A simple pendulum mounted on a car. The car starts to accelerate on horizontal surface …

Q: A simple pendulum mounted on a car. The car starts to accelerate on horizontal surface with constant acceleration a0 = g/√3 . The maximum deflection of pendulum from vertical is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Click to See Answer :
Ans: (a)

Sol: Let deflection of pendulum from vertical is θ

If T be the tension in the String

T cosθ = m g …(i)

T sinθ = m a0 …(ii)

On dividing (ii) by (i)

$\displaystyle tan\theta = \frac{a_0}{g} $

$\displaystyle tan\theta = \frac{g/\sqrt{3}}{g} = \frac{1}{\sqrt{3}}$

θ = 30°

 

A small block is placed at the top of sphere. It slides on the smooth surface of the sphere. What will be the angle made by the radius vector of the block with the horizontal, when it leaves the surface?

Q: A small block is placed at the top of sphere. It slides on the smooth surface of the sphere. What will be the angle made by the radius vector of the block with the horizontal, when it leaves the surface?

(a) 52°

(b) 48°

(c) 42°

(d) 38°

Click to See Answer :
Ans: (c)

 

The string of a pendulum of length l and bob of mass m is displaced through 90°….

Q:The string of a pendulum of length l and bob of mass m is displaced through 90°. Minimum strength of the string in order to withstand the tension should be.

(a) mg

(b) 2 mg

(c) 3 mg

(d) 4 mg

Click to See Answer :
Ans: (c)

At mean position body is in vertical equilibrium

$ T – \frac{m v^2}{l} = m g$ …(i)

$ v^2 = 2 g l $ …(ii)

from (i) & (ii)

$T – 2 m g = m g$

T = 3 m g

 

A ball of mass 0.6 kg attached to a light inextensible string rotates in a vertical circle …

Q: A ball of mass 0.6 kg attached to a light inextensible string rotates in a vertical circle of radius 0.75 m such that it has a speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is (g = 10 m s-2)

(a) 30 N

(b) 26 N

(c) 20 N

(d) 6 N

Click to See Answer :
Ans: (c)

Force at equilibrium ,

$T = \frac{m v^2}{r} $

$T = \frac{0.6 \times 5^2}{0.75} $

T = 20 N

 

A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m….

Q: A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical?

(a) 30°

(b) 45°

(c) 10.5°

(d) 26.5°

Click to See Answer :
Ans: (d)
$tan\theta = \frac{v^2}{r g}$

$tan\theta = \frac{9.8^2}{19.6 \times 9.8} $

$tan\theta = \frac{1}{2} $

$\theta = 26.5^o$