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1D PYQ — IISER Physics MCQs with Solutions
Free IISER Physics 1D PYQ MCQs with step-by-step solutions covering PYQ. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — PYQ · easy
A boy standing at the top of a tower of $20\,\text{m}$ height drops a stone. Assuming $g = 10\,\text{ms}^{-2}$, the velocity with which it hits the ground is
A. $10.0\,\text{m/s}$
B. $20.0\,\text{m/s}$ ✓ Correct
C. $40.0\,\text{m/s}$
D. $5.0\,\text{m/s}$
Solution: $v=\sqrt{2gh}=\sqrt{2\cdot10\cdot20}=20\,\text{m/s}$.
Q2 — PYQ · easy
A particle covers half of its total distance with speed $v_1$ and the rest half distance with speed $v_2$. Its average speed during the complete journey is
A. $\dfrac{v_1 + v_2}{2}$
B. $\dfrac{v_1 v_2}{v_1 + v_2}$
C. $\dfrac{2v_1 v_2}{v_1 + v_2}$ ✓ Correct
D. $\dfrac{v_1^2 v_2^2}{v_1^2 + v_2^2}$
Solution: For equal distances the average speed is the harmonic mean $\dfrac{2v_1 v_2}{v_1+v_2}$.
Q3 — PYQ · easy
A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $S_1$ and that covered in the first 20 seconds is $S_2$, then
A. $S_2 = 3S_1$
B. $S_2 = 4S_1$ ✓ Correct
C. $S_2 = S_1$
D. $S_2 = 2S_1$
Solution: From rest $S\propto t^2$, so $\dfrac{S_2}{S_1}=\left(\dfrac{20}{10}\right)^2=4$, i.e. $S_2=4S_1$.
Q4 — PYQ · easy
A particle moves in a straight line with a constant acceleration. It changes its velocity from $10\,\text{ms}^{-1}$ to $20\,\text{ms}^{-1}$ while passing through a distance $135\,\text{m}$ in $t$ second. The value of $t$ is
A. $12$
B. $9$ ✓ Correct
C. $10$
D. $1.8$
Solution: Average velocity $=\dfrac{10+20}{2}=15$ m/s, so $t=\dfrac{135}{15}=9$ s.
Q5 — PYQ · easy
The distance travelled by a particle starting from rest and moving with an acceleration $\dfrac{4}{3}\,\text{ms}^{-2}$, in the third second is
A. $\dfrac{10}{3}\,\text{m}$ ✓ Correct
B. $\dfrac{19}{3}\,\text{m}$
C. $6\,\text{m}$
D. $4\,\text{m}$
Solution: $s_n=\dfrac{a}{2}(2n-1)=\dfrac{4/3}{2}(5)=\dfrac{10}{3}\,\text{m}$.
Q6 — PYQ · easy
A car moves from $X$ to $Y$ with a uniform speed $v_u$ and returns to $Y$ with a uniform speed $v_d$. The average speed for this round trip is
A. $\sqrt{v_u v_d}$
B. $\dfrac{v_d v_u}{v_d + v_u}$
C. $\dfrac{v_u + v_d}{2}$
D. $\dfrac{2 v_d v_u}{v_d + v_u}$ ✓ Correct
Solution: Equal distances each way give the harmonic mean $\dfrac{2v_u v_d}{v_u+v_d}$.
Q7 — PYQ · easy
Two bodies $A$ (of mass $1\,\text{kg}$) and $B$ (of mass $3\,\text{kg}$) are dropped from heights of $16\,\text{m}$ and $25\,\text{m}$, respectively. The ratio of the time taken by them to reach the ground is
A. $4/5$ ✓ Correct
B. $5/4$
C. $12/5$
D. $5/12$
Solution: $t=\sqrt{2h/g}$, so $\dfrac{t_A}{t_B}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$.
Q8 — PYQ · easy
Motion of a particle is given by equation $s = (3t^3 + 7t^2 + 14t + 8)\,\text{m}$. The value of acceleration of the particle at $t = 1\,\text{sec}$ is
A. $10\,\text{m/s}^2$
B. $32\,\text{m/s}^2$ ✓ Correct
C. $23\,\text{m/s}^2$
D. $16\,\text{m/s}^2$
Solution: $v=9t^2+14t+14$, $a=18t+14$; at $t=1$, $a=32\,\text{m/s}^2$.
Q9 — PYQ · easy
A car moving with a speed of $40\,\text{km/h}$ can be stopped by applying brakes after at least $2\,\text{m}$. If the same car is moving with a speed of $80\,\text{km/h}$, what is the minimum stopping distance?
A. $4\,\text{m}$
B. $6\,\text{m}$
C. $8\,\text{m}$ ✓ Correct
D. $2\,\text{m}$
Solution: Stopping distance $\propto v^2$; doubling speed gives $2\times4=8\,\text{m}$.
Q10 — PYQ · easy
If a car at rest accelerates uniformly to a speed of $144\,\text{km/h}$ in $20\,\text{sec}$, it covers a distance of
A. $1440\,\text{cm}$
B. $2980\,\text{cm}$
C. $20\,\text{m}$
D. $400\,\text{m}$ ✓ Correct
Solution: $144\,\text{km/h}=40\,\text{m/s}$; distance $=\tfrac12(0+40)(20)=400\,\text{m}$.
Q11 — PYQ · easy
The velocity of a train increases uniformly from $20\,\text{km/h}$ to $60\,\text{km/h}$ in $4$ hours. The distance travelled by the train during this period is
A. $160\,\text{km}$ ✓ Correct
B. $180\,\text{km}$
C. $100\,\text{km}$
D. $120\,\text{km}$
Solution: Distance $=\dfrac{20+60}{2}\times4=160\,\text{km}$.
Q12 — PYQ · easy
A body starts from rest, what is the ratio of the distance travelled by the body during the $4^{th}$ and $3^{rd}$ second?
A. $\dfrac{7}{5}$ ✓ Correct
B. $\dfrac{5}{7}$
C. $\dfrac{7}{3}$
D. $\dfrac{3}{7}$
Solution: $s_n\propto(2n-1)$, so ratio $=\dfrac{2\cdot4-1}{2\cdot3-1}=\dfrac{7}{5}$.
Q13 — PYQ · easy
A car covers the first half of the distance between two places at 40 $km/h$ and another half at 60 $km/h$. The average speed of the car is
A. 40 $km/h$
B. 48 $km/h$ ✓ Correct
C. 50 $km/h$
D. 60 $km/h$
Solution: $v_{avg}=\dfrac{2\times40\times60}{40+60}=48\,km/h$.
Q14 — PYQ · easy
What will be the ratio of the distance moved by a freely falling body from rest in $4^{th}$ and $5^{th}$ seconds of journey?
A. 4:5
B. 7:9 ✓ Correct
C. 16:25
D. 1:1
Solution: Distance in $n$th second $\propto(2n-1)$: 4th $\to7$, 5th $\to9$, ratio $7:9$.
Q15 — PYQ · easy
A vehicle travels half the distance with speed $v$ and the remaining distance with speed $2v$. Its average speed is
A. $\dfrac{2v}{3}$
B. $\dfrac{4v}{3}$ ✓ Correct
C. $\dfrac{3v}{4}$
D. $\dfrac{v}{3}$
Solution: $v_{avg}=\dfrac{2\cdot v\cdot2v}{v+2v}=\dfrac{4v}{3}$.
(Source: NEET 2023)
Q16 — PYQ · easy
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second is
A. 1:4:9:16
B. 1:3:5:7 ✓ Correct
C. 1:1:1:1
D. 1:2:3:4
Solution: Distance in $n$th second $\propto(2n-1)$, giving $1:3:5:7$.
(Source: NEET 2022)
Q17 — PYQ · easy
The displacement-time graph of two moving particles makes angles of $30^\circ$ and $45^\circ$ with the X-axis as shown in the figure. The ratio of their respective velocity is
A. $1:1$
B. $1:2$
C. $1:\sqrt{3}$ ✓ Correct
D. $\sqrt{3}:1$
Solution: Velocity equals the slope, so the ratio is $\tan 30^\circ : \tan 45^\circ = \dfrac{1}{\sqrt{3}} : 1 = 1:\sqrt{3}$.
(Source: NEET 2022)
Q18 — PYQ · easy
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_1$. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_2$. The time taken by her to walk up on the moving escalator will be:
A. $\dfrac{t_1 + t_2}{2}$
B. $\dfrac{t_1 t_2}{t_2 - t_1}$
C. $\dfrac{t_1 t_2}{t_2 + t_1}$ ✓ Correct
D. $t_1 - t_2$
Solution: Rates add: $\dfrac{1}{t} = \dfrac{1}{t_1} + \dfrac{1}{t_2}$, giving $t = \dfrac{t_1 t_2}{t_1 + t_2}$.
Q19 — PYQ · easy
The speed of a swimmer in still water is $20$ m/s. The speed of river water is $10$ m/s and 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 stroke w.r.t. north is given by:
A. $45^\circ$ west
B. $30^\circ$ west ✓ Correct
C. $0^\circ$
D. $60^\circ$ west
Solution: To cancel the eastward current, $\sin\theta = \dfrac{10}{20} = \dfrac{1}{2}$, so $\theta = 30^\circ$ west of north.
Q20 — PYQ · easy
A body is moving with velocity $30$ m/s towards east. After $10$ seconds its velocity becomes $40$ m/s towards north. The average acceleration of the body is:
A. $1\ \text{m/s}^2$
B. $7\ \text{m/s}^2$
C. $\sqrt{7}\ \text{m/s}^2$
D. $5\ \text{m/s}^2$ ✓ Correct
Solution: $|\Delta v| = \sqrt{40^2 + 30^2} = 50$ m/s over $10$ s, so $a = 5\ \text{m/s}^2$.
Q21 — PYQ · easy
The motion of a particle along a straight line is described by equation: $x = 8 + 12t - t^3$ where $x$ is in meter and $t$ in second. The retardation of the particle when its velocity becomes zero, is:
A. $24\ \text{ms}^{-2}$
B. zero
C. $6\ \text{ms}^{-2}$
D. $12\ \text{ms}^{-2}$ ✓ Correct
Solution: $v = 12 - 3t^2 = 0 \Rightarrow t = 2$ s; $a = -6t = -12\ \text{m/s}^2$, so retardation is $12\ \text{m/s}^2$.
Q22 — PYQ · easy
A particle is moving such that its position coordinates $(x, y)$ are $(2\text{m}, 3\text{m})$ at time $t = 0$, $(6\text{m}, 7\text{m})$ at time $t = 2$ s and $(13\text{m}, 14\text{m})$ at time $t = 5$ s, Average velocity vector $(\vec{V}_{av})$ from $t = 0$ to $t = 5$ s is
A. $\dfrac{1}{5}\left(13\hat{i} + 14\hat{j}\right)$
B. $\dfrac{7}{3}\left(\hat{i} + \hat{j}\right)$
C. $2\left(\hat{i} + \hat{j}\right)$
D. $\dfrac{11}{5}\left(\hat{i} + \hat{j}\right)$ ✓ Correct
Solution: $\vec{V}_{av} = \dfrac{(13-2)\hat{i} + (14-3)\hat{j}}{5} = \dfrac{11}{5}(\hat{i} + \hat{j})$.
Q23 — PYQ · easy
The x and y coordinates of the particle at any time are $x = 5t - 2t^2$ and $y = 410t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2$ s is:
A. $0$
B. $5\ \text{m/s}^2$
C. $-4\ \text{m/s}^2$ ✓ Correct
D. $-8\ \text{m/s}^2$
Solution: $a_x = \dfrac{d^2x}{dt^2} = -4$, $a_y = 0$, so acceleration is $-4\ \text{m/s}^2$ (constant, independent of time).
Q24 — PYQ · easy
A ball is thrown vertically downward with a velocity of $20$ m/s from the top of a tower. It hits the ground after some time with a velocity of $80$ m/s. The height of the tower is $(g = 10\ \text{m/s}^2)$
A. $340$ m
B. $320$ m
C. $300$ m ✓ Correct
D. $360$ m
Solution: $v^2 = u^2 + 2gh \Rightarrow 80^2 = 20^2 + 2(10)h \Rightarrow 6000 = 20h \Rightarrow h = 300$ m.
(Source: NEET 2020)
Q25 — PYQ · hard
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v(x)=\beta x^{-2n}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ is given by
A. $-2\beta^2 x^{-2n+1}$
B. $-2n\beta^2 e^{-4n+1}$
C. $-2n\beta^2 x^{-2n-1}$
D. $-2n\beta^2 x^{-4n-1}$ ✓ Correct
Solution: $a=v\dfrac{dv}{dx}=\beta x^{-2n}\cdot(-2n\beta x^{-2n-1})=-2n\beta^2 x^{-4n-1}$.
Q26 — PYQ · hard
A particle moves a distance $x$ in time $t$ according to equation $x = (t+5)^{-1}$. The acceleration of particle is proportional to
A. $(\text{velocity})^{3/2}$ ✓ Correct
B. $(\text{distance})^{2}$
C. $(\text{distance})^{-2}$
D. $(\text{velocity})^{2/3}$
Solution: $v=-(t+5)^{-2}$, $a=2(t+5)^{-3}$; eliminating $(t+5)=v^{-1/2}$ gives $a\propto v^{3/2}$.
Q27 — PYQ · hard
The water drops fall at regular intervals from a tap $5\,\text{m}$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
A. $3.75\,\text{m}$ ✓ Correct
B. $4.00\,\text{m}$
C. $1.25\,\text{m}$
D. $2.50\,\text{m}$
Solution: First drop takes $1$ s to fall $5$ m (2 intervals), so each interval is $0.5$ s; second drop falls $\tfrac12\cdot10\cdot0.5^2=1.25$ m, leaving it $5-1.25=3.75\,\text{m}$ above ground.
Q28 — PYQ · hard
Two particles A and B, move with constant velocities $\vec{v}_1$ and $\vec{v}_2$. At the initial moment their position vectors are $\vec{r}_1$ and $\vec{r}_2$ respectively. The condition for particles A and B for their collision is:
A. $\vec{r}_1\cdot\vec{v}_1 = \vec{r}_2\cdot\vec{v}_2$
B. $\vec{r}_1\times\vec{v}_1 = \vec{r}_2\times\vec{v}_2$
C. $\vec{r}_1 - \vec{r}_2 = \vec{v}_1 - \vec{v}_2$
D. $\dfrac{\vec{r}_1 - \vec{r}_2}{|\vec{r}_1 - \vec{r}_2|} = \dfrac{\vec{v}_2 - \vec{v}_1}{|\vec{v}_2 - \vec{v}_1|}$ ✓ Correct
Solution: For collision the relative separation must be directed exactly opposite to the relative velocity, i.e. the two unit vectors in option (d) must be equal.
Q29 — PYQ · hard
A person sitting in the ground floor of a building notices through the window of height $1.5$ m, a ball dropped from the roof of the building crosses the window in $0.1$ s. What is the velocity of the ball when it is at the topmost point of the window? $(g = 10\ \text{m/s}^2)$
A. $15.5$ m/s
B. $14.5$ m/s ✓ Correct
C. $4.5$ m/s
D. $20$ m/s
Solution: Average speed across window $= 1.5/0.1 = 15$ m/s = speed at its middle. Using $15^2 = v^2 + 2(10)(0.75)$ gives $v^2 = 210$, so $v \approx 14.5$ m/s at the top.
(Source: NEET 2020)
Q30 — PYQ · hard
A toy car with charge $q$ moves on a frictionless horizontal plane surface under the influence of a uniform dielectric field E. Due to the force $qE$. Its velocity increases from $0$ to $6$ m/s in one second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between $0$ to $3$ second are respectively.
A. $1$ m/s, $3.5$ m/s
B. $1$ m/s, $3$ m/s ✓ Correct
C. $2$ m/s, $4$ m/s
D. $1.5$ m/s, $3$ m/s
Solution: With $a = 6$: after $1$ s, $x = 3$ m, $v = 6$; over the next $2$ s at $a = -6$, net displacement is $0$ (returns to $x = 3$) while distance covered is $6$ m. Net displacement $3$ m $\Rightarrow$ avg velocity $1$ m/s; total distance $9$ m $\Rightarrow$ avg speed $3$ m/s.
(Source: NEET 2018)