हिंदी

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.

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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.

विकल्प

  • Only the statement P is correct.

  • Only the statement Q is correct.

  • Both the statements are correct.

  • Both the statements are incorrect.

MCQ
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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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अध्याय 1: Rotational Dynamics - Exercises [पृष्ठ २३]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
अध्याय 1 Rotational Dynamics
Exercises | Q 1.2 | पृष्ठ २३

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