MHT CET · Physics · Laws of Motion

A stone of mass $1 k g$ is tied to a string $2 m$ long and is rotated at a constant speed of $40 {m s}^{- 1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 {m s}^{- 2}$ ]

A $\frac{81}{79}$
B $\frac{79}{81}$
C $\frac{19}{12}$
D $\frac{12}{19}$

Verified Solution

Answer & Solution

Correct Answer

(B) $\frac{79}{81}$

Step-by-step Solution

Detailed explanation

T_{t o p} + m g = \frac{m v^{2}}{R} \Rightarrow T_{t o p} = {\frac{(1) (40)}{2}}^{2} - (1) (10) = 790 N

T_{B o t} - m g = \frac{m v^{2}}{R} \Rightarrow T_{B o t} = {\frac{(1) (40)}{2}}^{2} + (1) (10) = 810

\frac{T_{t o p}}{T_{B o t}} = \frac{790}{810} = \frac{79}{81}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A stone of mass 1kg is tied to a string 2m long and is rotated at a constant speed of 40ms-1 in a vertical circle. The ratio of the tension at the top and the bottom is [Take g=10ms-2 ]

Step-by-step Solution

A stone of mass $1 k g$ is tied to a string $2 m$ long and is rotated at a constant speed of $40 {m s}^{- 1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 {m s}^{- 2}$ ]