JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 17)
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)
(Given bulk modulus of metal, B = 8 $$ \times $$ 1010 Pa)
0.6
20
1.67
5
Explanation
Bulk Modulus, B = $$\left( - \right){{\Delta P} \over {\Delta V/V}} $$
$$\Delta P = -\left( {{{\Delta V} \over V}} \right).B$$
$$ = -{{3\Delta L} \over L} \times B$$
$$ \therefore $$ $$|{{\Delta L} \over L}| = {{\Delta P} \over {3B}}$$
$$ \therefore $$ % change, $${{\Delta L} \over L} \times 100\% $$
= $${1 \over 3}{{\Delta P} \over B} \times 100$$
= $${{4 \times {{10}^9}} \over {8 \times {{10}^{10}}}} \times 100$$
= $${1 \over {60}} \times 100$$ = 1.67
$$\Delta P = -\left( {{{\Delta V} \over V}} \right).B$$
$$ = -{{3\Delta L} \over L} \times B$$
$$ \therefore $$ $$|{{\Delta L} \over L}| = {{\Delta P} \over {3B}}$$
$$ \therefore $$ % change, $${{\Delta L} \over L} \times 100\% $$
= $${1 \over 3}{{\Delta P} \over B} \times 100$$
= $${{4 \times {{10}^9}} \over {8 \times {{10}^{10}}}} \times 100$$
= $${1 \over {60}} \times 100$$ = 1.67
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