The quantity $\frac{1}{2}\varepsilon_0 E^2$ represents the energy density of an electric field, which is energy stored per unit volume.
Dimension of Energy $[E] = ML^2T^{-2}$. Dimension of Volume $[V] = L^3$.
Thus, the dimension of energy density is $\frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$.
Therefore, the dimension of $\frac{1}{2}\varepsilon_0 E^2$ is $ML^{-1}T^{-2}$.
As shown in the figure, the ratio of \(T_1\) and \(T_2\) is 
In the circuit shown below, find the voltage across the capacitor in steady state.
Velocity versus time graph is given. Find the magnitude of acceleration of the particle at t = 5 s.


Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
As shown in the figure, the ratio of \(T_1\) and \(T_2\) is 