A step down transformer converts transmission line voltage from 1100V to 220V. The primary of the transformer has 6000 turns and the efficiency of the transformer is 60 %. If the output power is 9kW, then the input power will be
$\text{A}\text{. }11kW$
$\text{B}\text{. }12kW$
$\text{C}\text{. }14kW$
$\text{D}\text{. }15kW$

Hint: A transformer has a primary coil and a secondary coil. The ratio of the power of the secondary coil to the power of primary coil is called efficiency of the transformer i.e. $\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$. With formula find the input power (power in the primary coil).
Formula used:
$\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$

Complete step by step solution:
A transformer is a device that can increase or decrease the voltage of the circuit as per requirement. It consists of a primary coil and secondary coil. The primary coil acts as an input and the secondary coil acts as an output.
A transformer is used in two ways. One is to get a voltage greater than the voltage of the primary coil. This type of transformer is called a set-up transformer.
Another is get the voltage of the secondary coil smaller than the voltage of the primary coil. This type of transformer is called a set-down transformer.
Both the coils have some power associated with them. The power of the input or primary coil passed to the secondary coil. However, there is some leakage of this power and hence the output power (${{P}_{s}}$) is less than the input power (${{P}_{p}}$).
Therefore, we define the efficiency of the transformer which is equal to $\eta =\dfrac{{{P}_{s}}}{{{P}_{p}}}$ …. (i).
It is given that the efficiency of the transformer is 60%, which means the ratio is equal to 0.6. And the output power is given to 9kW. This means ${{P}_{s}}=9kW=9000W$
Substitute the values of $\eta $ and output power in equation (i).
Therefore, this implies
$0.6=\dfrac{9000}{{{P}_{p}}}$
This further implies that
${{P}_{p}}=\dfrac{9000}{0.6}=15000W=15kW$.
Therefore, the power of the primary coil is 15kW.
Hence, the correct option is D.

Note: The ratio of the voltages across the primary and secondary coils depends on the number of turns in both coils. i.e. $\dfrac{{{V}_{s}}}{{{V}_{p}}}=\dfrac{{{N}_{s}}}{{{N}_{p}}}$.
Here, ${{N}_{s}}$ and ${{N}_{p}}$ are the number of turns in the secondary and the primary coils respectively.
When ${{N}_{s}}$>${{N}_{p}}$, the transformer acts as a set-up transformer.
When ${{N}_{s}}$<${{N}_{p}}$, the transformer acts as a set-down transformer.