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Three blocks of masses 10kg, 6kg and 4kg are connected by mass less strings and placed on a smooth horizontal surface. If forces of 10 N and 40 N are applied as shown, then tension $T_{1}$ is

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20 N

25 N

10 N

35 N

answer is C.

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Detailed Solution

For the whole system, $a = \frac{F n e t}{m_{1} + m_{2} + m_{3}} = \frac{30}{20} = 1.5 m / s^{2}$

For horizontal force acting on 10kg block

$T_{1} - 10 = 10 (1.5)$ .

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