A body of density $d_{1}$ is counterpoised by Mg of weights of density $d_{2}$ in air of density d. Then the true mass of the body is

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$M ⟮ 1 - \frac{d}{d_{2}} ⟯$

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$M ⟮ 1 - \frac{d}{d_{1}} ⟯$

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$\frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})}$

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Solution

The correct option is D
$\frac{M (1 - \frac{d}{d_{2}})}{(1 - \frac{d}{d_{1}})}$

Let $M_{O}$ =mass of body in vacuum.

Apparent weight of the body in air = Apparent weight of standard weights in air

Actual weight of the body - upthrust due to displaced air

= Actual weight of the weights- upthrust due to displaced air

$\Rightarrow M_{0} g - ⟮ \frac{M_{0}}{d_{1}} ⟯ d g = M g - ⟮ \frac{M}{d_{2}} ⟯ d g \Rightarrow M_{0} = \frac{M [1 - \frac{d}{d_{2}}]}{[1 - \frac{d}{d_{1}}]}$

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