NEET Kinematics MCQs & PYQs

Practice NEET Kinematics Questions

Question 281:

easy

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R).

Assertion (A): If two particles, moving along straight line with constant velocities have to meet, the relative velocity must be along the line joining the two particles.

Reason (R): Relative motion means motion of one particle as viewed from the other particle.

For two particles to meet, the relative velocity vector must align with the line joining them so that from one's frame, the other moves directly towards it.

💬 Discuss this Question

Question 282:

moderate

Equation of trajectory of a projectile is $y = \sqrt{3}x – 5x^2$. Then angle of projection with vertical is (Assume x-axis as horizontal and y-axis as vertical)

Comparing with $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$, we find $\tan\theta = \sqrt{3}$, so horizontal angle $\theta = 60^\circ$. The angle with vertical is $90^\circ - 60^\circ = 30^\circ$.

💬 Discuss this Question

Question 283:

easy

Two particles are projected with same initial velocity one makes angle $\theta$ with horizontal while other makes an angle $\theta$ with vertical. If their common range is R then product of their time of flight is directly proportional to:

(1999)

Concept: Time of flight for complementary angles and range formula.
Formula: $T = (2u sin \alpha) / g$, $R = (u^2 sin 2\alpha) / g$.
For angles $\theta$ and $90° - \theta$, times are $T_1 = (2u sin \theta) / g$ and $T_2 = (2u cos \theta) / g$.
Their product $T_1 T_2 = (4u^2 sin \theta cos \theta) / g^2 = (2u^2 sin 2\theta) / g^2 = (2/g) R$. Thus, $T_1 T_2 \propto R$.

💬 Discuss this Question

Question 284:

difficult

A point P consider at contact point of a wheel on ground which rolls on ground without slipping then value of displacement of point P when wheel completes half of rotation (If radius of wheel is $1\text{ m}$):

(2002)

Horizontal displacement is $\pi R$ and vertical displacement is $2R$. Total displacement is given by $\sqrt{(pi R)^2 + (2R)^2} = R\sqrt{\pi^2+4}$, which evaluates to $\sqrt{pi^2+4}\text{ m}$ for $R = 1\text{ m}$.

💬 Discuss this Question