For the given figure List - I gives acceleration of blocks of masses m1- m2- m3 - m4 and List - II gives their magnitudes- Here m1 - m2 - 2m and m3 - m4 - m- List - I List - II- I a1 P 2g- II a2 Q g- III a3 R g-2- IV a4 S g-4- T 0- U None- -Just after cutting string connecting m1 and m2 correct match for accelerations of blocks is

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Combination of Springs

For the given...

Question

For the given figure List - I gives acceleration of blocks of masses $m_{1}, m_{2}, m_{3} & m_{4}$ and List - II gives their magnitudes. Here $m_{1} = m_{2} = 2 m$ and $m_{3} = m_{4} = m$

$\begin{matrix} List - I & List - II (I) a_{1} & (P) 2 g (II) a_{2} & (Q) g (III) a_{3} & (R) \frac{g}{2} (IV) a_{4} & (S) \frac{g}{4} (T) 0 (U) None \end{matrix}$
Just after cutting string connecting $m_{1}$ and $m_{2}$ correct match for accelerations of blocks is

I \to Q; I I \to Q; I I I \to P; I V \to Q

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I \to Q; I I \to P; I I I \to R; I V \to T

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I \to Q; I I \to Q; I I I \to T; I V \to T

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I \to P; I I \to Q; I I I \to R; I V \to U

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Solution

The correct option is C $I \to Q; I I \to Q; I I I \to T; I V \to T$
$I \to Q; I I \to Q; I I I \to T; I V \to T$
$m_{1}$ falls freely after the cut, so its acceleration is $g$ .
The tension in the right side spring before cutting is $(m_{1} + m_{2}) g = 4 m g$ . After cutting, the net force on $m_{2}$ is $4 m g - m_{2} g = 2 m g$ . So, its acceleration is $\frac{2 m g}{m_{2}} = g$ .

Just after cutting, all the springs still maintain the tension in them. Therefore, the blocks $m_{3}$ and $m_{4}$ remain in equilibrium.

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