Relative Velocity — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Relative Velocity MCQs with step-by-step solutions (6 questions). Part of Motion in a Plane. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Relative Velocity · medium
The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path the angle at which he should make his strokes w.r.t. north is given by
A. $0^\circ$
B. $60^\circ$ west
C. $45^\circ$ west
D. $30^\circ$ west ✓ Correct
Solution: Given speed of river $v_R=10$ m/s and speed of swimmer in still water $v_{SN}=20$ m/s. For the shortest path he should swim at an angle $(90^\circ+\theta)$ with the stream flow, where $\sin\theta=\dfrac{|\mathbf{v}_R|}{|\mathbf{v}_{SN}|}=\dfrac{10}{20}=\dfrac{1}{2}$, so $\theta=30^\circ$. As the river flows East, he should swim towards West.
Q2 — Relative Velocity · medium
Two particles $A$ and $B$, move with constant velocities $\mathbf{v}_1$ and $\mathbf{v}_2$. At the initial moment, their position vectors are $\mathbf{r}_1$ and $\mathbf{r}_2$ respectively. The condition for particles $A$ and $B$ for their collision is
A. $\dfrac{\mathbf{r}_1-\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}=\dfrac{\mathbf{v}_2-\mathbf{v}_1}{|\mathbf{v}_2-\mathbf{v}_1|}$ ✓ Correct
B. $\mathbf{r}_1\cdot\mathbf{v}_1=\mathbf{r}_2\cdot\mathbf{v}_2$
C. $\mathbf{r}_1\times\mathbf{v}_1=\mathbf{r}_2\times\mathbf{v}_2$
D. $\mathbf{r}_1-\mathbf{r}_2=\mathbf{v}_1-\mathbf{v}_2$
Solution: For the two particles to collide, the direction of the relative velocity of one with respect to the other must be directed towards the relative position of the other particle. Here $\dfrac{\mathbf{r}_1-\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}$ gives the direction of the relative position of $A$ w.r.t. $B$, and $\dfrac{\mathbf{v}_2-\mathbf{v}_1}{|\mathbf{v}_2-\mathbf{v}_1|}$ gives the direction of the relative velocity. So for collision $\dfrac{\mathbf{r}_1-\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}=\dfrac{\mathbf{v}_2-\mathbf{v}_1}{|\mathbf{v}_2-\mathbf{v}_1|}$.
Q3 — Relative Velocity · medium
A person swims in a river aiming to reach exactly opposite point on the bank of a river. His speed of swimming is 0.5 m/s at an angle $120^\circ$ with the direction of flow of water. The speed of water in stream is
A. $1.0$ m/s
B. $0.5$ m/s
C. $0.25$ m/s ✓ Correct
D. $0.43$ m/s
Solution: Let $u$ be the speed of the stream and $v=0.5$ m/s the speed of the person, making $120^\circ$ with the flow $u$. For the resultant to be along the straight line across (opposite point), $u=v\sin\theta=v\sin30^\circ=\dfrac{v}{2}=\dfrac{0.5}{2}=0.25$ m/s.
Q4 — Relative Velocity · medium
The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest possible path in 15 min. The velocity of the river water is (in km/h)
A. $5$
B. $1$
C. $3$ ✓ Correct
D. $4$
Solution: Let $v_r$ be the velocity of the river, $v_{br}=5$ km/h the boat's speed in still water and $w=1.0$ km the width. Time to cross $=15$ min $=\dfrac{1}{4}$ h. Along the shortest path $t=\dfrac{w}{\sqrt{v_{br}^2-v_r^2}}$, so $\dfrac{1}{4}=\dfrac{1}{\sqrt{5^2-v_r^2}}$, giving $5^2-v_r^2=16\Rightarrow v_r^2=9\Rightarrow v_r=3$ km/h.
Q5 — Relative Velocity · medium
A boat is sent across a river with a velocity of $8\,\text{km h}^{-1}$. If the resultant velocity of boat is $10\,\text{km h}^{-1}$, then velocity of river is
A. $12.8\,\text{km h}^{-1}$
B. $6\,\text{km h}^{-1}$ ✓ Correct
C. $8\,\text{km h}^{-1}$
D. $10\,\text{km h}^{-1}$
Solution: Let $v_b$ be the boat's velocity, $v_r$ the river's velocity and $v_{rb}$ the resultant. Since they are perpendicular, $v_{rb}^2=v_r^2+v_b^2$, so $v_r=\sqrt{v_{rb}^2-v_b^2}=\sqrt{10^2-8^2}=6\,\text{km h}^{-1}$.
Q6 — Relative Velocity · medium
A bus is moving on a straight road towards North with a uniform speed of 50 km/h. If the speed remains unchanged after turning through $90^\circ$, the increase in the velocity of bus in the turning process is
A. $70.7$ km/h along South-West direction ✓ Correct
B. zero
C. $50$ km/h along West
D. $70.7$ km/h along North-West direction
Solution: Here $\mathbf{v}_1=50$ km/h due North and $\mathbf{v}_2=50$ km/h due West, with the angle between them $90^\circ$. Change in velocity $=|\mathbf{v}_2-\mathbf{v}_1|=|\mathbf{v}_2+(-\mathbf{v}_1)|=\sqrt{v_2^2+v_1^2}=\sqrt{50^2+50^2}=70.7$ km/h. Since $-\mathbf{v}_1$ is due South and $\mathbf{v}_2$ is due West, the direction is South-West.