A particle of mass$m$ moving horizontally with a velocity$v_0$ collides and sticks to the bottom of the smoothly pivoted hanging rod of mass $M (= 3 m)$and length$l$.

The fraction loss of $KE$ of the system (rod + particle) during collision is:

A rod of length $\text{'}L\text{'}$ has variable mass density, $\lambda \left(x\right)={\lambda }_0\left(1+\frac{x}{L}\right).$ There exists an axis parallel to $AB$ about which the moment of inertia of the rod is minimum. If the distance of such an axis from $AB$ is of the form $\frac{5 L}{ N},$ then find $N.$

Let $I$ be the moment of inertia of a uniform square plate about an axis $AB$ that passes through its centre and is parallel to two of its sides. $CD$ is a line in the plane of the plate that passes through the centre of the plate and makes an angle $\mathrm{\theta }$ with $AB$ as shown in figure. The moment of inertia of the plate about the axis $CD$ is then equal to

A uniform rod $OB$ of length $1 \mathrm{m}\text{,}$ cross-sectional area $0.012 {\mathrm{m}}^2$ and relative density $2.0$ is free to rotate about $\mathrm{O}$ in vertical plane. The rod is held with a horizontal string $AB$ which can withstand a maximum tension of $45 \mathrm{N}.$ The rod and string system is kept in water as shown in figure. The maximum value of angle $\alpha$ which the rod can make with vertical without breaking the string is