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The rotational kinetic energy of a body is E. In the absence of external torque, if mass of the body is halved and radius of gyration is doubled, then its rotational kinetic energy will be:-

Answer
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Hint: Whenever there is an absence of external torque, always remember angular momentum L is conserved. Therefore for this question we can use conservation of angular momentum where moment of inertia can be written in the terms of radius of gyration I=mk2 where m is mass of the body and k is radius of gyration.
 L=Iω , I=mk2 , K.E=12Iω2

Complete step by step answer:
The initial rotational kinetic energy of the body is given in the question as K.E1=E .then it was said that there was an absence of external torque which means the total angular momentum is constant which implies the angular momentum is conserved.
CONSERVATION OF ANGULAR MOMENTUM-
 L=Iω , where I is moment of inertia and ω is angular velocity and L is angular momentum
 L1=I1ω1 ………initial conditions (1)
 L2=I2ω2 ………in the absence of torque (2)
As angular momentum is conserved we write L1=L2
 I1ω1=I2ω2 …...... (3)
 I=mk2 , where I is moment of inertia and m is mass of body and k is radius of gyration.
Now we can write
 I1=m1k12 ……….. (4)
 I2=m2k22 ……….. (5)
UNDER ABSENCE OF TORQUE,
Given mass of body is halved so we can write m2=m12
Given radius of gyration is double so we can write k2=2k1
Now, substituting these in equation 4 and 5 and taking ratios we get,
  I1I2=m1k12m2k22=m1k12m12(2k1)2=2×(14)=12
From equation 3 we can write
  I1I2=ω2ω1=12 2I1=I2 , ω1=2ω2
Now initial rotational kinetic energy is K.E1=12I1ω12 =E and final rotational kinetic energy can be written as K.E2=12I2ω22
 K.E1K.E2=12I1ω1212I2ω22=I1ω12I2ω22=I1(2ω2)22I1(ω22)=2
 K.E1K.E2=EK.E2=2
 K.E2=E2
Therefore, in the absence of external torque, if mass of the body is halved and radius of gyration is doubled, then its rotational kinetic energy will be halved.

Note:
Torque can be defined as the rate of change of angular momentum, analogous to force. The net external torque on any system is always equal to the total torque on the system; in other words, the sum of all internal torques of any system is always 0 (this is the rotational analogue of Newton's Third Law). Here we need to understand that angular momentum is always conserved or total angular momentum is constant in the absence of external torque.
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