A cube of metal is subjected to a hydrostatic pressure of 4 GPa. The percentage change in the
length of the side of the cube is close to :
(Given bulk modulus of metal, B = 8 $\times$ 1010 Pa)
Solution
Bulk Modulus, B = $\left( - \right){{\Delta P} \over {\Delta V/V}}$<br><br>$\Delta P = -\left( {{{\Delta V} \over V}} \right).B$<br><br>$= -{{3\Delta L} \over L} \times B$<br><br>$\therefore$ $|{{\Delta L} \over L}| = {{\Delta P} \over {3B}}$
<br><br>$\therefore$ % change, ${{\Delta L} \over L} \times 100\%$
<br><br>= ${1 \over 3}{{\Delta P} \over B} \times 100$
<br><br>= ${{4 \times {{10}^9}} \over {8 \times {{10}^{10}}}} \times 100$
<br><br>= ${1 \over {60}} \times 100$ = 1.67
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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