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JEE Main 202325 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2023 · 25 January · Evening Shift · Physics · Mechanical Properties of Fluids

A spherical drop of liquid splits into 1000 identical spherical...

A spherical drop of liquid splits into 10001000 identical spherical drops. If ui{u}_{i} is the surface energy of the original drop and uf{u}_{f} is the total surface energy of the resulting drops, the (ignoring evaporation), ufui=(10x)\dfrac{{u}_{f}}{{u}_{i}} = \left(\dfrac{10}{x}\right). Then value of xx is ______:

Correct answer

0

Step-by-step solution

Let RR be the radius of bigger drop and rr be the radius of smaller drop.

Here, volume will remain conserved, so VR=Vr{V}_{R} = {V}_{r}.

⇒43πR3=1000×43πr3⇒ \dfrac{4}{3}\pi{R}^{3} = 1000 \times \dfrac{4}{3}\pi{r}^{3}

⇒R=10r⇒ R = 10r

As surface energy is stated as u=SAu = SA, here, SS is surface tension and AA is surface area.

Thus, ui=S{u}_{i} = S\left(4\pi{R}^{2}\right)anduf=S(4πr2)×1000and {u}_{f} = S\left(4\pi{r}^{2}\right) \times 1000

The ratio is ufui=1000r2R2\dfrac{{u}_{f}}{{u}_{i}} = \dfrac{1000{r}^{2}}{{R}^{2}}

⇒ufui=101⇒ \dfrac{{u}_{f}}{{u}_{i}} = \dfrac{10}{1}

Hence, the value of x=1x = 1.

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