A solid sphere rolls down without slipping on an inclined plane at angle 60° over a distance of 10 m. The acceleration (in ms -2 ) is

Question

A solid sphere rolls down without slipping on an inclined plane at angle 60° over a distance of 10 m. The acceleration (in ms -2 ) is (A) 4 (B) 5 (C) 6.06 (D) 7

Tardigrade Admin · Accepted Answer

Here, θ=60°, l=10 m , a =? For solid sphere, K2=(2/5) R2 ∴ a=(9.8 sin 60°/1+(2)5) (∵ a =(g sin θ/1+(K2)R2)) Or a=(5/7) × 9.8 × (√3/2)=6.06 ms -2

Q. A solid sphere rolls down without slipping on an inclined plane at angle $6 0^{\circ}$ over a distance of $10 m$ . The acceleration (in $m s^{- 2}$ ) is

4

5

6.06

7

Here, $θ = 6 0^{\circ}, l = 10 m, a = ?$
For solid sphere, $K^{2} = \frac{2}{5} R^{2}$
$∴ a = \frac{9.8 s i n 6 0 ^{\circ}}{1 + \frac{2}{5}}$
$(∵ a = \frac{g s i n θ}{1 + \frac{K ^{2}}{R ^{2}}})$
Or $a = \frac{5}{7} \times 9.8 \times \frac{3}{2} = 6.06 m s^{- 2}$

Q. A solid sphere rolls down without slipping on an inclined plane at angle 60∘ over a distance of 10m. The acceleration (in ms−2 ) is

4

5

6.06

7

Here, θ=60∘,l=10m,a=? For solid sphere, K2=52​R2 ∴a=1+52​9.8sin60∘​ (∵a=1+R2K2​gsinθ​) Or a=75​×9.8×23​​=6.06ms−2

Q. A solid sphere rolls down without slipping on an inclined plane at angle $6 0^{\circ}$ over a distance of $10 m$ . The acceleration (in $m s^{- 2}$ ) is

Here, $θ = 6 0^{\circ}, l = 10 m, a = ?$
For solid sphere, $K^{2} = \frac{2}{5} R^{2}$
$∴ a = \frac{9.8 s i n 6 0 ^{\circ}}{1 + \frac{2}{5}}$
$(∵ a = \frac{g s i n θ}{1 + \frac{K ^{2}}{R ^{2}}})$
Or $a = \frac{5}{7} \times 9.8 \times \frac{3}{2} = 6.06 m s^{- 2}$