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The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 $kg - m^{2}$ . It is rotating with an angular velocity 100 radian/sec. Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilojoules is:

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Detailed Solution

$I_{1} = 1 kg m^{2}, ω_{1} = 100 rad / \sec$

Since the mass is doubled,

Final moment of inertia together, $I_{2} = \frac{M_{2} R^{2}}{2} = \frac{2 M . R^{2}}{2} = 2 I_{1} = 2 {kgm}^{2}$

Conservation of angular momentum,

$I_{1} ω_{1} = I_{2} ω_{2} or 1 \times 100 = 2 \times ω_{2}$

or $ω_{2} = 50 rad / \sec$

$E_{1} = \frac{1}{2} I_{1} {ω_{1}}^{2} = \frac{1}{2} \times 1 \times {(100)}^{2} = 5 \times 10^{3} J$

$E_{2} = \frac{1}{2} I_{2} {ω_{2}}^{2} = \frac{1}{2} \times 2 \times {(50)}^{2} = 2.5 \times 10^{3} J .$

Loss in KE = $E_{1} - E_{2} = 5 \times 10^{3} - 2.5 \times 10^{3} = 2.5 kJ$

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