Question

If the radius of the circular path and frequency of revolution of a particle of mass m are doubled, then the change in its kinetic energy will be (E_i and E_f are the initial and final kinetic energies of the particle respectively.)

Options

• 12 E_f

• 16 E_i

• 8 E_f

• 15 E_i

Answer 1

15 E_i 

Explanation:

Initial kinetic energy of body,

`"E"_"i" = 1/2 "mv"^2 = 1/2 "m"((2pi"r"_1)/"T"_1)^2     ....[because "v" = (2pi"r")/"T"]`

`"E"_"i" = 2pi^2 "mr"_1^2 "t"_1^2`   ...(i)

Where, f₁ = frequency of revolution of the body.

When, r₂ = 2r₁ and f₂ = 2f₁, then

`"E"_"f" = 2pi^2"m" * "r"_2^2 * "f"_2^2`

`= 2pi^2 "m" (2"r"_1)^2 (2"f"_1)^2`

`"E"_"f" = 32pi^2 "m" "r"_1^2 "f"_1^2 = 16 * 2pi^2 "mr"_1^2 "f"_1^2`

`"E"_"f" = 16 "E"_"i"`

∴ Change in kinetic energy,

ΔE = E_f - E_i

= 16E_i - E_i = 15 E_i