A uniform rod AB, 12 m long weighing 24 kg, is supported at end B by a flexible light string and a lead weight (of very small size) of 12 kg attached at end A.The rod floats in water with one-half of its length submerged. Find the volume of the rod in ×10−3 m3 Take g=10 m/s2, density of water =1000 kg/m3
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answer is 64.
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Detailed Solution
T+U=W1+W2=360N …(i)U=V2ρwg=(V/2)103(10)or U=0.5×104V …(ii)Taking torque about M = 0∴ 120l4+T3l4=240l4or T=240−1203=40 NNow from Eqs. (i) and (ii), we get V=6.4×10−2 m3=64×10−3 m3