SATHEE: Work Energy And Power Question 212

Work Energy And Power Question 212

Question: Two identical beads of $m = 100$ gram are connected by an inextensible massless string can slide along the two arms AC and BC of a rigid smooth wire frame in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of 0.1 m is $8 \times 10^{- 3} J$ . Find the value of $x . (g = 10 m / s^{2})$

Options:

A) 8

B) 6

C) 9

D) 11

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Let m be the mass of the beads.

By energy conservation $m g h = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$ ….(i)

$\frac{v_{1}}{v_{2}} = \frac{4}{3}$ -(ii) $v_{1} = \frac{4 \sqrt{2}}{5}, v_{2} = \frac{3 \sqrt{2}}{5} .,$

$K . E . = \frac{1}{2} \times 100 \times 10^{- 3} \times \frac{16 \times 2}{25} = 64 \times 10^{- 3} J$ x=8.

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Work Energy And Power Question 212

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