Two small balls ‘A’ and ‘B’ of masses 0.3 kg and 0.6 kg respectively are rigidly attached to the ends of a light rigid rod of length $R \sqrt{2}$ and placed vertically, inside a fixed smooth spherical cavity of radius R = 0.5 m with the ball ‘B’ located at the lowest point of cavity as shown in the figure. Then choose the correct option(s) immediately after the rod is released from rest. (Take $g = 10 m / s^{2}$ )

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The force acting on the ball ‘B’ due to light rod is $2 \sqrt{2} N .$

The normal force on the ball ‘A’ due to surface of cavity is 2N.

The normal force on the ball ‘B’ due to surface of cavity is 8N.

The normal force on the ball ‘A’ due to surface of cavity is 4N.

answer is A, C, D.

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Detailed Solution

Radial acceleration is zero for both A and B initially as their speeds at the initial moments are zero.

$∴ a_{A} = a_{B} = a_{\tan g e n t r a l} = a m_{A} = m = 0.3 k g m_{b} = 2 m = 0.6 k g B : N_{1} = 2 m g + \frac{F}{\sqrt{2}} ...... (i) \frac{F}{\sqrt{2}} = 2 m a ...... (i i) A : N_{2} = \frac{F}{\sqrt{2}} ...... (i i i) m g - \frac{F}{\sqrt{2}} = m a ...... (i v) F r o m (i i) a n d (i v) m g = 3 m a \Rightarrow a = \frac{g}{3} from (ii) = \frac{F}{\sqrt{2}} = 2 m a = \frac{2 m g}{3} = 2 ∴ F = 2 \sqrt{2} N ∴ N_{2} = \frac{F}{\sqrt{2}} = 2 N F r o m (i) N_{1} = 2 m g + \frac{F}{\sqrt{2}} = 2 m g + \frac{2 m g}{3} = \frac{8 m g}{3} = 8 N$