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The maximum and minimum tension in the s...

The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio `5:3` then its velocity is

A

`10(m)/(s)`

B

`7(m)/(s)`

C

`15(m)/(s)`

D

`5(m)/(s)`

Text Solution

Verified by Experts

The correct Answer is:
A
In this problem it is assumed that particle although moving in a vertical loop, its speed remain constant.
Tension at lowest point `T_(max)=(mv^2)/(r )+mg`
Tension at highest point `T_(min)=(mv^2)/(r )-mg`
`(T_max)/(T_min)=((mv^2)/(r)+mg)/((mv^2)/(r )-mg)=(5)/(3)`

`v=sqrt(4gr)=sqrt(4xx10xx2.5)=10(m)/(s)`
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Knowledge Check

  • For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio 5 : 3. If the radius of vertical circle is 2 m, the speed of revolving body is ( g=10m//s^(2) )

    A
    `sqrt(5)m//s`
    B
    `4sqrt(5)m//s`
    C
    5 m/s
    D
    10 m/s

  • A particle of mass m attached to an inextensible light string is moving in a vertical circle of radius r. The critical velocity at the highest point is v_( 0) to complete the vertical circle. The tension in the string when it becomes horizontal is

    A
    ` (3mv_(0)^(2))/(r)`
    B
    `(9 mv_(0)^(2))/(r)`
    C
    3mg
    D
    both (A) and (C) are correct

  • A stone of mass 1kg is tied with a string and it is whirled in a vertical circle of radius 1 m. If tension at the highest point is 14 N, then velocity at lowest point will be

    A
    `3m//s`
    B
    `4m//s`
    C
    `6m//s`
    D
    `8m//s`

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