Current Electricity 1 Question 4
4. 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 $60^{\circ}$ at the centre, the equivalent resistance between these two points will be
(Main 2019, 9 April II)
(a) $\frac{7}{2} \Omega$
(b) $\frac{5}{2} \Omega$
(c) $\frac{12}{5} \Omega$
(d) $\frac{5}{3} \Omega$
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Solution:
- Initial resistance of wire is $3 \Omega$. Let its length is $l$ and area is $A$.
Then, $\quad R _{\text {initial }}=\rho \frac{l}{A}=3 \Omega$
When wire is stretched twice its length, then its area becomes $A^{\prime}$, on equating volume, we have
$$ A l=A^{\prime} 2 l \Rightarrow A^{\prime}=\frac{A}{2} $$
So, after stretching, resistance of wire will be
$$ R^{\prime}=R _{\text {final }}=\rho \frac{l^{\prime}}{A^{\prime}}=4 \rho \frac{l}{A}=12 \Omega \quad \text { [using Eq. (i)] } $$
Now, this wire is made into a circle and connected across two points $A$ and $B$ (making $60^{\circ}$ angle at centre) as
Now, above arrangement is a combination of two resistances in parallel,
$$ \begin{aligned} & R _1=\frac{60 \times R^{\prime}}{360}=\frac{1}{6} \times 12=2 \Omega \\ & \text { and } \quad R _2=\frac{300}{360} \times R^{\prime}=\frac{5}{6} \times 12=10 \Omega \end{aligned} $$
Since, $R _1$ and $R _2$ are connected in parallel.
So, $\quad R _{A B}=\frac{R _1 R _2}{R _1+R _2}=\frac{10 \times 2}{12}=\frac{5}{3} \Omega$
