$A$ particle of mass $m$ performs uniform circular motion of radius $r$ with linear speed $v$ under the application of force $F$. If $m$,$v$,and $r$ are all increased by $20 \%$,the necessary change in force required to maintain the particle in uniform circular motion is: (in $\%$)

$A$ ball of mass $160 \, g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 \, m/s$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $\left(g=10 \, m/s^{2}\right)$ (in $kg \cdot m^{2}/s$).

$A$ hill is $500 \, m$ high. Supplies are to be sent across the hill,using a cannon that can hurl packets at a speed of $125 \, m/s$ over the hill. The cannon is located at a distance of $800 \, m$ from the foot of the hill and can be moved on the ground at a speed of $2 \, m/s$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach the ground on the other side of the hill? Take $g = 10 \, m/s^2$.

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

$A$ car travels north with a uniform velocity. It goes over a piece of mud which sticks to the tyre. The particles of the mud,as they leave the ground,are thrown: