Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A gas filled freely collapsible balloon is pushed from the surface level of lake to a depth of 100 metre . Calculate what per cent of its original volume , the balloon finally have ? Assume ideal gas nature .

$(a)\;9\% \;of\; V_i\qquad(b)\;9.37\% \;of\; V_i\qquad(c)\;0.93\% \;of\; V_i\qquad(d)\;10\% \;of\; V_i$

Can you answer this question?

1 Answer

0 votes
Pressure at 100 metre depth
= Atmospheric pressure + pressure of water
= $76\times13.6\times981 + 100\times100\times1\times981$
= 1013961.6 + 9810000
$= 10823961.6 \;dyne\;cm^{-2}$
Let volume of balloon at the surface level be $V_i \;cm^3$ and the volume of balloon at the depth of 100 metre is $V_f$ then
$P_1V_1 = P_2V_2$
$\therefore 76\times13.6\times981\times V_i = 10823961.6\times V_f$
$V_f = \large\frac{76\times13.6\times981}{10823961}\times V_i$
$V_f = 0.0937 V_i$
$V_f = 9.37 \%\; of\; V_i$
Hence answer is (B)
answered Apr 3, 2014 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App