A body of mass 5kg is suspended by the strings making angles 600 and 300 with the horizontal as shown in figure (g =10m / s2)

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A body of mass 5kg is suspended by the strings making angles 60⁰ and 30⁰ with the horizontal as shown in figure (g =10m / s²)

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$T_{2} = 25 \sqrt{3} N$

$T_{2} = 25 N$

T₁ = 25N

$T_{1} = 25 \sqrt{3} N$

answer is A, D.

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Detailed Solution

The FBD of the block is as shown in fig

For equilibrium along X-axis
$\begin{array}{l} T_{2} \cos 60 = T_{1} \cos 30 \\ \frac{T_{2}}{2} = \frac{T_{1} \sqrt{3}}{2} \Rightarrow T_{2} = \sqrt{3} T_{1} \end{array}$
For equilibrium along Y-axis,
$\begin{array}{l} T_{2} \sin 60 = T_{1} \sin 30 = 5 g \\ \frac{T_{2} \sqrt{3}}{2} + \frac{T_{1}}{2} = 50 \Rightarrow T_{2} \sqrt{3} + T_{1} = 100 \end{array}$
Solving above equations, we get
$T_{1} = 25 N; T_{2} = 25 \sqrt{3 N}$