MHTCET 19 Apr 2026 Shift 1 Paper

Concept:For a point on the axial line of an electric dipole, the field is $E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}$.At the zero-field point, the field magnitudes of the two dipoles must cancel each other.Explanation:Let the point be at distance $x$ from the dipole of moment $P$.Since the dipoles are 24 cm apart, the distance from the other dipole is $(24-x)$ cm.The dipoles are in opposite directions, so their fields are opposite at the point between them.Equating their magnitudes:$$\frac{2KP}{x^3} = \frac{2K(nP)}{(24-x)^3}$$Cancelling the common terms:$$\frac{1}{x^3} = \frac{n}{(24-x)^3}$$Taking the cube root on both sides:$$24-x = \sqrt[3]{n}\,x$$Solving for $x$:$$x = \frac{24}{1+\sqrt[3]{n}}$$For $n = 8$ (dipoles $P$ and $8P$):$$x_1 = \frac{24}{1+\sqrt[3]{8}} = \frac{24}{1+2} = 8\ \text{cm}$$For $n = 27$ (dipoles $P$ and $27P$):$$x_2 = \frac{24}{1+\sqrt[3]{27}} = \frac{24}{1+3} = 6\ \text{cm}$$Thus,$$\left|x_2 - x_1\right| = |6-8| = 2\ \text{cm}$$Answer:The required difference is $2\ \text{cm}$, so the correct option is A.