In the system of two particles of masses ' $\mathrm{m}_1$ ' and ' $\mathrm{m}_2$ ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

In projectile motion two particles of masses $\mathrm{m}_1$ and $m_2$ have velocities $\vec{V}_1$, and $\vec{V}_2$ respectively at time $t=0$. Their velocities become $\overline{V_1^{\prime}}$ and $\overrightarrow{V_2^{\prime}}$ at time 2 t while still moving in air. The value of $\left[\left(m_1 \overrightarrow{V_1^{\prime}}+m_2 \overrightarrow{V_2^{\prime}}\right)-\left(m_1 \vec{V}_1+m_2 \vec{V}_2\right)\right]$ is ( $\mathrm{g}=$ acceleration due to gravity)

A meter scale is supported on a wedge at its centre of gravity. A body of weight ' $w$ ' is suspended from the 20 cm mark and another weight of 25 gram is suspended from 74 cm mark balances it and the meter scale remains perfectly horizontal. Neglecting the weight of the meter scale, the weight of the body is

A person with machine gun can fire 50 g bullets with a velocity of $$240 \mathrm{~m} / \mathrm{s}$$. A $$60 \mathrm{~kg}$$ tiger moves towards him with a velocity of $$12 \mathrm{~m} / \mathrm{s}$$. In order to stop the tiger in track, the number of bullets the person fires towards the tiger is