The initial speed of a bullet fired from a rifle is $630 \mathrm{~m} / \mathrm{s}$. The rifle is fired at the centre of a target 700 m away at the same level as the target. How far above the centre of the target should the rifle be aimed so that the bullet hits the centre of the target? (Take $g = 10 \mathrm{~m/s}^2$) (1) 1.0 m (2) 4.2 m (3) 6.1 m (4) 9.8 m
A body of mass 5 kg under the action of constant force $\vec{F} = F_x \hat{i} + F_y \hat{j}$ has velocity at $\mathrm{t} = 0 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}} = (6\hat{\mathrm{i}} - 2\hat{\mathrm{j}}) \mathrm{m/s}$ and at $\mathrm{t} = 10 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}} = +6\hat{\mathrm{j}} \mathrm{m/s}$. The force $\overrightarrow{\mathrm{F}}$ is: (1) $(-3\hat{\mathrm{i}} + 4\hat{\mathrm{j}}) \mathrm{N}$ (2) $\left(-\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}\right) \mathrm{N}$ (3) $(3\hat{\mathrm{i}} - 4\hat{\mathrm{j}}) \mathrm{N}$ (4) $\left(\frac{3}{5}\hat{\mathrm{i}} - \frac{4}{5}\hat{\mathrm{j}}\right) \mathrm{N}$
A small ball of mass $m$ starts at a point $A$ with speed $v_o$ and moves along a frictionless track $AB$ as shown. The track BC has coefficient of friction $\mu$. The ball comes to stop at C after travelling a distance $L$ which is: (1) $\frac{2\mathrm{~h}}{\mu} + \frac{\mathrm{v}_\mathrm{o}^2}{2\mu\mathrm{~g}}$ (2) $\frac{\mathrm{h}}{\mu} + \frac{\mathrm{v}_0^2}{2\mu\mathrm{~g}}$ (3) $\frac{\mathrm{h}}{2\mu} + \frac{\mathrm{v}_\mathrm{o}^2}{\mu\mathrm{g}}$ (4) $\frac{\mathrm{h}}{2\mu} + \frac{\mathrm{v}_\mathrm{o}^2}{2\mu\mathrm{g}}$
A thin bar of length L has a mass per unit length $\lambda$, that increases linearly with distance from one end. If its total mass is $M$ and its mass per unit length at the lighter end is $\lambda_\mathrm{O}$, then the distance of the centre of mass from the lighter end is: (1) $\frac{\mathrm{L}}{2} - \frac{\lambda_0 \mathrm{~L}^2}{4\mathrm{M}}$ (2) $\frac{\mathrm{L}}{3} + \frac{\lambda_0 \mathrm{~L}^2}{8\mathrm{M}}$ (3) $\frac{\mathrm{L}}{3} + \frac{\lambda_0 \mathrm{~L}^2}{4\mathrm{M}}$ (4) $\frac{2\mathrm{~L}}{3} - \frac{\lambda_0 \mathrm{~L}^2}{6\mathrm{M}}$