Question

a body of density d1 is counterpoised by Mg of weights of density d2 in air of density d. Then the true mass of the body is​

Answer 1

The true mass of the body is $m\frac{(1-\frac{d}{d_1})}{(1-\frac{d}{d_2})}$

Explanation:

=> Apparent weight of body is equals to the apparent weight of mass.

Apparent weight of mass = mg - Vdg

But V = m/d₁

∴ Apparent weight of mass =$mg-\frac{m}{d_1}*dg$

=$m(1-\frac{d}{d_1})g$ ...(1)

Similarly apparent weight of body = M$(1-\frac{d}{d_2})g$ ...(2)

=> Apparent weight of body = Apparent weight of mass.

M$(1-\frac{d}{d_2})g$ = $m(1-\frac{d}{d_1})g$

M =  m $\frac{(1-\frac{d}{d_1})}{(1-\frac{d}{d_2})}$

Thus, the true mass of the body is $m\frac{(1-\frac{d}{d_1})}{(1-\frac{d}{d_2})}$.

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