Question

A solid metal sphere of volume 0.31m3
is dropped in an ocean where water
pressure is 2x10²N/m². Calculate change
in volume of the sphere if bulk modulus
of the metal is 6.1x1010 N/m2
​

Answer 1

There is Decrement of $- 10^-^9\ m^3$

Explanation:

Here,  

B = $\frac {- \Delta P}{\Delta V V^-^1}$

$B \times \frac {\Delta V}{V} = - \Delta P$

So, $\Delta V = \frac {\Delta P \times V}{B}$

or $\Delta V = \frac {- 2 \times 10^2 \times 0.31}{6.1 \times 10^1^0}$

So, $\Delta V = \frac {- 6.2 \times 10}{6.1 \times 10^1^0}$

hence,$\Delta V = - 10^-^9\ m^3$

And this Negative Sign indicates the Decrease in Volume

Answer 2

Answer:1*10raise to power -4 m cube

Explanation:

V=0.31m3

=31*10raise to power -2

dp=2*10raise to power 7

K(bulk modulus) =6.1*10raise to power 9 N/m2

=61*10raise to power 9

dv=?

K=V.dp/dv

dv=V.dp/k

31*10