A light string passes over a frictionless pulley. To one of its ends a mass of $6 k g$ is attached to the other end a mass of $10 k g$ is attached. The tension of the thread is:

24.5 N

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2.45 N

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79 N

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73.5 N

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Solution

The correct option is D $73.5 N$
Suppose the $10 k g$ is moving
downwards unit acceleration
so amking $F . B . D$ of $10 k g$
$10 g - T = 100$
i.e., $F = m a$
$\Rightarrow 10 g - T = 10 a - - - (1)$
making $F . B . D$ of $6 k g$
$T - 69 = 6 a - - - (2)$
$(1) \times 3 - (2) \times 5$
we get
$30 g - 3 T - (5 T - 30 g) = 30 a - 30 a$
or $30 g - S T + 30 g = 0$
$\Rightarrow T = \frac{60 g}{8} = \frac{60 \times 9.8}{8} = 73.5 N$

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