A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from the battery becomes 10I. The value of 'n' is?

When the resistors are connected in series, the total resistance is nR + R = R(n+1). The current drawn is given by Ohm's law:
I = E / (R(n+1))

When the resistors are connected in parallel, the total resistance is R/n. The current drawn is:
10I = E / (R/n + R) = E / (R(1/n + 1)) = nE / (R(1+n))

We have two equations:

1. I = E / (R(n+1))
2. 10I = nE / (R(1+n))

Substitute equation (1) into equation (2):
10 * [E / (R(n+1))] = nE / (R(1+n))
10 = n
Therefore, the value of n is 10.
Option 2 is correct