Two blocks of masses 4 kg and 1 kg are connected by a massless string that passes over an ideal pulley- If4 kg block reaches the ground isA- 2 secB- 1-1 secC- 3 secD- 4 sec

The correct option is A 2 seca = 4g sin 30 1g/5 = 40 ×1/2 10/5 =10/5 = 2 m/s^2 s =0 t + 1/2 at^2 4 = 1/2× 2 t^2 t^2 = 4 t =2 sec ...

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Standard XII

Physics

Power Factor

Two blocks of...

Question

Two blocks of masses 4 $k g$ and 1 $k g$ are connected by a massless string that passes over an ideal pulley. If the system is released from rest, the time when 4 $k g$ block reaches the ground is

1.1 sec

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2 sec

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3 sec

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4 sec

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Solution

The correct option is B 2 sec
$a = \frac{4 g s i n 30 - 1 g}{5} = \frac{40 \times \frac{1}{2} - 10}{5} = \frac{10}{5} = 2 m / s^{2}$
$s = 0 t + \frac{1}{2} a t^{2}$
$4 = \frac{1}{2} \times 2 t^{2}$
$t^{2} = 4$
$t = 2$ sec

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Two blocks of masses 4 kg and 1 kg are connected by a massless string that passes over an ideal pulley. If the system is released from rest, the time when 4 kg block reaches the ground is

The correct option is B 2 seca=4gsin30−1g5=40×12−105=105=2 m/s2 s=0 t+12at2 4=12×2t2 t2=4 t=2 sec

Two blocks of masses 4 $k g$ and 1 $k g$ are connected by a massless string that passes over an ideal pulley. If the system is released from rest, the time when 4 $k g$ block reaches the ground is

The correct option is B 2 sec
$a = \frac{4 g s i n 30 - 1 g}{5} = \frac{40 \times \frac{1}{2} - 10}{5} = \frac{10}{5} = 2 m / s^{2}$
$s = 0 t + \frac{1}{2} a t^{2}$
$4 = \frac{1}{2} \times 2 t^{2}$
$t^{2} = 4$
$t = 2$ sec