The correct option is (A) : 100%
P1 = \(\sqrt{2}\)mE1 ; P2 = \(\sqrt{2}\)mE2
=\(\sqrt{2}\)m(\(\frac{E_1+300}{100E_1}\)) =√2m(4E1)=2P1
∴ % change = \(\frac{P_2-P_1}{P_1}\times100\) = \(\frac{2P_2-P_1}{P_1}\times100\)=100%
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec:
Read More: Work and Energy