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प्रश्न
A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform the vertical circular motion, under gravity. The minimum speed of a particle is 5 m/s. Consider the following statements.
P) Maximum speed must be `5sqrt5` m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N.
Select the correct option.
पर्याय
Only the statement P is correct.
Only the statement Q is correct.
Both the statements are correct.
Both the statements are incorrect.
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उत्तर
Only the statement Q is correct.
Explanation:
The minimum speed of the particle = `v_A` = 5 m/s
The length of the string (h) = 2.4 m
The radius (l) = 1.2 m
The mass of the particle, m = 1 kg
Conserving energy between A and B
⇒ `1/2 mv_A^2 + mgh = 1/2mv_B^2 + 0`
⇒ `v_A^2 + 2gh = v_B^2`
⇒ `5^2 + 2 xx 9.8 xx 2.4 = v_B^2`
⇒ `v_B^2 = 72`
⇒ `v_B = sqrt72` m/s
At point B,
`T_B - mg = (mv_B^2)/r`
`T_B = (mv_B^2)/r + mg` ...(1)
At point A,
`T_A + mg = (mv_A^2)/r`
`T_A = (mv_A^2)/r - mg` ...(2)
`T_B - T_A = ((mv_B^2)/r + mg) - ((mv_A^2)/r - mg)`
= `m/r (v_B^2 - v_A^2) + 2mg`
= `1/1.2 (72 - 25) + 2(1)(9.8)`
= 60 N
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