Question 89 of 210

As per given figure, a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $g=10m{s}^{-2}$)

A$\left(4\sqrt{3}+1\right)N$B$4\left(\sqrt{3}+1\right)N$C$4\left(\sqrt{3}-1\right)N$D$\left(4\sqrt{3}-1\right)N$

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Answer: (B) $4\left(\sqrt{3}+1\right)N$

The magnitude of acceleration of both the blocks will be same and is given by
$a=\dfrac{4g\sin {60}^{^{\circ}}-g\sin {30}^{^{\circ}}}{4+1}$
$=\dfrac{20\sqrt{3}-5}{5}$
$a=\left(4\sqrt{3}-1\right)$ $m{s}^{-1}$
Applying Newton's second law for second block (along the direction parallel to the incline on which it is placed)
$T-5=1\times a$
$\Rightarrow T=5+4\sqrt{3}-1$
$=4\left(\sqrt{3}+1\right)N$

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