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A horizontal massless rod of length $4l$ is suspended on two vertical strings of negligible mass. Two particles of masses $m$ and $2 \mathrm{~m}$ are in equilibrium where their distances from the left end being $2l$ and $3l$ respectively. Then the tension $T$ of the left string at the instant when the right string snaps will be:-

Option: 1
$\frac{mg}{11}$

Option: 2
$\frac{2mg}{11}$

Option: 3
$\frac{4mg}{33}$

Option: 4
$\frac{mg}{33}$

Answers (1)

best_answer

F.B.D of the rod just after snapping the right string.

$T+N_2-N_1=0$ - - - (i)

$\begin{gathered} N_1(2 \ell)=N_2(3 \ell) \\ N_1=\frac{3 N_2}{2} \end{gathered}$

From equation (i)

$\begin{gathered} T+N_2-\frac{3 N_2}{2}=0 \\ T=\frac{N_2}{2} \end{gathered}$

Let, accelerations of $m$ and $2m$ be $a_1$ and $a_2$ respectively.

$\begin{aligned} \frac{a_1}{2 l} & =\frac{a_2}{3 \ell} \\ a_1 & =\frac{2}{3} a_2 \\ m g-N_1^{\prime} & =m a_1 \\ 2 m g+N_2^{\prime} & =(2 m) a_2 \end{aligned}$

From newton's third law of motion

$\begin{aligned} & N_1=N_1^{\prime}, N_2=N_2^{\prime} \\ & m g-\frac{3 N_2}{2}=m a_1, 2 m g+N_2=(2 m)\left(\frac{3}{2}\right) a_1 \end{aligned}$

$m g-\frac{3 {N_2}}{2}=m a_1 \\$

${\frac{3}{2}( 2 m g)+\frac{3}{2} N_2}=\frac{3}{2}(2 m)\left(\frac{3}{2}\right) a_1 \\$

___________________________________

$\begin{gathered} 4 m g=m a_1+\frac{9}{2} m a_1 \\ 4 g=\frac{11}{2} a_1 \\ a_1=\frac{8 g}{11} \end{gathered}$

Using,

$m g-\frac{3 {N_2}}{2}=m \left ( \frac{8g}{11} \right ) \\$

$\begin{aligned} & 3 \frac{N_2}{2}=m g-\frac{8 m g}{11}=\frac{3 m g}{11} \\ & N_2=\frac{2}{11} m g \\ & T=\frac{N_2}{2}=\frac{m g}{11} \end{aligned}$

Posted by

Devendra Khairwa

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