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A spherical liquid drop of radius R is divided into 8 equal droplets. If the surface tension is T, then work done in the process will be
Options
(a) 2πR²T
(b) 3πR²T
(c) 4πR²T
(d) 2πRT²
Correct Answer:
4πR²T
Explanation:
Given: Surface tension = T
Radius of liquid drop = R
Let r be the radius of each small droplet
.·. Volume of big drop = volume of small droplets
(4/3) πR³ = 8 x (4/3) πr²
or R = 2r or r = R/2 ——(i)
Surface area of big drop = 4πR²
Surface are of 8 small droplets = 8 x 4πr²
8 x 4π x (R/2)² = 2(4πR²)
Increase in surface area = 2(4πR²) – 4πR² = 4πR²
Work done = T x increase in surface area
= T x 4πR² = 4πR²T.
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Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The input signal given to a CE amplifier having a voltage gain of 150 is
- Three solids of masses m₁,m₂ and m₃ are connected with weightless string
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- At what angle should the two forces 2P and √2 P act,so that the resultant force is P√10?
- B is doped in Si or Ge, then we will get
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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