Question

# A solid sphere weight $$40\ N$$ in water and $$45\ N$$ in a liquid of relative density $$0.75$$. Relative density of sphere will be

A
3
B
2
C
1
D
4

Solution

## The correct option is A $$3$$If apparent weight is $$w'$$ in a liquid It means $$w'$$ is the weight of the sphere minus the weight of the liquid the sphere's volume displaces.So $$w'=\rho_sV_sg-\rho_lV_sg$$for water $$w_1=\rho_sV_sg-\rho_wV_sg$$$$40=\rho_sV_sg-1\times V_sg$$                                   (1)for liquid $$w_1=\rho_sV_sg-\rho_wV_sg$$$$45=\rho_sV_sg-0.75\times V_sg$$                              (2)Solving equations we get   $$\rho_s=3$$Hence relative density of sphere is 3.Physics

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