JEE MAIN - Physics (2004 - No. 23)
A material $$'B'$$ has twice the specific resistance of $$'A'.$$ A circular wire made of $$'B'$$ has twice the diameter of a wire made of $$'A'$$. Then for the two wires to have the same resistance, the ratio $${l \over B}/{l \over A}$$ of their respective lengths must be
$$1$$
$${l \over 2}$$
$${l \over 4}$$
$$2$$
Explanation
$${\rho _B} = 2{\rho _A}$$
$${d_B} = 2{d_A}$$
$${R_B} = {R_A} \Rightarrow {{{\rho _B}{\ell _B}} \over {{A_B}}} = {{{P_A}{\ell _A}} \over {{A_A}}}$$
$$\therefore$$ $${{{\ell _B}} \over {{\ell _A}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{d_B^2} \over {d_A^2}}$$ $$ = {{{\rho _A}} \over {2{\rho _A}}} \times {{4d_d^2} \over {d_A^2}} = 2$$
$${d_B} = 2{d_A}$$
$${R_B} = {R_A} \Rightarrow {{{\rho _B}{\ell _B}} \over {{A_B}}} = {{{P_A}{\ell _A}} \over {{A_A}}}$$
$$\therefore$$ $${{{\ell _B}} \over {{\ell _A}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{d_B^2} \over {d_A^2}}$$ $$ = {{{\rho _A}} \over {2{\rho _A}}} \times {{4d_d^2} \over {d_A^2}} = 2$$
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