A box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $\mathbf{F}_1$ is 10 N , what is the magnitude of $\mathbf{F}_3$ ?

A moving body with a mass $m_1$ and velocity $u$ strikes a stationary body of mass $m_2$. The masses $m_1$ and $m_2$ should be in the ratio $\frac{m_1}{m_2}$, so as to decrease the velocity of the first body to $\frac{2 u}{3}$ and giving a velocity of $v$ to $m_2$ assuming a perfectly elastic impact. Then, the ratio $\frac{m_1}{m_2}$ is

A force acts on a 30 g particle in such a way that the position of the particle as a function of time is given by $x=\alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha=1 \mathrm{m} / \mathrm{s}^2$. The work done during the first 4 s is

A straight rod of length $L$ is made of a material having mass per unit length $m(x)=\lambda|x|$, where $x$ is measured from the centre of rod. The moment of inertia about an axis perpendicular to the rod and passing through one end of the rod will be $L=1 \mathrm{~m}$ and $\lambda=16 \mathrm{~kg} / \mathrm{m}^2$.