$F _{ ex }=\frac{ d P }{ dt }=0 $
$\Rightarrow dP =0 $
$\Rightarrow P =$ constant
$\vec{ P }_{ i }=\vec{ P }_{ f }$
$0=\vec{ P }_{ Nu }+\vec{ P }_{ Ph }$
$\left|\vec{ P }_{ Nu }\right|=\left|\vec{ P }_{ Ph }\right|=\frac{ h }{\lambda}=\frac{ hv }{ c }$
Recoil $K.E$. of nucleus $K . E _{ Nu }=\frac{ P _{ Nu }^{2}}{2 M _{ Nu }}$
$K.E _{\cdot Nu }=\frac{( hv / c )^{2}}{2 M }=\frac{ h ^{2} v^{2}}{2 Mc ^{2}}$
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: