The reading of a spring balance when a block | vedclass

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(A) Let the weight of the block in air be $W_{actual} = 60\,N$.Let the apparent weight of the block in water be $W_{app} = 40\,N$.The buoyant force $F

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$3$

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(A) Let the weight of the block in air be $W_{actual} = 60\,N$.Let the apparent weight of the block in water be $W_{app} = 40\,N$.The buoyant force $F_B$ acting on the block is the difference between the actual weight and the apparent weight:$F_B = W_{actual} - W_{app} = 60\,N - 40\,N = 20\,N$.According to Archimedes' principle,the buoyant force is equal to the weight of the water displaced:$F_B = V \cdot ho_w \cdot g = 20\,N$,where $V$ is the volume of the block and $ho_w$ is the density of water.The actual weight of the block is $W_{actual} = V \cdot ho_s \cdot g = 60\,N$,where $ho_s$ is the density of the block.Dividing the two equations:$\frac{W_{actual}}{F_B} = \frac{V \cdot ho_s \cdot g}{V \cdot ho_w \cdot g} = \frac{ho_s}{ho_w} = 	ext{Specific Density}$.$	ext{Specific Density} = \frac{60}{20} = 3$.

The reading of a spring balance when a block is suspended from it in air is $60\,N$. This reading changes to $40\,N$ when the block is immersed in water. The specific density (relative density) of the block is:

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