Verified Solution • 2025–26 Session moderate

QNEET-2017-Ph-986: NEET 2017 Physics - Class 11 NEET-2017-Ph-986 Solution: Two discs of same moment of inertia rotating about their…

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1\omega_1 and ω2\omega_2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is 14I(ω1−ω2)2\frac{1}{4}I(\omega_1-\omega_2)^2 I(ω1−ω2)2I(\omega_1-\omega_2)^2 18I(ω1−ω2)2\frac{1}{8}I(\omega_1-\omega_2)^2 12I(ω1+ω2)2\frac{1}{2}I(\omega_1+\omega_2)^2

Multiple Choice Options

  • A 14I(ω1−ω2)2{1 \over 4}I{\left( {{\omega _1} - {\omega _2}} \right)^2} 4 1 _ I ( _ 1 _ _ _ 2 _ ) 2
  • B I(ω1−ω2)2I{\left( {{\omega _1} - {\omega _2}} \right)^2} I ( _ 1 _ _ _ 2 _ ) 2
  • C 18I(ω1−ω2)2{1 \over 8}I{\left( {{\omega _1} - {\omega _2}} \right)^2} 8 1 _ I ( _ 1 _ _ _ 2 _ ) 2
  • D 12I(ω1+ω2)2{1 \over 2}I{\left( {{\omega _1} + {\omega _2}} \right)^2} 2 1 _ I ( _ 1 _ + _ 2 _ ) 2

Step-by-Step Verified Solution

According to the problem, I _ 1 + I _ 2 = 2I _ 0 _ 0 = ( _ 1 _ _ 2 ) 2 K i = 1 2 I ( _ 1 2 + _ 2 2 ) K f = 1 4 ( _ 1 + _ 2 ) 2 Loss _ K = I [ _ 1 2 2 + _ 2 2 2 _ _ 1 2 4 _ _ 2 2 4 _ 2 _ 1 _ 2 4 ] = I [ _ 1 2 4 + _ 2 2 4 _ 2 _ 1 _ 2 4 ] = I 4 [ _ 1 _ _ 2 ] 2

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