Question

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. 18 m
C. 180 m
D. 90 m

Answer 1

Correct Answer - d
Bulk modulus, `B = ("Normal stress")/("Volume strain")`
`:. B = (p)/(Delta V//V)`
Given, `(Delta V)/(V) = (0.1)/(100), B = 9 xx 10^(8) Nm^(-2)`
`p = h d g = h xx 1000 xx 10`
`:. (B Delta V)/(V) = p = hdg`
`implies h = (B Delta V)/(Vdg) = (9 xx 10^(8) xx 0.1)/(1000 xx 100 xx 10) = 90 m`