Question:

A large number of liquid drops each of radius a coalesce to form a single spherical drop of radius bb. The energy released in the process is converted into the kinetic energy of the big drop formed, the speed of the big drop is

Updated On: Jul 29, 2022
  • 4Tρ(1a1b) \sqrt{\frac{4T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}
  • 2Tρ(1a1b) \sqrt{\frac{2T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}
  • Tρ(1a1b) \sqrt{\frac{T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}
  • 6Tρ(1a1b) \sqrt{\frac{6T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)}
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The Correct Option is D

Solution and Explanation

According to question 43πa3\frac{4}{3} \pi a^{3} n=43πb3 n=\frac{4}{3} \pi b^{3} n=(ba)3 n=\left(\frac{b}{a}\right)^{3} W=T.4π[na2n2]WW=T .4 \pi\left[n a^{2}-n^{2}\right] W =12mv2=1243πb3pv2=\frac{1}{2} m v^{2}=\frac{1}{2} \cdot \frac{4}{3} \pi b^{3} p v^{2} 12×43πb3pv2=T.4π[na2b2] \therefore \frac{1}{2} \times \frac{4}{3} \pi b^{3 p v^{2}=} T .4 \pi\left[n a^{2}-b^{2}\right] or v=6Tρ(na2b3b2b3)v=\sqrt{\frac{6 T}{\rho}\left(\frac{n a^{2}}{b^{3}}-\frac{b^{2}}{b^{3}}\right)} v=6Tρ(1a1b)\therefore v=\sqrt{\frac{6 T}{\rho}\left(\frac{1}{a}-\frac{1}{b}\right)}
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Concepts Used:

Kinetic energy

Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.