1.

A chain of length `l = pi R//4` is placed. On a smooth hemispherical surface of radius `R` with one of its ends fixed at the top of the sphere. Mass of chain is `sqrt(pi kg)` and `R = 1m.(g = 10 m//s^(2))`. The tangential acceleration of the chain when its starts sliding down.A. `(40)/(pi)(1-(1)/(sqrt(2)))`B. `(20)/(pi)(1-(1)/(sqrt(2)))`C. `10(1-(1)/(sqrt(2)))`D. zero

Answer» Correct Answer - A
Tangential force on differential element
`dF_(t) = dmg sin theta`
`F_(t)=int_(0)^(pi//4) ((m)/(l) R d theta)g sin theta = (gR)/(l)[1 -cos 45^@)`.


Discussion

No Comment Found

Related InterviewSolutions