Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The energy of a bubble is E. If one thousand such bubbles coalesce to form a large bubble, its energy is E1. If the ratio of E and E1 will be equal to  <img alt="\frac{1}{x}" src="ht

The energy of a bubble is E. If one thousand such bubbles coalesce to form a large bubble, its energy is E1. If the ratio of E and E1 will be equal to  \frac{1}{x}, the value of x is:

Option: 1

200


Option: 2

100


Option: 3

400


Option: 4

300


Answers (1)

best_answer

As volume remains constant, therefore

R = n^{1/3}r

Energy of one small drop / Energy of one big drop = \frac{E}{E1}

        = \frac{(4\prod r^2) T}{4\prod R^2) T}

\implies \frac{E}{E1} = \frac{r^2}{R^2}

\frac{E}{E1} = \frac{r^2}{(n^{1/3 r})^2}

\frac{E}{E1} = \frac{1}{(n^{2/3 })}

Given, n=100, the ratio of \frac{E}{E1} becomes:

\frac{E}{E1} = \frac{1}{(1000)^{2/3}}

\frac{E}{E1} = \frac{1}{(10^3)^{2/3}}                        \because 10^3 = 1000

\frac{E}{E1} = \frac{1}{10^2}

\frac{E}{E1} = \frac{1}{100}

From the question, the ratio is:

\frac{E}{E1} = \frac{1}{x}

\implies \frac{1}{100} = \frac{1}{x}

x = 100

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA Exam Calendar 2025 Live: JEE Main, NEET, CUET schedule to be announced soon; download at nta.ac.in
anam-khan

NTA Exam Calendar 2025: JEE Main, NEET, CUET exam dates soon
anam-khan

JEE Main, NEET, CUET 2025 Dates Live: NTA to announce exam calendar soon on nta.ac.in; latest updates

Latest Question