Two blocks of masses 400 g and 200 g are connected through a light string going

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Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia

1.6 \times 10^{- 4} k g - m^{2}

and a radius 2.0 cm. Find 'x', which is the kinetic energy of the system as the 400 g block falls through 50 cm and also find 'y', the speed of the blocks at this instant.

Options:

A . x = 10.5 J, y = 2.6 m/s

B . x = 9.8 J, y = 2.6 m/s

C . x = 9.8 J, y = 1.4 m/s

D . x = 0.98 J, y = 1.4 m/s

Answer: Option D
:
D
According to the question

Two Blocks Of Masses 400 G And 200 G Are Connected Through A...

0.4 g - T_{1} = 0.4 a

...(1)

T_{2} - 0.2 g = 0.2 a

....(2)

(T_{1} - T_{2}) r = \frac{I a}{r}

...(3)
From equation 1,2 and 3

\Rightarrow a = \frac{(0.4 - 0.2) g}{(\frac{0.4 + 0.2 + 1.6}{0.4})} = \frac{g}{5}

Therefore (y) V=

\sqrt{2 a h} = \sqrt{(2 \times g l^{5} \times 0.5)}

\Rightarrow \sqrt{(\frac{g}{5})} = \sqrt{(\frac{9.8}{5})} = \frac{1.4 m}{s}

.
(x) total kinectic energy of the system

= \frac{1}{2} m_{1} V^{2} + \frac{1}{2} m_{2} V^{2} + \frac{1}{2} 18^{2}

= (\frac{1}{2} \times 0.4 \times {1.4}^{2}) + (\frac{1}{2} \times 0.2 \times {1.4}^{2}) + (\frac{1}{2} \times (\frac{1.6}{4}) \times {1.4}^{2}) = 0.98 j o u l e

.

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