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A block of mass $100 kg$ is hanging with three vertical strings $A, B, C$ of equal lengths and equal radius but Young's modulus $Y_{1}, Y_{2} a n d Y_{3}$ . Strings A and C are at the ends while B is attached at the middle. For $Y_{2} = 2 Y_{1}$ and if the block remains horizontal mark the correct option (s).

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Tension in A and C will be equal

Tension in A and C could be different

$Y_{1}$ and $Y_{3}$ should be equal

Tension in B is $50 kg .$

answer is A, C, D.

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Detailed Solution

For all three wires, $l, ∆ l$ and A are same, So, $Y = \frac{\frac{F}{A}}{\frac{Δ!}{l}}$

$\Rightarrow F \propto Y .$ For translation equilibrium, $F_{1} + F_{2} + F_{3} = 100 k g$ and for rotational equilibrium,

$\begin{array}{l} F_{1} \cdot \frac{l}{2} - F_{3} \cdot \frac{l}{2} = 0 \\ \Rightarrow F_{1} = F_{3} \\ \Rightarrow Y_{1} = Y_{3} and Y_{2} = 2 Y_{1}, gives F_{2} = 2 F_{1} . \end{array}$

Thus, $(F_{1} / 2) + F_{2} + (F_{1} / 2) = 100 k g$

$\Rightarrow F_{1} = 25 k g and F_{2} = 50 K g .$

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