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The M.I of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 kg m². It is rotating with an angular velocity of 100 rad s^-1. A second disc of same mass and radius is joined to it coaxially. Now these two discs together continue to rotate about the same axis. Then the lose in kinetic energy in kilo joules is _____

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answer is 2.5.

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Detailed Solution

In this case the total angular momentum is conserved

$I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = 50 rad / s$

Final Kinetic energy= $\frac{1}{2} \times 2 \times 50^{2} = 2500 J$

Initial kinetic energy = $\frac{1}{2} \times 1 \times 100^{2} = 5000 J$

Loss in Kinetic energy = 2500J=2.5 KJ

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