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A slender homogeneous rod of length $2 L$ floats partly immersed in water, being supported by a string fastened to one of its ends, as shown. The specific gravity of the rod is $0.75$ . The length of rod that extends out of water is

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$L$

$\frac{1}{4} L$

$\frac{1}{2} L$

$3 L$

answer is A.

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Detailed Solution

$W = (2 L A) \times 0.75 g, F_{b} = (A y) \times 1 \times g$

As rod is in equilibrium, so $Σ τ_{P} = 0$

or $W \times L \cos θ - F_{b} \times (2 L - y / 2) \cos θ = 0$

After substituting and simplifying, we get

$y = L$

So the length of rod out of water is $L$ .

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