JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 18)
When the temperature of a metal wire is
increased from 0oC to 10oC, its length
increases by 0.02%. The percentage change in
its mass density will be closest to :
0.008
0.06
0.8
2.3
Explanation
Given, $${{\Delta L} \over L}$$ = 0.02%
We know, $$\Delta $$L = L$$\alpha $$$$\Delta $$T
$$ \Rightarrow $$ $${{\Delta L} \over L} = \alpha \Delta T$$ = 0.02
Also, $$\beta $$ = 2$$\alpha $$
$$ \Rightarrow $$ $$\beta \Delta T = 2\alpha \Delta T$$ = 0.04
Density($$\rho $$) = $${M \over {AL}}$$
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = {{\Delta M} \over M} - {{\Delta A} \over A} - {{\Delta L} \over L}$$ ( $${{\Delta M} \over M}$$ = 0 as M = constant)
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = {{\Delta A} \over A} + {{\Delta L} \over L}$$
= $$\beta \Delta T + \alpha \Delta T$$
= 0.04 + 0.02 = 0.06 %
We know, $$\Delta $$L = L$$\alpha $$$$\Delta $$T
$$ \Rightarrow $$ $${{\Delta L} \over L} = \alpha \Delta T$$ = 0.02
Also, $$\beta $$ = 2$$\alpha $$
$$ \Rightarrow $$ $$\beta \Delta T = 2\alpha \Delta T$$ = 0.04
Density($$\rho $$) = $${M \over {AL}}$$
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = {{\Delta M} \over M} - {{\Delta A} \over A} - {{\Delta L} \over L}$$ ( $${{\Delta M} \over M}$$ = 0 as M = constant)
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = {{\Delta A} \over A} + {{\Delta L} \over L}$$
= $$\beta \Delta T + \alpha \Delta T$$
= 0.04 + 0.02 = 0.06 %
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