A rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at angle of 30° with the horizontal. Assuming that g=9.81 ms -2, time taken is

Question

A rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at angle of 30° with the horizontal. Assuming that g=9.81 ms -2, time taken is (A) 0.68 s (B) 0.8 s (C) 0.5 s (D) 0.7 s

Tardigrade Admin · Accepted Answer

Here, l=1 m, θ=30°, g =9.81 ms -2, t= ? ∵ t=√(2 l(I+K2 / R2)/g sin θ) For a rupee coin, K2=(1/2) R2 ⇒ t=√(2 × 1(1+1 / 2)/9.81 sin 30°) =√(6/9.81)=0.78 s

Q. A rupee coin starting from rest rolls down a distance of $1 m$ on an inclined plane at angle of $3 0^{\circ}$ with the horizontal. Assuming that $g = 9.81 m s^{- 2}$ , time taken is

0.68 s

0.8 s

0.5 s

0.7 s

Here, $l = 1 m, θ = 3 0^{\circ}, g = 9.81 m s^{- 2}, t =$ ?
$∵ t = \frac{2 l ( I + K ^{2} / R ^{2} )}{g s i n θ}$
For a rupee coin, $K^{2} = \frac{1}{2} R^{2}$
$\Rightarrow t = \frac{2 \times 1 ( 1 + 1/2 )}{9.81 s i n 3 0 ^{\circ}}$
$= \frac{6}{9.81} = 0.78 s$

Q. A rupee coin starting from rest rolls down a distance of 1m on an inclined plane at angle of 30∘ with the horizontal. Assuming that g=9.81ms−2, time taken is