SSC JE Electrical Engineering 2010 Paper

$\mathit{E}=\mathit{P}\mathit{\varphi }\frac{\mathit{Z}}{\mathit{A}}\frac{\mathit{N}}{60}$
P = Number of Pole = 4
Z = Number of conductor = 1500
N = Speed in RPM = 1200
Q = Flux per pole in weber = .01
A = Number of Parallel path = P = 4
In lap would generator A = Pole = 4
$\mathit{E}=4\times .01\times \frac{1500}{4}$ $\times \frac{1200}{60}$ $=304\mathrm{volts}$