SATHEE: Kinematics Question 488

Kinematics Question 488

Question: A stone of mass 1 kg tied to a light inextensible string of length $L = \frac{10}{3} m$ it’s whirling in a circular path of radius $L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if $g$ it’s taken to be $10 m / \sec^{2}$ , the speed of the stone at the highest point of the circle it’s [CBSE PMT 1990]

Options:

A) $20 m / \sec$

B) $10 \sqrt{3} m / \sec$

C) $5 \sqrt{2} m / \sec$

D) $10 m / \sec$

Show Answer

Answer:

Correct Answer: D

Solution:

Since the maximum tension $T_{B}$ in the string moving in the vertical circle is at the bottom and minimum tension $T_{T}$ is at the top.

$T_{B} = \frac{m v_{B}^{2}}{L} + m g$ and $T_{T} = \frac{m v_{T}^{2}}{L} - m g$

$\frac{T_{B}}{T_{T}} = \frac{\frac{m v_{B}^{2}}{L} + m g}{\frac{m v_{T}^{2}}{L} - m g} = \frac{4}{1}$ or $\frac{v_{B}^{2} + g L}{v_{T}^{2} - g L} = \frac{4}{1}$ or $v_{B}^{2} + g L = 4 v_{T}^{2} - 4 g L$

but $v_{B}^{2} = v_{T}^{2} + 4 g L$

$v_{T}^{2} + 4 g L + g L = 4 v_{T}^{2} - 4 g L$

therefore $3 v_{T}^{2} = 9 g L$

$v_{T}^{2} = 3 \times g \times L = 3 \times 10 \times \frac{10}{3}$

or $v_{T} = 10 m / \sec$

Kinematics Question 488

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Kinematics Question 488

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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