JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 27)
A metal wire of resistance 3 $$\Omega $$ is elongated to
make a uniform wire of double its previous
length. This new wire is now bent and the ends
joined to make a circle. If two points on this
circle make an angle 60° at the centre, the
equivalent resistance between these two points
will be :-
5 /2 $$\Omega $$
12/5 $$\Omega $$
7/2 $$\Omega $$
5 / 3 $$\Omega $$
Explanation
$$R = {{\rho l} \over A} = {{\rho l} \over {\left( {V/l} \right)}} = {{\rho {l^2}} \over V}\left( {V{\rm{ }} \to {\rm{ }}Volume{\rm{ }}\,of{\rm{ }}\,wire} \right)$$
$$ \Rightarrow $$ Final resistance = 3 × (B)2 = 12 $$\Omega $$
$${R_{eq}} = 2\Omega \parallel 10\Omega = {5 \over 3}\Omega $$
$$ \Rightarrow $$ Final resistance = 3 × (B)2 = 12 $$\Omega $$
$${R_{eq}} = 2\Omega \parallel 10\Omega = {5 \over 3}\Omega $$
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