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A cylindrical rod of uniform cross section, is attached at O in a water tank. The linear mass density of rod is $λ_{0} x,$ where x is distance of the element of the rod, from end O as shown in figure. If the tension in string is given by $\frac{10^{4}}{P} N$ then P is (Length of rod 1m, radius of area of cross section is $\frac{1}{\sqrt{π}} m, ρ_{w a t e r} = 1000 k g m^{- 3}, g = 10 m s^{- 2}, λ_{0} = 10^{3}$ in S.I. unit)

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answer is 6.

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Detailed Solution

$X_{C M} = \frac{\int d m . x}{\int d m} = \frac{2 L}{3} .$

Mass of rod; $M = \int_{0}^{L} λ_{0} x d x = \frac{10^{3}}{2} .$

Torque about ‘O’; $F_{B} \frac{L}{2} \cos 37^{0} - M g \frac{2 L}{3} \cos 37^{0} = T L \cos 37^{0} 10^{3} π (\frac{1}{π}) \times 10 \frac{1}{2} - \frac{10^{4}}{2} \times \frac{2}{3} = T T = \frac{10^{4}}{6} N .$

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