Question

Please answer this question.
I will mark you as brainliest​

Answer 1

In the case of the block,

$\sf{T_2=f}$

That is,

$\sf{T_2\leq 12\ N}$

And in the case of the weight W,

$\sf{W=T_3}\quad\quad\dots(1)$

At the point P,

$\sf{T_1\cos 30^{\circ}=T_2}\\\\\\\sf{\dfrac {T_1\sqrt3}{2}\leq12\ N}\\\\\\\sf{T_1\leq8\sqrt3\ N}$

And,

$\sf{T_3=T_1\sin 30^{\circ}}\\\\\\\sf{T_3\leq4\sqrt3\ N}$

Then, in (1),

$\sf{W\leq4\sqrt3\ N}\\\\\\\sf{\underline {\underline {W\leq6.92\ N}}}$

For maximum value,

$\sf{\underline {\underline {W=6.92\ N}}}$

Hence (b) is the answer.