If a pressure of 8 times 10^8 N m^{-2} is a | vedclass

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(C) The pressure applied on the block is $p = 8 	imes 10^8 \ N \ m^{-2}$.Let the initial volume be $V_1$. The final volume $V_2$ is reduced by $20 \%

Vedclass · Accepted Answer

$4 	imes 10^9 \ N \ m^{-2}$

Vedclass · Answer

(C) The pressure applied on the block is $p = 8 	imes 10^8 \ N \ m^{-2}$.Let the initial volume be $V_1$. The final volume $V_2$ is reduced by $20 \%$,so $V_2 = V_1 - 0.20 V_1 = 0.8 V_1 = \frac{4}{5} V_1$.The change in volume is $\Delta V = V_1 - V_2 = V_1 - 0.8 V_1 = 0.2 V_1 = \frac{V_1}{5}$.The Bulk modulus $B$ is defined as $B = \frac{p}{\Delta V / V_1}$.Substituting the values,$B = \frac{8 	imes 10^8}{(0.2 V_1) / V_1} = \frac{8 	imes 10^8}{0.2} = 40 	imes 10^8 = 4 	imes 10^9 \ N \ m^{-2}$.

If a pressure of $8 \times 10^8 \ N \ m^{-2}$ is applied to a lead block,so that its volume reduces by $20 \%$. The Bulk modulus of the lead block is:

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