Figure shows a weigh 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. lt 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

Question

Figure shows a weigh 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. lt 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 (A) same as R (B) 40 g less than R (C) 40 g more than R (D) 80 g more than R

Tardigrade Admin · Accepted Answer

Upthrust on Q=V ⋅ ρ ⋅ g=(40 cc )((1 g / cc )) g=40 g Reaction of Q on P= Upthrust =40 g Hence, R prime=R+40 g

same as R

40 g less than R

40 g more than R

80 g more than R

Upthrust on Q=V⋅ρ⋅g=(40 cc )(cc1g​)g=40g Reaction of Q on P= Upthrust =40g Hence, R′=R+40g

Upthrust on $Q = V \cdot ρ \cdot g = (40$ cc $) (\frac{1 g}{cc}) g = 40 g$
Reaction of $Q$ on $P =$ Upthrust $= 40 g$
Hence, $R^{'} = R + 40 g$