The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is $2 k g m^{- 2}$ .. It is rotating with an angular velocity $100 r a d s^{- 1}$ . A second identical disc rotating opposite to that of the first but with same angular velocity is joined to it coaxially. Now these two discs together continue to rotate about the same axis. Then the loss in KE in KJ is

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answer is B.

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Detailed Solution

1st body; $I_{c} = 2 k g m^{2}$
$L_{i} = 2 \times 100 = 200 k g m^{2} s^{- 1} f o r 2^{n d} b o d y$
$L_{2} = 2 \times (- 100) = - 200 k g m^{2} s^{- 1}$
( $∵$ rotating in opposite direction).
from L.O.C.A.M.; $L_{i} = L_{f}$
$L_{1} + L_{2} = (I_{1} + I_{2}) ω_{c} \Rightarrow 0 = 4. ω_{c} \Rightarrow ω_{c} = 0$
Loss in K.E. $= \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} I_{2} ω_{2}^{2} - \frac{1}{2} (I_{1} + I_{2}) ω_{c}^{2} = 20 J$

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