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Two discs of same moment of inertia rotating about their regular axis passing through centre and $⊥ er$ to the plane of disc has angular velocities $ω_{1} and ω_{2}$ . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during his process is

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$\frac{1}{2} I {(ω_{1} - ω_{2})}^{2}$

$I {(ω_{1} - ω_{2})}^{2}$

$\frac{1}{8} I {(ω_{1} - ω_{2})}^{2}$

$\frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

answer is A.

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Detailed Solution

Loss in $KE = \frac{1}{2} \frac{I_{1} I_{2}}{I_{1} + I_{2}} {(ω_{1} - ω_{2})}^{2}$

here I₁ = I₂

$Loss in K . E = \frac{1}{2} \times \frac{I \times I}{2 I} {(ω_{1} - ω_{2})}^{2} = \frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

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