a solid weighs 6kg in air. if its density is 2000kg/m3, what will be its apparent weight in water?

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Solution

$V o l u m e o f s o l i d = m a s s / d e n s i t y V = 6 / 2000 = 0.003 m^{3} w e i g h t i n w a t e r = a c t u a l w e i g h t - l o s s i n w e i g h t w e i g h t i n w a t e r = m g - v \times d e n s i t y o f w a t e r \times g = 6 \times 10 - 0.003 \times 1000 \times 10 60 - 30 = 30 N$

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a solid weighs 6kg in air. if its density is 2000kg/m3, what will be its apparent weight in water?

Volume of solid = mass/densityV= 6/2000 = 0.003 m3weight in water = actual weight - loss in weight weight in water = mg-v×density of water ×g= 6×10 - 0.003×1000×1060-30= 30 N

$V o l u m e o f s o l i d = m a s s / d e n s i t y V = 6 / 2000 = 0.003 m^{3} w e i g h t i n w a t e r = a c t u a l w e i g h t - l o s s i n w e i g h t w e i g h t i n w a t e r = m g - v \times d e n s i t y o f w a t e r \times g = 6 \times 10 - 0.003 \times 1000 \times 10 60 - 30 = 30 N$