JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 10)
In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :
3.9
8.4
7.5
3.0
Explanation
$$V = I \times \rho {l \over A}$$
$$ \Rightarrow \rho = {{VA} \over {Il}} = {\pi \over 4}{{V{d^2}} \over {Il}}$$
$${{\Delta \rho } \over \rho } = {{2\Delta d} \over d} + {{\Delta V} \over V} + {{\Delta I} \over I} + {{\Delta l} \over l}$$
$$ = 2\left( {{{0.01} \over 5}} \right) + {{0.1} \over 5} + {{0.01} \over 2} + {{0.1} \over {10}}$$
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = 0.039 = 3.9\% $$
$$ \Rightarrow \rho = {{VA} \over {Il}} = {\pi \over 4}{{V{d^2}} \over {Il}}$$
$${{\Delta \rho } \over \rho } = {{2\Delta d} \over d} + {{\Delta V} \over V} + {{\Delta I} \over I} + {{\Delta l} \over l}$$
$$ = 2\left( {{{0.01} \over 5}} \right) + {{0.1} \over 5} + {{0.01} \over 2} + {{0.1} \over {10}}$$
$$ \Rightarrow $$ $${{\Delta \rho } \over \rho } = 0.039 = 3.9\% $$