Fluid MechanicsMedium
Question
Figure shows a weighing-bridge, with a beaker P with water on one pan and a balancing weight R on the other. A solid ball Q is hanging with a thread outside water. It has volume 40 cm3 and weighs 80 g. If this solid is lowered to sink fully in water, but not touching the beaker anywhere, the balancing weight R′ will be
Options
A.same as R
B.40
C.40 g more than R
D.80 g more than R
Solution
Since not touching,
So R = Fb = ρl(vg) = 40g.
R′ − R = 80g − 40g = 40g
Hence R′ will be 40g more than R
So R = Fb = ρl(vg) = 40g.
R′ − R = 80g − 40g = 40g
Hence R′ will be 40g more than R
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