A cylindrical iron rod 8cm height and 6cm in diameter stands in a cylindrical tin 12cm in diameter water is poured into the tin unfit it's depth is 8cm how far until the level drop when the rod is removed
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the volume of water displaced by the rod is
v = πr^2 h = π * 3^2 * 8 = 72π cm^3
The drop in water level is the rod volume divided by the cross-section area of the tin, or
72π/(π*6^2) = 2 cm
v = πr^2 h = π * 3^2 * 8 = 72π cm^3
The drop in water level is the rod volume divided by the cross-section area of the tin, or
72π/(π*6^2) = 2 cm
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