The reading of a spring balance when a block | vedclass

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(A) Let $V$ be the volume of the block,$\rho$ be the density of the block,and $\rho_w$ be the density of water.In air,the weight of the block is $W_{a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let $V$ be the volume of the block,$\rho$ be the density of the block,and $\rho_w$ be the density of water.In air,the weight of the block is $W_{air} = V \rho g = 60 \, N$.When submerged in water,the apparent weight is $W_{water} = V \rho g - V \rho_w g = 40 \, N$.Substituting $V \rho g = 60$,we get $60 - V \rho_w g = 40$,which implies the buoyant force $F_B = V \rho_w g = 20 \, N$.Now,the specific gravity of the block is defined as $SG = \frac{\rho}{\rho_w} = \frac{V \rho g}{V \rho_w g}$.Substituting the values,$SG = \frac{60}{20} = 3$.Therefore,the specific gravity of the block is $3$.

The reading of a spring balance when a block is suspended from it in air is $60 \, N$. This reading is changed to $40 \, N$ when the block is submerged in water. The specific gravity of the block must be therefore ............

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