A circular ring of diameter 20cm has a resistance of $0.01\Omega $. How much charge will flow through the ring if it is turned in a uniform magnetic field of 2.0T from an initial position perpendicular to the field to a position parallel to the field?
A. 63C
B. 0.63C
C. 6.3C
D. 0.063C

Hint: As a very first step, one could read the question well and hence note down the important values from the given question. Then one could recall the relation for induced charge in terms of the given quantities and then carry out the substitution accordingly and hence find the answer.
Formula used:
Charge induced,
$Q=\dfrac{d\phi }{R}$

Complete step by step answer:
In the question, we are given a circular ring which has a diameter of 20cm with a resistance of $0.01\Omega $. This ring is being turned in a uniform magnetic field of 2.0T from an initial position that is perpendicular to the field to a certain position that is parallel to the field. We are supposed to find the charge flow through the ring.
Radius of the ring would be,
$r=\dfrac{20}{2}cm=10cm=0.1m$
Now the area of the ring would be,
$A=\pi {{r}^{2}}=\pi {{\left( 0.1 \right)}^{2}}=0.0314{{m}^{2}}$
Change flux would be given by,
$d\phi =BA\left( \cos 0{}^\circ -\cos 90{}^\circ \right)=2\times 0.0314\left( 1-0 \right)=0.0628Wb$
Now the amount of charge that is induced corresponding change in magnetic flux would be given by,
$Q=\dfrac{d\phi }{R}$
Substituting the values,
$Q=\dfrac{0.0628}{0.01}$
$\therefore Q=6.28C\approx 6.3C$
Therefore, we found the induced charge due to the mentioned rotation to be 6.3C.

So, the correct answer is “Option C”.

Note: Many may not be familiar with the direct relation used to find the charge induced in the above case in terms of change in flux and resistance. We know the expression for induced emf given by,
$\varepsilon =-\dfrac{d\phi }{dt}$
We know that,
$\varepsilon =IR=\dfrac{q}{t}R$
Equating the RHS of the above relations, we get,
$q=\dfrac{d\phi }{R}$