If the length of a simple pendulum is increased by a factor of 4, by what factor does its period of oscillation increase?

The period ($T$) of a simple pendulum is given by:

Where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

From the formula, the period $T$ is directly proportional to the square root of the length $L$:

Let the original length be $L_1$ and the original period be $T_1$.

Where $C = 2\pi/\sqrt{g}$ is a constant.
If the length is increased by a factor of 4, the new length $L_2 = 4L_1$.
The new period $T_2$ will be:

The new period is twice the original period, meaning it is increased by a factor of 2.

The period will be $\boxed{\text{c) increased by 2}}$.

Internal resistance of the cell, $r = 2 \Omega$.
The terminal voltage ($V$) is reduced to $\frac{3}{5}$ of its nominal value (EMF, $E$). So, $V = \frac{3}{5} E$.

The terminal voltage ($V$) across an external resistance ($R$) connected to a cell with EMF ($E$) and internal resistance ($r$) is given by:

We have $V = \frac{3}{5} E$ and $r = 2 \Omega$.

Divide both sides by $E$ (assuming $E \neq 0$):

Cross-multiply:

Subtract $3R$ from both sides:

Divide by 2:

The value of R is $\boxed{\text{c) } 3 \Omega}$.