Let R_1 and R_2 be the radii of two mercury d | vedclass

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(B) When two mercury drops combine to form a single larger drop,the total volume of the mercury remains conserved.Let $R$ be the radius of the resulta

Vedclass · Accepted Answer

$\left(R_1^3+R_2^3\right)^{\frac{1}{3}}$

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(B) When two mercury drops combine to form a single larger drop,the total volume of the mercury remains conserved.Let $R$ be the radius of the resultant big drop.The volume of the first drop is $V_1 = \frac{4}{3} \pi R_1^3$.The volume of the second drop is $V_2 = \frac{4}{3} \pi R_2^3$.The volume of the resultant drop is $V = \frac{4}{3} \pi R^3$.Since the total volume is conserved,$V = V_1 + V_2$.$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_1^3 + \frac{4}{3} \pi R_2^3$.Dividing both sides by $\frac{4}{3} \pi$,we get $R^3 = R_1^3 + R_2^3$.Therefore,the radius of the resultant drop is $R = (R_1^3 + R_2^3)^{1/3}$.

Let $R_1$ and $R_2$ be the radii of two mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

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