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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
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B
2
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C
1
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D
4
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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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