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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

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)

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...