Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is :
Solution
${P_{in}} = {P_0} + {{4T} \over {{R_1}}}$<br><br>
$\Rightarrow 1.01 = 1 + {{4T} \over {{R_1}}}$<br><br>
$\Rightarrow {{4T} \over {{R_1}}} = 0.01$<br><br>
$1.02 = 1 + {{4T} \over {{R_2}}}$<br><br>
$\Rightarrow {{4T} \over {{R_2}}} = 0.02$<br><br>
$\therefore {{{R_2}} \over {{R_1}}} = {1 \over 2}$<br><br>
$\Rightarrow {R_1} = 2{R_2}$<br><br>
$${{{V_1}} \over {{V_2}}} = {{R_1^3} \over {R_2^3}} = {{8R_2^3} \over {R_2^3}} = {8 \over 1}$$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
This question is part of PrepWiser's free JEE Main question bank. 183 more solved questions on Properties of Solids and Liquids are available — start with the harder ones if your accuracy is >70%.