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Answer 1

Solution:-

1) Using Lami's Theorem

$\rm \to \: \dfrac{p}{ \sin \theta_{1} } = \dfrac{Q}{ \sin \theta _{2} } = \dfrac{R}{ \sin\theta _{ 3}}$

$\rm \to \: \dfrac{p}{ \sin \theta_{1} } = \dfrac{R}{ \sin\theta _{3}}$

Now put the given value

$\rm \to \: \dfrac{p}{ \sin \theta_{1} } \times { \sin\theta _{3}}= {R}$

$\rm \: R = \sin150 \times \dfrac{p}{ \sin \theta_{1}}$

$\rm \to \: R = \dfrac{1}{2} \times \dfrac{1.9318}{0.9659}$

$\rm \to \: R = \dfrac{1}{2} \times 2 = 1$

Ans = option :- b is correct

2)

:- FBD is attached see fig (ii)

By taking horizontal and vertical components we get

$\rm \to \: T \times \sin60 \degree = w \: \: \: \: ...(i)eq$

$\rm \to \: T \cos60 = 30 N\: \: \: \: \: .....(ii)eq$

$\to \rm \: T \times \dfrac{ \sqrt{3} }{2} = w \: \: \: \: ...(i)eq$

$\rm \to \: \dfrac{1}{2} \times T = 30N\: \: \: \: \: ..(ii)$

From second equation we get T

$\rm \: T = 60N$

Now put the value of T on first equation

$\rm \: 60 \times \dfrac{ \sqrt{3} }{2} = w$

So

$\rm \to \: w = 30 \sqrt{3}$

So option :- A is correct

3)FBD is attached see fig (iii)

By taking horizontal and vertical components we get

$\to \rm \: T \cos30 \degree = 0.05g \: \: \: \: \: \: .....(i)eq$

$\rm \to \: T \sin 30 \degree = F \: \: \: \: \: ....(ii)eq$

Now

$\to \rm \: T \times \dfrac{ \sqrt{3} }{2} = 0.05 \times 10 \: \: \: \: ....(i)eq$

$\rm \to \: T \times \dfrac{1}{2} = F \: \: \: \: \: \: ...(ii)eq$

Now take (i)st equation

$\to \rm \: T \times \dfrac{ \sqrt{3} }{2} = 0.5 \: \: \: \: ....(i)eq$

$\to \rm \: T \times { \sqrt{3} }{} = 1 \: \: \: \: ....(i)eq$

$\to \rm \: T {} = \dfrac{1}{ \sqrt{3} } \: \: \: \: ....(i)eq$

Now put the value of T on (ii) eq

$\rm \to \: \dfrac{1}{ \sqrt{3} } \times \dfrac{1}{2} = F \: \: \: \: \: \: ...(ii)eq$

$\rm \to \: F = 0.28 N\: \: \: \: \: \: ...$

Ans :- option D is correct