JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 12)
A conducting wire of length 'l', area of cross-section A and electric resistivity $$\rho$$ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.
If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :
If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :
$$4{{VA} \over {\rho l}}$$
$${3 \over 4}{{VA} \over {\rho l}}$$
$${1 \over 4}{{VA} \over {\rho l}}$$
$${1 \over 4}{{\rho l} \over {VA}}$$
Explanation
We know that
$$R = \rho {l \over A}$$
Now, new length : $$l' = 2l$$
new area of cross section : $$A' = A/2$$
$$ \therefore $$ New resistance : $$R' = \rho .{{2l} \over {A/2}}$$
$$ \Rightarrow R' = 4{{\rho l} \over A}$$
$$ \Rightarrow R' = 4R$$
$$ \therefore $$ Resultant current : $$I = {V \over {4R}}$$
$$ \Rightarrow $$ $$I = {1 \over 4}{{VA} \over {\rho l}}$$
$$R = \rho {l \over A}$$
Now, new length : $$l' = 2l$$
new area of cross section : $$A' = A/2$$
$$ \therefore $$ New resistance : $$R' = \rho .{{2l} \over {A/2}}$$
$$ \Rightarrow R' = 4{{\rho l} \over A}$$
$$ \Rightarrow R' = 4R$$
$$ \therefore $$ Resultant current : $$I = {V \over {4R}}$$
$$ \Rightarrow $$ $$I = {1 \over 4}{{VA} \over {\rho l}}$$
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