Question

please answer this question with proper explanation and i will mark the answer as brainliest​

Answer 1

see rough free body diagram of both bodies.

for body of mass m,

at equilibrium,

Tsin45° + Tsin45° = mg

or, 2Tsin45° = mg

or, Tsin45° = mg/2 ......(1)

and also T = mg/√2 ......(2)

for body of mass M,

at equilibrium,

Tsin45° = $T_1cos\theta$

from equation (1)

mg/2 = $T_1cos\theta$....(3)

again, $T_1sin\theta=Tcos45°+Mg$

from equation (2),

$T_1sin\theta=\frac{mg}{2}+Mg$....(4)

dividing equation (4) by equation (3),

$\frac{T_1sin\theta}{Tcos\theta}=\frac{mg/2+Mg}{mg/2}$

or, $tan\theta=\frac{mg/2}{mg/2}+\frac{Mg}{mg/2}$

or, $tan\theta=1+\frac{2M}{m}$

hence, option (A) is correct choice.