Question

If the volume of a block of aluminum is decreased by $1\%$, the pressure (stress) in its surface is increased by (Bulk modulus of $A1=7.5\times {10}^{10}N{m}^{-2})$

• A. $7.5\times {10}^{10}N{m}^{-2}$
• B. $7.5\times {10}^{8}N{m}^{-2}$
• C. $7.5\times {10}^{6}N{m}^{-2}$
• D. $7.5\times {10}^{4}N{m}^{-2}$
• E. $7.5\times {10}^{2}N{m}^{-2}$

Answer 1

Answer: (B)

$7.5\times {10}^{8}N{m}^{-2}$

$\begin{array}{c}\text{Given,}\\ \text{volume of block decreased by}1\%\\ \text{Buk modulus}=7.5\times {10}^{10}{Nm}^{-2}\end{array}$

$\text{we know,}$

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

$\begin{array}{cc}\Delta p& =-\beta \left(\frac{\Delta v}{v}\right)\\ & =+7.5\times {10}^{10}\times 1\times {10}^{-2}\\ \Delta p& =7.5\times {10}^{8}N/m^2\end{array}$