A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination 30° with horizontal. The acceleration of train up the plane is a=g / 2. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is

Question

A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination 30° with horizontal. The acceleration of train up the plane is a=g / 2. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is (A) 30° (B)  tan -1(2 / √3) (C)  tan -1(√3 / 2) (D)  tan -1(2)

Tardigrade Admin · Accepted Answer

T sin θ-m g sin 30°=m a ⇒ T sin θ=m g sin 30°+m g / 2 ... (i) T cos θ=m g cos 30° ...(ii) Dividing Eqs. (i) by (ii), we get tan θ=(2/√3)

Q. A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination $3 0^{\circ}$ with horizontal. The acceleration of train up the plane is $a = g /2$ . The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is

$3 0^{\circ}$

$tan^{- 1} (2/ 3)$

$tan^{- 1} (3 /2)$

$tan^{- 1} (2)$

$T sin θ - m g sin 3 0^{\circ} = ma$
$\Rightarrow T sin θ = m g sin 3 0^{\circ} + m g /2$ ... (i)
$T cos θ = m g cos 3 0^{\circ}$ ...(ii)
Dividing Eqs. (i) by (ii), we get
$tan θ = \frac{2}{3}$

Q. A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination 30∘ with horizontal. The acceleration of train up the plane is a=g/2. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is

30∘

tan−1(2/3​)

tan−1(3​/2)

tan−1(2)