

$E_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8}}{(2 R)^2} $
$E_1-E_2=\frac{\rho R}{4 \varepsilon_0}-\frac{\rho . R}{\varepsilon_0 \cdot 24 \times 4} $
$=\frac{\rho R}{4 \varepsilon_0}\left[1-\frac{1}{24}\right] $
$=\frac{23 \rho R }{96 \varepsilon_0}=\frac{23 \rho R }{16 K \varepsilon_0} $
$\Rightarrow \quad K =6 $
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

The magnitude of the magnetic field $(B)$ due to the loop $ABCD$ at the origin $(O)$ is :

$F = 600 - 2 \times 10^5\ t$
Where $F$ is in newton and $t$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?
