In the shown figure a block of mass $m = \frac{1}{2} k g$ is suspended in equilibrium with the help of ideal string and two ideal spring of force constant $k = 100 N / m .$ Now spring AP is cut. Pullies are massless, frictionless, AP is vertical and other spring is horizontal. Just after cutting the spring (g= $10 m / s^{2}$ )

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Tension in string BC is 3N

Acceleration of the suspended block is $8 m / s^{2}$

Acceleration of the suspended block is $\frac{32}{7} m / s^{2}$

Tension in string BC is $\frac{37}{7} N$

answer is B, C.

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Detailed Solution

Initial tension in spring $k x = \frac{4 m g}{7}$
FBD of block just after string is cut Since, v = 0
$∴$ acceleration along string is zero
$T = k \cos 37^{0} + m g \sin 37^{0} = \frac{37}{7} N$
Acceleration is only perpendicular to spring
$\begin{array}{l} m a = m g \cos 37^{0} - k x \sin 37^{0} \\ \Rightarrow a = \frac{32}{7} m / s^{2} \end{array}$