Question

Pls solve this question
The correct answer-
T1=100/(root 3), T2=200/(root 3), T3=200/(root 3), T4=200

Answer 1

Answer:

here, T2 cos30° = mg

therefore, T2 is 200/ ( root3)

also T1= T2 sin30° so,

T1 is 100/( root3)

also, T4cos60° = T2cos30°

so, T4 is 200

also, T2cos60° = T3

therefore T3 is 100/ ( root 3)

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{T_{1}=\frac{100}{\sqrt{3}}\:N}}}$

$\green{\tt{\therefore{T_{2}=\frac{200}{\sqrt{3}}\:N}}}$

$\green{\tt{\therefore{T_{3}=\frac{200}{\sqrt{3}}\:N}}}$

$\green{\tt{\therefore{T_{4}=200\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: block = 10 \: kg \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Tension( T_{1}) = ?\\ \\ \tt: \implies Tension( T_{2}) = ?\\ \\ \tt: \implies Tension( T_{3}) = ?\\ \\ \tt: \implies Tension( T_{4}) = ?$

• According to given question :

$\bold{For \:component \: of\: tension( T_{4}) \: at \: point \: B}\\ \tt: \implies Horizontal \: component = T_{4} \: sin \: \theta \\ \\ \tt: \implies Horizontal \: component = T_{4} \:sin \: 60 \degree \\ \\ \tt: \implies Horizontal \: component = \frac{ T_{4} \sqrt{3} }{2} \\ \\ \tt: \implies Vertical\: component = T_{4} \: cos \: \theta \\ \\ \tt: \implies Vertical\: component = T_{4} \:cos \: 60 \degree \\ \\ \tt: \implies Vertical\: component = \frac{ T_{4}}{2} \\ \\ \bold{For \: Component \: of\: tension( T_{2} ) \: at \: point \: B}\\ \tt: \implies Horizontal \: component = T_{2} \: cos \: \theta \\ \\ \tt: \implies Horizontal \: component = T_{2} \:cos \: 60 \degree \\ \\ \tt: \implies Horizontal \: component = \frac{ T_{2} }{2} \\ \\ \tt: \implies Vertical\: component = T_{2} \: sin \: \theta$

$\tt: \implies Vertical\: component = T_{2} \:sin \: 60 \degree \\ \\ \tt: \implies Vertical\: component = \frac{ T_{2} \sqrt{3} }{2} \\ \\ \bold{For \:component \: of\: tension( T_{2}) \:at \: point \: A } \\ \tt: \implies Horizontal \: component = T_{2} \: sin \: \theta \\ \\ \tt: \implies Horizontal \: component = T_{2} \:sin \: 30 \degree \\ \\ \tt: \implies Horizontal \: component = \frac{ T_{2} }{2} \\ \\ \tt: \implies Vertical\: component = T_{2} \: cos \: \theta \\ \\ \tt: \implies Vertical\: component = T_{2} \:cos \: 30 \degree \\ \\ \tt: \implies Vertical\: component = \frac{ T_{4} \sqrt{3} }{2}$

$\bold{At \: point \: A\: in \: horizontal : } \\ \tt: \implies T_{1} = \frac{ T_{2} }{2} \\ \\ \tt: \implies 2T_{1} = T_{2} - - - - - (1) \\ \\ \bold{At \: point \: A \: in \: Vertical : } \\ \tt: \implies \frac{ T_{2} \sqrt{3} }{2} = mg \\ \\ \tt: \implies \frac{ T_{2} \sqrt{3} }{2} = 100 \\ \\ \green{\tt: \implies T_{2} = \frac{200}{ \sqrt{3} } \: N } \ \\ \\ \bold{At \: point \: B \: in \: vertical : } \\ \tt: \implies \frac{ T_{4} }{2} = \frac{ T_{2} \sqrt{3} }{2} \\ \\ \tt: \implies T_{2} = \frac{T_{4} }{ \sqrt{3} } \\ \\ \tt: \implies \frac{200}{ \sqrt{3} } = \frac{ T_{4}}{ \sqrt{3}}$

$\green{\tt: \implies T_{4} = 200 \: N} \\ \\ \bold{At \: point \: B \: in\: horizontal : } \\ \tt: \implies \frac{ T_{4} \sqrt{3} }{2} = T_{3} + \frac{ T_{2} }{2} \\ \\ \tt: \implies T_{4} \sqrt{3} - T_{2} = 2 T_{3} \\ \\ \tt: \implies 200 \sqrt{3} - \frac{200}{ \sqrt{3} } = 2 T_{3} \\ \\ \tt: \implies \frac{600 - 200}{ 2\sqrt{3} } = T_{3} \\ \\ \green{\tt: \implies T_{3} = \frac{200}{ \sqrt{3} } \: N} \\ \\ \text{Putting \: value \: of \: }T_{2} \: \text{in \: (1)} \\ \tt: \implies 2 T_{1} = \frac{200}{ \sqrt{3} } \\ \\ \tt: \implies T_{1} = \frac{200}{ 2\sqrt{3} } \\ \\ \green{\tt: \implies T_{1} = \frac{100}{ \sqrt{3} } \: N }$