The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02\% is m. (Take density of sea water =103 kgm-3, Bulk modulus of rubber =9 108 Nm-2, and g=10 ms-2)
Answer
18
Solution
B= P | VV |, extra pressure at depth h: P= gh
gh=B | VV |
103 10 h=9 108 0.02100
104h=1.8 105
h=18 m
Answer: 18