The position vector of a moving body at any instant of time is given as $\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 s$ is,

A photograph of a landscape is captured by a drone camera at a height of 18 km . The size of the camera film is $2 \mathrm{~cm} \times 2 \mathrm{~cm}$ and the area of the landscape photographed is $400 \mathrm{~km}^2$. The focal length of the lens in the drone camera is :

In the first configuration (1) as shown in the figure, four identical charges $\left(q_0\right)$ are kept at the corners A, B, C and D of square of side length ' $a$ '. In the second configuration (2), the same charges are shifted to mid points $G, E, H$ and $F$, of the square. If $K=\frac{1}{4 \pi \epsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by :

A particle oscillates along the $x$-axis according to the law, $x(\mathrm{t})=x_0 \sin ^2\left(\frac{\mathrm{t}}{2}\right)$ where $x_0=1 \mathrm{~m}$. The kinetic energy $(\mathrm{K})$ of the particle as a function of $x$ is correctly represented by the graph