rim increases with θ for 0° < θ < 180° and decreases with θ for 180° < θ < 30°.