Question
A wire of resistance $$12\,\Omega {m^{ - 1}}$$ is bent to form a complete circle of radius $$10\,cm.$$ The resistance between its two diametrically opposite points $$A$$ and $$B$$ as shown in the figure is
A wire of resistance $$12\,\Omega {m^{ - 1}}$$ is bent to form a complete circle of radius $$10\,cm.$$ The resistance between its two diametrically opposite points $$A$$ and $$B$$ as shown in the figure is
A.
$$0.6\,\pi \,\Omega $$
B.
$$3\,\Omega $$
C.
$$6\,\pi \,\Omega $$
D.
$$6\,\Omega $$
Answer :
$$0.6\,\pi \,\Omega $$
Solution :
Circumference of circle $$ = 2\pi r = 2 \times \pi \frac{{10}}{{100}}$$
$$ = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$$
Resistance of wire $$ = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}$$
Resistance of each section $$ = \frac{{12\pi }}{{10}}\Omega $$
$$\therefore $$ Equivalent resistance $$ = \frac{{\frac{{12\pi }}{{10}} \times \frac{{12\pi }}{{10}}}}{{\frac{{12\pi }}{{10}} + \frac{{12\pi }}{{10}}}}$$
$$ = \frac{{6\pi }}{{10}} = 0.6\,\pi \,\Omega $$
Circumference of circle $$ = 2\pi r = 2 \times \pi \frac{{10}}{{100}}$$
$$ = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$$
Resistance of wire $$ = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}$$
Resistance of each section $$ = \frac{{12\pi }}{{10}}\Omega $$
$$\therefore $$ Equivalent resistance $$ = \frac{{\frac{{12\pi }}{{10}} \times \frac{{12\pi }}{{10}}}}{{\frac{{12\pi }}{{10}} + \frac{{12\pi }}{{10}}}}$$
$$ = \frac{{6\pi }}{{10}} = 0.6\,\pi \,\Omega $$
