A six pole generator with fixed field excitation develops an emf of 100v, when operating at 1500rpm. At what speed must it rotate to develop 120v?
1) 1200rpm
2) 1800rpm
3) 1500rpm
4) 400rpm

Hint:A generator has two poles North and South and in between the poles there is a coil which when rotated generates an EMF. There would be two equations formed one for the emf that is given and another one that is not given. Equate the two equations and find the speed.

Formula used:
$e = NBA\omega $;
Where:
e = Emf;
N = Number of turns;
B = Magnetic Field;
A = Area;
$\omega $= Angular velocity;

Complete step-by-step answer:
Write the equation and solve.
$e = NBA\omega $;
Keep B to the RHS and take the rest of the variables to the LHS
$\dfrac{e}{{NA\omega }} = B$;
Put the given value of e and $\omega $
$\dfrac{{100}}{{1500 \times NA}} = B$;
Write the equation for the second emf, e= 120v
$e = NBA\omega $;
Put the values in the above equation:
$120 = NBA\omega $;
Put value of B in the above equation ($\dfrac{{100}}{{1500 \times NA}} = B$):
$120 = N \times \dfrac{{100}}{{1500 \times NA}} \times A \times \omega $;
Cancel out the common numerator and denominator,
$120 = \dfrac{1}{{15}} \times \omega $;
Solve for$\omega $:
$\omega = 1800rpm$;

Final Answer:Option “2” is correct. A six pole generator with fixed field excitation develops an emf of 100V, when operating at 1500rpm. The speed it must rotate to develop 120V is 1800rpm.

Note:Here we have to compare one emf to another emf to find out the speed i.e. the number of rotations. The common factors in both of the emf will cancel each other out. Go step by step, write the first equation and then the second.