JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 18)
Two wires of same length and thickness having specific resistances 6$$\Omega$$ cm and 3$$\Omega$$ cm respectively are connected in parallel. The effective resistivity is $$\rho$$$$\Omega$$ cm. The value of $$\rho$$, to the nearest integer, is ____________.
Answer
4
Explanation
Let length of each wire is l and area A. When they are connected in parallel then their effective area 2A.
From formula we know,
$${R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$$
$$ \Rightarrow {{\rho l} \over {2A}} = {{{\rho _1}{l \over A} \times {\rho _2}{l \over A}} \over {{\rho _1}{l \over A} + {\rho _2}{l \over A}}}$$
$$ \Rightarrow {\rho \over 2} = {{{\rho _1} \times {\rho _2}} \over {{\rho _1} + {\rho _2}}}$$
$$ \Rightarrow {\rho \over 2} = {{6 \times 3} \over {6 + 3}} = 2$$
$$ \Rightarrow \rho = 4$$
From formula we know,
$${R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$$
$$ \Rightarrow {{\rho l} \over {2A}} = {{{\rho _1}{l \over A} \times {\rho _2}{l \over A}} \over {{\rho _1}{l \over A} + {\rho _2}{l \over A}}}$$
$$ \Rightarrow {\rho \over 2} = {{{\rho _1} \times {\rho _2}} \over {{\rho _1} + {\rho _2}}}$$
$$ \Rightarrow {\rho \over 2} = {{6 \times 3} \over {6 + 3}} = 2$$
$$ \Rightarrow \rho = 4$$
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