An increases in pressure required to decreases the $200\,litres$ volume of a liquid by $0.004\%$ in container is ............ $kPa$  (Bulk modulus of the liquid $= 2100\,MPa$ )
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$B=\frac{\Delta P}{\left(-\frac{\Delta V}{V}\right)}$

$\Rightarrow \Delta \mathrm{P}=\mathrm{B}\left(-\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)$

$\Rightarrow \Delta \mathrm{P}=\left(2100 \times 10^{6}\right)\left(\frac{0.004}{100}\right)$

$\Rightarrow\left(\text { As }-\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=0.004\right)$

$\Rightarrow \Delta \mathrm{P}=84 \times 10^{3} \mathrm{P_a}$

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