A solid cylinder on moving with constant speed v0 reaches the bottom of an incline of 30° . A hollow cylinder of same mass and radius moving with the same constant speed v0 reaches the bottom of a different incline of inclination θ. There is no slipping and both of them go through the same distance in the same time; θ is then equal to

Question

A solid cylinder on moving with constant speed v0 reaches the bottom of an incline of 30° . A hollow cylinder of same mass and radius moving with the same constant speed v0 reaches the bottom of a different incline of inclination θ. There is no slipping and both of them go through the same distance in the same time; θ is then equal to (A) 37° (B) 30° (C) 42° (D) 45°

Tardigrade Admin · Accepted Answer

For solid cylinder, θ=30°, K2=(1/2) R2 For hollow cylinder, θ=?, K2=R2 Hence, ((1+(1/2))/ sin 30°)=(1+1/ sin θ) ∴ sin θ =(2/3)=0.6667 θ =42°

$3 7^{\circ}$

$3 0^{\circ}$

$4 2^{\circ}$

$4 5^{\circ}$

For solid cylinder, $θ = 3 0^{\circ}, K^{2} = \frac{1}{2} R^{2}$
For hollow cylinder, $θ = ?, K^{2} = R^{2}$
Hence,
$\frac{( 1 + \frac{1}{2} )}{s i n 3 0 ^{\circ}} = \frac{1 + 1}{s i n θ}$
$∴ sin θ = \frac{2}{3} = 0.6667$
$θ = 4 2^{\circ}$

37∘

30∘

42∘

45∘

For solid cylinder, θ=30∘,K2=21​R2 For hollow cylinder, θ=?,K2=R2 Hence, sin30∘(1+21​)​=sinθ1+1​ ∴sinθ=32​=0.6667 θ=42∘

$3 7^{\circ}$

$3 0^{\circ}$

$4 2^{\circ}$

$4 5^{\circ}$

For solid cylinder, $θ = 3 0^{\circ}, K^{2} = \frac{1}{2} R^{2}$
For hollow cylinder, $θ = ?, K^{2} = R^{2}$
Hence,
$\frac{( 1 + \frac{1}{2} )}{s i n 3 0 ^{\circ}} = \frac{1 + 1}{s i n θ}$
$∴ sin θ = \frac{2}{3} = 0.6667$
$θ = 4 2^{\circ}$