MHTCET 15 May 2024 Shift 1 Paper

Concept:Gauss's law determines the electric field from the charge enclosed by a symmetric Gaussian surface.Explanation:Consider a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the charged cylinder.For a point inside ($r < R$), the electric field $\vec{E}$ is radial and constant over the curved surface.The electric flux through the curved surface is $\phi = E(2\pi rL)$.The end caps contribute zero flux because $\vec{E}$ is parallel to them.The charge enclosed is $Q_{\text{enc}} = \lambda (\pi r^2 L)$, using the given volume charge density $\lambda$.Applying Gauss's law:$$E(2\pi rL) = \frac{\lambda \pi r^2 L}{\varepsilon_0}$$Cancelling $\pi$ and $L$ from both sides:$$E(2r) = \frac{\lambda r^2}{\varepsilon_0}$$$$E = \frac{\lambda r}{2\varepsilon_0}$$This shows that $E$ is directly proportional to $r$.Answer:The electric field inside the cylinder is directly proportional to $r$, so the correct option is B. $r$.