Find the tension in the string connected to the block of mass $4 kg$ . Assume the strings and pulleys to be ideal. (Take $g = 10 m / s^{2})$

\frac{100}{9} N

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\frac{40}{3} N

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\frac{15}{2} N

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\frac{50}{3} N

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Solution

The correct option is B $\frac{40}{3} N$
The given system can be reframed as shown

From constraint condition, we have
(i) ${. . L}_{1} + {. . L}_{2} = 0 \Rightarrow - a_{2} + a_{p} = 0 \Rightarrow a_{p} = a_{2}$
(ii) ${. . L}_{3} + {. . L}_{4} = 0 \Rightarrow - a_{p} - a_{p} + a_{1} = 0 \Rightarrow a_{1} = 2 a_{p}$
Hence, $a_{1} = 2 a_{2}$

Hence, let acceleration of $2 k g$ block is $a$ upwards, and acceleration of $4 k g$ block is $2 a$ downwards.

Applying Newton's Second Law,

For $2 kg$ body:
$2 T - 20 = 2 a$ ... (3)

For $4 kg$ body:
$40 - T = 4 \times 2 a$ ... (4)
From (3) and (4)
$4 \times (2 T - 20) = 40 - T$
$\Rightarrow 9 T = 120 \Rightarrow T = \frac{120}{9}$
Thus, the tension in the string connecting $4 k g$ mass is $T = \frac{120}{9} = \frac{40}{3} N$

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