Find the tension (in $N$ ) in the string connecting the $2 m$ mass. (Strings are massless and inextensible)

\frac{15 m g}{11}

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\frac{18 m g}{13}

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\frac{18 m g}{11}

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\frac{15 m g}{13}

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Solution

The correct option is C $\frac{18 m g}{11}$
The given system can be reframed as shown below as per intercept method

So, we have
a) ${. . L}_{1} + {. . L}_{2} = 0$
$\Rightarrow a_{1} - a_{p} = 0 \Rightarrow a_{p} = a_{1} \dots \dots (1)$

b) ${. . L}_{3} + {. . L}_{4} = 0 \Rightarrow a_{p} + a_{p} - a_{2} = 0$
$\Rightarrow a_{2} = 2 a_{p} \dots \dots (2)$
From (1) and (2) we get,
$a_{2} = 2 a_{1}$
From FBDs of the masses we have force equations as

$3 m g - T_{1} = 3 m a_{1} \dots \dots (3)$
$T_{2} - 2 m g = 2 m a_{2} = 2 m (2 a_{1}) = 4 m a_{1} \dots (4)$
and $T_{1} = 2 T_{2} \dots \dots (5)$
So, from above three equations we have
$- m g = 11 m a_{1} \Rightarrow a_{1} = \frac{- g}{11}$
Here, $- v e$ sign shows that the assumption that we made about the direction of acceleration of mass $3 m$ is wrong i.e. the mass $3 m$ will move upwards rather than downwards.
$∴$ $T_{1} = 3 m (g - a_{1}) \Rightarrow T_{1} = \frac{36 m g}{11}$
As we have $T_{1} = 2 T_{2}$
$\Rightarrow T_{2} = \frac{T_{1}}{2} = \frac{18 m g}{11} N$

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