PYQ — IISER Physics MCQs with Solutions
Free IISER Physics PYQ MCQs with step-by-step solutions (67 questions). Part of 1D PYQ. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — PYQ · medium
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. $\dfrac{t_1 t_2}{t_2 - t_1}$
B. $\dfrac{t_1 t_2}{t_2 + t_1}$ ✓ Correct
C. $t_1 - t_2$
D. $\dfrac{t_1 + t_2}{2}$
Solution: Walking speed adds to escalator speed: $\dfrac{1}{t}=\dfrac{1}{t_1}+\dfrac{1}{t_2}$, giving $t=\dfrac{t_1 t_2}{t_1+t_2}$.
Q2 — PYQ · medium
Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $x_P(t)=(at+bt^2)$ and $x_Q(t)=(ft-t^2)$. At what time do the cars have the same velocity?
A. $\dfrac{a-f}{1+b}$
B. $\dfrac{a+f}{2(b-1)}$
C. $\dfrac{a+f}{2(1+b)}$
D. $\dfrac{f-a}{2(1+b)}$ ✓ Correct
Solution: $v_P=a+2bt$, $v_Q=f-2t$; equating gives $t=\dfrac{f-a}{2(1+b)}$.
Q3 — PYQ · medium
If the velocity of a particle is $v = At + Bt^2$, where $A$ and $B$ are constants, then the distance travelled by it between $1\,\text{s}$ and $2\,\text{s}$ is
A. $\dfrac{3}{2}A + \dfrac{7}{3}B$ ✓ Correct
B. $\dfrac{A}{2} + \dfrac{B}{3}$
C. $\dfrac{3}{2}A + 4B$
D. $3A + 7B$
Solution: $\int_1^2 (At+Bt^2)\,dt = \dfrac{A}{2}(3)+\dfrac{B}{3}(7)=\dfrac{3}{2}A+\dfrac{7}{3}B$.
Q4 — 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: $a=v\dfrac{dv}{dx}=\beta x^{-2n}\cdot(-2n\beta x^{-2n-1})=-2n\beta^2 x^{-4n-1}$.
Q5 — PYQ · medium
A stone falls freely under gravity. It covers distances $h_1$, $h_2$ and $h_3$ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between $h_1$, $h_2$ and $h_3$ is
A. $h_2 = 3h_1$ and $h_3 = 3h_2$
B. $h_1 = h_2 = h_3$
C. $h_1 = 2h_2 = 3h_3$
D. $h_1 = \dfrac{h_2}{3} = \dfrac{h_3}{5}$ ✓ Correct
Solution: Distances in successive equal intervals from rest go as $1:3:5$, so $h_2=3h_1$, $h_3=5h_1$, i.e. $h_1=\dfrac{h_2}{3}=\dfrac{h_3}{5}$.
Q6 — PYQ · medium
The motion of a particle along a straight line is described by equation $x = 8 + 12t - t^3$ where $x$ is in metre 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: $v=12-3t^2=0\Rightarrow t=2$; $a=-6t=-12\,\text{ms}^{-2}$, so retardation is $12\,\text{ms}^{-2}$.
Q7 — 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: $v=\sqrt{2gh}=\sqrt{2\cdot10\cdot20}=20\,\text{m/s}$.
Q8 — 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 $\dfrac{2v_1 v_2}{v_1+v_2}$.
Q9 — 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 $a\propto v^{3/2}$.
Q10 — PYQ · medium
A ball is dropped from a high-rise platform at $t = 0$ starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t = 18\,\text{s}$. What is the value of $v$? (Take $g = 10\,\text{m/s}^2$)
A. $75\,\text{m/s}$ ✓ Correct
B. $55\,\text{m/s}$
C. $40\,\text{m/s}$
D. $60\,\text{m/s}$
Solution: First ball at $t=18$: $\tfrac12\cdot10\cdot18^2=1620$ m. Second (falls 12 s): $12v+\tfrac12\cdot10\cdot12^2=12v+720=1620\Rightarrow v=75\,\text{m/s}$.
Q11 — 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. $S_2=4S_1$.
Q12 — PYQ · medium
A bus is moving with a speed of $10\,\text{ms}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in $100\,\text{s}$. If the bus is at a distance of $1\,\text{km}$ from the scooterist, with what speed should the scooterist chase the bus?
A. $40\,\text{ms}^{-1}$
B. $25\,\text{ms}^{-1}$
C. $10\,\text{ms}^{-1}$
D. $20\,\text{ms}^{-1}$ ✓ Correct
Solution: Scooterist must cover $1000+10\cdot100=2000$ m in 100 s, so $v=20\,\text{ms}^{-1}$.
Q13 — 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.
Q14 — 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: $s_n=\dfrac{a}{2}(2n-1)=\dfrac{4/3}{2}(5)=\dfrac{10}{3}\,\text{m}$.
Q15 — 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 $\dfrac{2v_u v_d}{v_u+v_d}$.
Q16 — PYQ · medium
The position $x$ of a particle with respect to time $t$ along $x$-axis is given by $x = 9t^2 - t^3$ where $x$ is in metres and $t$ in seconds. What will be the position of this particle when it achieves maximum speed along the $+x$ direction?
A. $54\,\text{m}$ ✓ Correct
B. $81\,\text{m}$
C. $24\,\text{m}$
D. $32\,\text{m}$
Solution: Max speed when $a=18-6t=0\Rightarrow t=3$; $x=9\cdot9-27=54\,\text{m}$.
Q17 — 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 $\dfrac{t_A}{t_B}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$.
Q18 — PYQ · medium
A car runs at a constant speed on a circular track of radius $100\,\text{m}$, taking $62.8$ seconds for every circular lap. The average velocity and average speed for each circular lap respectively is
A. $10\,\text{m/s},\,0$
B. $0,\,0$
C. $0,\,10\,\text{m/s}$ ✓ Correct
D. $10\,\text{m/s},\,10\,\text{m/s}$
Solution: Displacement over a full lap is zero so average velocity $=0$; average speed $=\dfrac{2\pi\cdot100}{62.8}=10\,\text{m/s}$.
Q19 — PYQ · medium
A particle moves along a straight line $OX$. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?
A. $16\,\text{m}$ ✓ Correct
B. $24\,\text{m}$
C. $40\,\text{m}$
D. $56\,\text{m}$
Solution: $v=12-3t^2=0\Rightarrow t=2$; distance $=x(2)-x(0)=56-40=16\,\text{m}$.
Q20 — PYQ · medium
A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? (Given $g = 9.8\,\text{m/s}^2$)
A. more than $19.6\,\text{m/s}$ ✓ Correct
B. at least $9.8\,\text{m/s}$
C. any speed less than $19.6\,\text{m/s}$
D. only with speed $19.6\,\text{m/s}$
Solution: For 3 balls aloft, time of flight $\dfrac{2u}{g}>4$ s, giving $u>19.6\,\text{m/s}$.
Q21 — PYQ · medium
If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is
A. $ut$
B. $\dfrac{1}{2}gt^2$ ✓ Correct
C. $ut - \dfrac{1}{2}gt^2$
D. $(u+gt)t$
Solution: The last $t$ s of ascent mirror the first $t$ s of a fall from rest, so the distance is $\dfrac{1}{2}gt^2$.
Q22 — PYQ · medium
A particle is thrown vertically upward. Its velocity at half of the height is $10\,\text{m/s}$, then the maximum height attained by it is $(g = 10\,\text{m/s}^2)$
A. $8\,\text{m}$
B. $20\,\text{m}$
C. $10\,\text{m}$ ✓ Correct
D. $16\,\text{m}$
Solution: $u^2=2gH$ and at half height $10^2=u^2-gH=gH$, so $H=\dfrac{100}{10}=10\,\text{m}$.
Q23 — 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 $t=1$, $a=32\,\text{m/s}^2$.
Q24 — 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 $2\times4=8\,\text{m}$.
Q25 — PYQ · medium
A rubber ball is dropped from a height of $5\,\text{m}$ on a plane. On bouncing it rises to $1.8\,\text{m}$. The ball loses its velocity on bouncing by a factor of
A. $\dfrac{3}{5}$ ✓ Correct
B. $\dfrac{2}{5}$
C. $\dfrac{16}{25}$
D. $\dfrac{9}{25}$
Solution: Speed before $=\sqrt{2g\cdot5}=10$, after $=\sqrt{2g\cdot1.8}=6$; fraction of velocity lost $=\dfrac{10-6}{10}=\dfrac{2}{5}$ (option b). The key lists (a) $3/5$, which is the fraction retained, not lost.
Q26 — PYQ · medium
The position $x$ of a particle varies with time $(t)$ as $x = at^2 - bt^3$. The acceleration will be zero at time $t$ equal to
A. $\dfrac{a}{3b}$ ✓ Correct
B. zero
C. $\dfrac{2a}{3b}$
D. $\dfrac{a}{b}$
Solution: $a_{acc}=2a-6bt=0\Rightarrow t=\dfrac{a}{3b}$.
Q27 — 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 $=\tfrac12(0+40)(20)=400\,\text{m}$.
Q28 — PYQ · medium
A body dropped from a height $h$ with initial velocity zero, strikes the ground with a velocity $3\,\text{m/s}$. Another body of same mass dropped from the same height $h$ with an initial velocity of $4\,\text{m/s}$. The final velocity of second mass, with which it strikes the ground is
A. $5\,\text{m/s}$ ✓ Correct
B. $12\,\text{m/s}$
C. $3\,\text{m/s}$
D. $4\,\text{m/s}$
Solution: $2gh=3^2=9$; second body: $v^2=4^2+2gh=16+9=25\Rightarrow v=5\,\text{m/s}$.
Q29 — PYQ · medium
The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from origin with an initial velocity $v_0$. The distance travelled by the particle in time $t$ will be
A. $v_0 t + \dfrac{1}{3}bt^2$
B. $v_0 t + \dfrac{1}{2}bt^2$
C. $v_0 t + \dfrac{1}{6}bt^3$ ✓ Correct
D. $v_0 t + \dfrac{1}{3}bt^3$
Solution: $v=v_0+\tfrac12 bt^2$; integrating, $x=v_0 t+\tfrac16 bt^3$.
Q30 — 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.