Volume of water, \(V \)= \(1 \,L\)
It is given that water is to be compressed by \(0.10\) %.
\(\therefore\) Fractional change, \(\frac{\triangle V }{ V }\)
= \(\frac{0.1 }{ 100 \times 1}\) = \(10 - 3\)
Bulk modulus, \(B\) = \(\frac{p }{ \frac{\triangle V }{ V}}\)
\(P\) = \(B × \frac{\triangle V }{ V}\)
Bulk modulus of water, \(B\) = \(2.2 × 10^ 9 \,Nm^{ -2}\)
\(p\) = \(2.2 × 10^ 9 × 10^{ - 3}\)
= \(2.2 × 10^ 6 \,Nm^{ - 2}\)
Therefore, the pressure on water should be \(2.2 × 10^ 6 \,Nm^{ - 2}\).