Figure shows four solid spheres, each with charge Q uniformly distributed through its volume. The figure also shows a point P for each sphere, all at the same distance from the center of the sphere. Rank the spheres according to the magnitude of the electric field they produce at point P.

1. $III\to II\to I\to IV$
2. All tie
3. $II\to I\to III\to IV$
4. $I\text{\&}II\text{tie}\to III\to IV$

The correct option is D $I\text{\&}II\text{tie}\to III\to IV$
We know that
$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{1}{{\epsilon }_0}\left(Q_{enc}\right)$



Let us take a spherical Gaussian surface passing through P and enclosing the charge.
For (I) and (II) charge enclosed is the same and equal to Q
Since point P is lying outside the surface. Hence the magnitude of the electric field will be equal according to the above formula.

The charge density in case (III) is greater than case (IV) as the volume of sphere (III) is lesser than volume of sphere(IV). Hence the charge enclosed in case(III) is greater than case(IV), it means the magnitude of electric field in case(III) is greater than in case(IV)
Also charge enclosed by (III) and (IV) will be less than (I) and (II) since point P lying inide the former spheres due to which the charge enclosed will be less.
Thus, the magnitude of electric field at point.
$\left(I\right)=\left(II\right)\to \left(III\right)\to \left(IV\right)$
Hence, option (a) is correct.