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MHT CET202013 Oct 2020Morning ShiftPhysicsMotion in Two DimensionsActual

A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 ~cm . If the angular speed is doubled, the elongation in the spring is 6 ~cm . The original length of the spring is

Options

  1. A1. 3 Cm
  2. B9 Cm
  3. C6 Cn
  4. D12 Cm

Correct answer

B. 9 Cm

Step-by-step solution

Let be the original length of the spring. Let the initial angular velocity be and the corresponding elongation e ₁=1 ~cm . When the angular velocity is doubled the elongation e₂=6 ~cm . If k is the spring constant then we have array l m ( +e₁ ) ²= ke ₁ and m ( +e₂ )(2 )²= ke ₂ or m ( +e₂ ) 4 ²= ke ₂ array Dividing Eq(1) by Eq(2), we get +e₁ 4 ( +e₂ ) = e₁ e₂ solving we get =9 ~cm

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