Q 12-03-266JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
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 center, the equivalent resistance between these two points will be
Answer: (A) $\dfrac53\ \Omega$
Stretching to double length at constant volume multiplies resistance by $2^2$: $R = 12\ \Omega$.
A $60^\circ$ arc is $\frac16$ of the circle: $2\ \Omega$; the rest is $10\ \Omega$. In parallel:
$$R_{eq} = \frac{2\times10}{12} = \frac53\ \Omega$$
Solution by Sreeraj P, M.Sc Physics