A solid sphere rolls without slipping down a plane inclined at 30- to the horizontal- The distance traveled in -7 seconds after starting from rest is g -10 ms -2 A- 12-5 mB- 6-25 mC- 5-5 mD- 15 m

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Acceleration Due to Gravity

A solid spher...

Question

A solid sphere rolls (without slipping) down a plane inclined at 30^o to the horizontal. The distance traveled in √7 seconds after starting from rest is (g = 10 ms^-2)

12.5 m

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6.25 m

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15 m

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5.5 m

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Solution

The correct option is A
12.5 m

Acceleration of a body rolling down an incline of angle θ is

a = (g sinθ)/[1 + (k²R²)]

For a sphere, k²/R² = 2/5

Hence for a sphere, a = (g sinθ)/[1 + 2/5]

Or, a = (10 x 1/2)/(7/5)

Or, a = 25/7

We have, s = ut +1/2 at²= 0 + 1/2 x (25/7) x 7 = 12.5m

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A solid sphere rolls (without slipping) down a plane inclined at 30o to the horizontal. The distance traveled in √7 seconds after starting from rest is (g = 10 ms-2)

The correct option is A 12.5 m Acceleration of a body rolling down an incline of angle θ is a = (g sinθ)/[1 + (k2R2)] For a sphere, k2/R2 = 2/5 Hence for a sphere, a = (g sinθ)/[1 + 2/5] Or, a = (10 x 1/2)/(7/5) Or, a = 25/7 We have, s = ut +1/2 at2= 0 + 1/2 x (25/7) x 7 = 12.5m

A solid sphere rolls (without slipping) down a plane inclined at 30^o to the horizontal. The distance traveled in √7 seconds after starting from rest is (g = 10 ms^-2)

The correct option is A
12.5 m

Acceleration of a body rolling down an incline of angle θ is

a = (g sinθ)/[1 + (k²R²)]

For a sphere, k²/R² = 2/5

Hence for a sphere, a = (g sinθ)/[1 + 2/5]

Or, a = (10 x 1/2)/(7/5)

Or, a = 25/7

We have, s = ut +1/2 at²= 0 + 1/2 x (25/7) x 7 = 12.5m