Question

Kindly provide me the solution the 13th question

Answer 1

Answer:

$(2) \frac{mg}{2\sqrt{2}}$

Explanation:

1) Let the Hinge reaction of the pulley be $N_1$ and $N_2$along the strings connected to both the masses.

By string constraint,

$=>N_1=N_2=Tension =T(say)$

2) Applying Newton's Laws on both the masses,

Let  'a' be acceleration of mass 'm' as shown in figure.

For the right mass 'm',

$mg-T=ma$

For left mass 'm',

$T+mgsin(30^{\circ})=ma$

3) Using above two equations,

$T+mgsin(30^{\circ})=mg-T\\ \\=>2T=mg-mgsin(30^{\circ})=mg-\frac{mg}{2}\\ \\=> 2T=\frac{mg}{2}\\ \\=>T=\frac{mg}{4}$

4) Now, Magnitude of reaction given by hinge is given by

$Reaction = \sqrt{N_1^2+N_2^2}\\ \\=\sqrt{T^2+T^2}=\sqrt{2}T\\ \\=\sqrt{2}*\frac{mg}{4}\\ \\=\frac{mg}{2\sqrt{2}}$

Hence, Option (2) is correct answer.