The maximum and minimum tension in the string whirling in a vertical circle of the radius 2.5 m with constant speed are in the ratio 5:3 then its velocity is (.g=9.8 ms- 2.)

Question

The maximum and minimum tension in the string whirling in a vertical circle of the radius 2.5 m with constant speed are in the ratio 5:3 then its velocity is (.g=9.8 ms- 2.) (A) √98 m s- 1 (B) 7 m s- 1 (C) √490 m s- 1 (D) √4.9 m s- 1

Tardigrade Admin · Accepted Answer

In this problem, it is assumed that particle although moving in a vertical loop, but its speed remains constant
Tension at lowest point

T_{ma x} = \frac{m v ^{2}}{r} + m

g
Tension at highest point

T_{min} = \frac{m v ^{2}}{r} - m g

\frac{T _{ma x}}{T _{min}} = \frac{\frac{m v ^{2}}{r} + m g}{\frac{m v ^{2}}{r} - m g} = \frac{5}{3}

\frac{3 m v ^{2}}{r} + 3 m g = \frac{5 m v ^{2}}{r} - 5 m g

\frac{2 m v ^{2}}{r} = 8 m g

v = 4 r g

= 4 \times 2.5 \times 9.8

= 98 m s^{- 1}

Q. The maximum and minimum tension in the string whirling in a vertical circle of the radius $2.5 m$ with constant speed are in the ratio $5 : 3$ then its velocity is $(g = 9.8 m s^{- 2})$

$98 m s^{- 1}$

$7 m s^{- 1}$

$490 m s^{- 1}$

$4.9 m s^{- 1}$

Q. The maximum and minimum tension in the string whirling in a vertical circle of the radius 2.5m with constant speed are in the ratio 5:3 then its velocity is (g=9.8ms−2)

98​ms−1

7ms−1

490​ms−1

4.9​ms−1

Q. The maximum and minimum tension in the string whirling in a vertical circle of the radius $2.5 m$ with constant speed are in the ratio $5 : 3$ then its velocity is $(g = 9.8 m s^{- 2})$

$98 m s^{- 1}$

$7 m s^{- 1}$

$490 m s^{- 1}$

$4.9 m s^{- 1}$