Three block of masses $m_{1} = 2.0, m_{2} = 4.0$ and $m_{3} = 6.0 k g$ are connected by strings on a frictionless inclined plane of $60 °$ , as shown in figure. A force $F = 120 N$ is applied upward along the incline to the uppermost block, causing an upward movement of the blocks. The connecting cords are light. The values of tensions $T_{1}$ and $T_{2}$ in the cords are

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$T_{1} = 20 N, T_{2} = 60 N$

$T_{1} = 60 N, T_{2} = 60 N$

$T_{1} = 30 N, T_{2} = 50 N$

$T_{1} = 20 N, T_{2} = 100 N$

answer is A.

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Detailed Solution

If $a$ is the acceleration of blocks, then

$a = \frac{120 - 12 g \sin 60 °}{12} = \frac{(120 - 120 \frac{\sqrt{3}}{2}) m / s^{2}}{12} = (10 - 5 \sqrt{3}) m / s^{2}$

Now $T_{1} = m_{1} a + m_{1} g \sin 60 °$

$= 2 \times (10 - 5 \sqrt{3}) + 2 \times 10 \times \frac{\sqrt{3}}{2}$

Similarly, $T_{2} = 60 N$

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