MHT CET · Physics · Mechanical Properties of Fluids
A spherical drop of liquid splits into 1000 identical spherical drops. If ' \(\mathrm{E}_1\) ' is the surface energy of the original drop and ' \(\mathrm{E}_2\) ' is the total surface energy of the resulting drops, then \(\frac{E_1}{E_2}=\frac{x}{10}\). Then value of ' \(x\) ' is
- A 9
- B 7
- C 3
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(r=\text { Radius of small drop } \)
\( R=\text { Radius of bigger drop } \)
\( \therefore \frac{4}{3} \pi R^3=\frac{4}{3} \pi(1000) r^3 \)
\( \therefore R=10 r \)
\( E_1=T_1^2=T\left(4 \pi R^2\right) \)
\( E_2=n A_2^2=1000 \times T\left(4 \pi r^2\right) \)
\( \therefore \frac{E_1}{E_2}=\frac{R^2}{1000 r^2}=\frac{(10 r)^2}{1000 r^2} \)
\( \therefore \frac{E_1}{E_2}=\frac{1}{10} \)
\( \therefore x=1\)
\( R=\text { Radius of bigger drop } \)
\( \therefore \frac{4}{3} \pi R^3=\frac{4}{3} \pi(1000) r^3 \)
\( \therefore R=10 r \)
\( E_1=T_1^2=T\left(4 \pi R^2\right) \)
\( E_2=n A_2^2=1000 \times T\left(4 \pi r^2\right) \)
\( \therefore \frac{E_1}{E_2}=\frac{R^2}{1000 r^2}=\frac{(10 r)^2}{1000 r^2} \)
\( \therefore \frac{E_1}{E_2}=\frac{1}{10} \)
\( \therefore x=1\)
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