Question

A cube of metal is subjected to a hydrostatic pressure of $4GPa$. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $B=8\times {10}^{10}Pa$)

• A. $0.6$
• B. $20$
• C. $1.67$
• D. $5$

Answer 1

The correct option is C

$1.67$

Step 1; Given Data:

Applied pressure

$\begin{array}{rcl}P& =& 4GPa\\ & =& 4\times {10}^{9}Pa\end{array}$

Bulk modulus $B=8\times {10}^{10}Pa$

Step 2: Formula Usd:

$B=-\frac{P}{∆V/V}$

Where, $B=$ Bulk modulus

$P=$ Applied pressure

$∆V=$ Change in volume

$V=$ Volume

Step 3: Calculating the percentage change:

$\begin{array}{rcl}B& =& -\frac{P}{∆V/V}\\ 8\times {10}^9& =& -\frac{4\times {10}^{10}}{∆V/V}\\ \frac{∆V}{V}& =& -\frac{4\times {10}^9}{8\times {10}^{10}}\\ \frac{3∆L}{L}& =& -\frac{1}{20}\\ \frac{∆L}{L}& =& -\frac{1}{60}\end{array}$

Negative sign indicates compression means decrease in volume due to applied force.

$\begin{array}{rcl}\%change& =& \frac{∆L}{L}\times 100\\ & =& \frac{1}{60}\times 100\\ & =& 1.67\end{array}$

Hence, option C is the correct answer.