In the system shown in figure pulley is smooth. String is massless and inextensible. Find acceleration of the system and tensions $T_{1}$ and $T_{2}$ . $(g = 10 m / s^{2})$ .

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$\frac{10}{7} m / s^{2}, \frac{40}{3} N,$ $\frac{40}{3} N$

$\frac{20}{3} m / s^{2}, \frac{100}{3} N,$ $\frac{50}{3} N$

$\frac{20}{3} m / s^{2}, \frac{100}{3} N,$ $\frac{60}{3} N$

$10 m / s^{2}, \frac{70}{9} N,$ $\frac{30}{3} N$

answer is B.

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Detailed Solution

Here, net pulling force will be

Weight of $4 kg$ and $6 kg$ blocks on one side - weight of $2 kg$ block on the other side. Therefore,

a =

$\frac{Net pulling force}{Total mass} = \frac{(6 \times 10) + (4 \times 10) - (2) (10)}{6 + 4 + 2} = \frac{20}{3} m / s^{2}$

For $T_{1}$ , let us consider FBD of $2 kg$ block. Writing equation of motion, we get

$T_{1} - 20 = 2 a or T_{1} = 20 + 2 \times \frac{20}{3} = \frac{100}{3} N$

For $T_{2}$ , we may consider FBD of $6 kg$ block. Writing equation of motion, we get

$60 - T_{2} = 6 a T_{2} = 60 - 6 a = 60 - 6 (\frac{20}{3}) = \frac{60}{3} N$

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