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To what depth below the surface of sea s...

To what depth below the surface of sea should a rubber ball be taken as to decreases its volume by 0.1% (Given denisty of sea water = 1000 kg `m^(-3)`, Bulk modulus of rubber `= 9 xx 10^(8) Nm^(-2)`, acceleration due to gravity `= 10 ms^(-2)`)

A

9m

B

18m

C

180m

D

90m

Text Solution

Verified by Experts

The correct Answer is:
D
`(Delta V)/(V)= 0.001, rho = 1000kg//m^(3) , B= 9 xx 10^(8) N//m^(2)`
`g= 10 m//s^(2)`
`B= (P)/((-Delta V)/(V)) rArr B= (rho hg)/(0.1 xx 10^(-2))`
`rArr h= (9 xx 10^(8) xx 10^(-3))/(10^(3) xx 10) = 90m`
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