A solid sphere rolls down without slipping on an inclined plane at angle `60^(@)` over a distance of 5 m. The acceleration (in `ms^(-2)`) is

Options

A. 5.23
B. 7.07
C. 6.06
D. 3.23

Answer

B

Explanation

Given, `theta=60^(@), l=5m, a=?`
The acceleration, `a=(g sin theta)/(1+(K^(2))/(R^(2)))`
For solid sphere, `K^(2)=(2)/(5)R^(2)=(g sin theta)/(1+(2)/(5))=(5 g sin theta)/(7)`
`rArr" "a=(5)/(7)xx9.8xxsin 60^(@)=(5)/(7)xx9.8xx(sqrt3)/(2)=6.06ms^(-1)`