Question

A horizontal bar of mass `m_(1)` Prism of mass `m_(2)` can move as shown. There is no friction at any contact point. During the motion the length of the rod is always horizontal Now, magnitude valuse of
.
A. Acceleration of `m_(1)` is `g//(1 + eta cot^(2)theta)` where `eta =m_(2)//m_(1)` .
B. Acceleration of `m_(1)` is `(g tan theta)/(eta[1+tan^(2)theta]` where `eta =m_(2)//m_(1)` .
C. Acceleration of `m_(2)` is `g//(tan theta + eta cot theta)` where `eta =m_(2)//m_(1)` .
D. Acceleration of `m_(2)` is`(g tan^(2) theta)/(eta[1+tan^(2)theta]` where `eta =m_(2)//m_(1)` .

Answer 1

Correct Answer - A::C
From `F.B.D m_(1) g - N cos theta = m_(1) a -----(1)`
`N sin theta = m_(2) -----(2)`
also by constrain equation `A sin theta = a cos theta---(3)`
on solving `(1),(2)` and (3)
`a=(g)/(1+etacot^(2)theta)` where `eta=(m_(2))/(m_(1))`
`A = (g)/(tan theta + eta cot theta)`, where `eta=(m_(2))/(m_(1))`
.