JEE Main 2014 · 19 Apr · Shift 1 · Physics Q10

← JEE Main 2014 · 19 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 10 of 90

A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of liquid T, density ρ)

\sqrt{\dfrac T\rho\left(\dfrac1r-\dfrac1R\right)}

\sqrt{\dfrac{2T}\rho\left(\dfrac1r-\dfrac1R\right)}

\sqrt{\dfrac{4T}\rho\left(\dfrac1r-\dfrac1R\right)}

\sqrt{\dfrac{6T}\rho\left(\dfrac1r-\dfrac1R\right)}

Answer and worked solution

Official answer: D

With $nr^3=R^3$ , the surface energy released is
$4\pi T(nr^2-R^2)=4\pi TR^3\left(\frac1r-\frac1R\right).$
Setting it equal to $\frac12\cdot\frac43\pi R^3\rho v^2$ :
$v=\sqrt{\frac{6T}\rho\left(\frac1r-\frac1R\right)}.$

Topics:Surface tension

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 10