JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Work, Power & Energy

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A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio

5 : 1

. The velocity of the bob at the highest position is ____

m / s . (

Take

g = 10 m / s^{2})

Answer: 5

Solution

The string pulls hardest at the lowest point and least at the highest point:

T_{m a x} = \frac{m v _{L}^{2}}{r} + m g, T_{m i n} = \frac{m v _{H}^{2}}{r} - m g

Energy is conserved between the two points, which are

2 r

apart in height, so

v_{L}^{2} = v_{H}^{2} + 4 g r

. Putting this into

T_{m a x} = 5 T_{m i n}

and cancelling

\frac{m}{r}

v_{H}^{2} + 4 g r + g r = 5 (v_{H}^{2} - g r) \Rightarrow 4 v_{H}^{2} = 10 g r

With

g = 10

m/s

^{2}

and

r = 1

m this gives

v_{H}^{2} = 25

, so the bob passes the highest position at

5

m/s.

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Work, Power & Energy

Solution

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