A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is in the direction shown, which one of the following options is correct ( and are any highest and lowest points on the wheel, respectively)?

All 50 Physics questions we hold from this paper, of the 200 on the site for this sitting — every option and the answer key. 4 marks each, −1 for a wrong answer.
45 questions, one correct option each. +4 for a correct answer, −1 for a wrong one, nothing for a blank.
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively)?

Solution
For a wheel rolling without slipping, the point of contact Q is momentarily at rest and the topmost point P moves at 2v. So P moves faster than Q.
Match List I with List II.
| List I (Spectral lines of hydrogen for transitions from) | List II (Wavelength, nm) |
|---|---|
| A. n2=3 to n1=2 | I. 410.2 |
| B. n2=4 to n1=2 | II. 434.1 |
| C. n2=5 to n1=2 | III. 656.3 |
| D. n2=6 to n1=2 | IV. 486.1 |
Choose the correct answer from the options given below:
Solution
Balmer wavelengths follow λ1∝(41−n21): the smallest jump (n=3→2) gives the longest wavelength, 656.3 nm (H−α), then 486.1, 434.1 and 410.2 nm as n rises. Hence A-III, B-IV, C-II, D-I.
A thermodynamic system is taken through the cycle abcda shown in the P–V diagram. The work done by the gas along the path bc is:

Solution
Path bc runs vertically on the P–V diagram: the volume stays at 400 cm3 while the pressure rises. With ΔV=0, the work done by the gas ∫PdV is zero.
The terminal voltage of a battery whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω, is:
Solution
Current I=R+rε=4+110=2 A. Terminal voltage V=ε−Ir=10−2=8 V.
In an ideal transformer, the turns ratio is NSNP=21. The ratio VS:VP is equal to (the symbols carry their usual meaning):
Solution
For an ideal transformer VPVS=NPNS. With NSNP=21, VPVS=2, i.e. 2:1.
A light ray enters through a right-angled prism at point P with the angle of incidence 30∘ as shown in the figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is:

Solution
Let the prism angles at B and C be B and C with B+C=90∘. The ray inside is parallel to BC, so its angle of refraction at AB is r=90∘−B and its angle of incidence at AC is 90∘−C=B. Emerging along AC means B is the critical angle: nsinB=1. At P, sin30∘=nsinr=ncosB, so ncosB=21. Squaring and adding: n2=1+41⇒n=25.
The quantities which have the same dimensions as those of solid angle are:
Solution
Solid angle is a ratio of area to (length)2 — dimensionless. Strain (length/length) and plane angle (arc/radius) are the two dimensionless quantities listed; stress has dimensions of pressure.
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If the surface tension of water is 0.07 Nm−1, then the excess force required to take it away from the surface is:
Solution
A disc lifted off a liquid surface is held by surface tension along its full circumference: F=T×2πr=0.07×2π×0.045≈1.98×10−2 N =19.8 mN.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential V at any axial point, at 2 m distance r from the centre of a dipole of dipole moment p of magnitude 4×10−6 C m, is ±9×103 V. (Take 4πε01=9×109 SI units.)
Reason R: V=4πε01r22p, where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
Solution
Axial potential of a dipole is V=4πε01r2p, not r22p (the factor 2 belongs to the axial field). Numerically V=9×109×44×10−6=9×103 V, so A is true but R is false.
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8×10−6 kgm2. If the magnitude of the magnetic moment of the needle is x×10−5 Am2, then the value of x is:
Solution
Oscillation period T=205=0.25 s. For a magnetic needle, T=2πMBI, so M=T2B4π2I=0.0625×0.0494π2×9.8×10−6=1280π2×10−5 Am2. Hence x=1280π2.
If the monochromatic source in Young's double slit experiment is replaced by white light, then:
Solution
With white light every wavelength produces a bright fringe at the centre (zero path difference), so the central fringe is white; on either side the different colours spread at different fringe widths and only a few coloured fringes are seen before they wash out.
Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
Solution
Atoms are neutral because they carry equal positive and negative charge — Statement I is correct. Atoms emit their characteristic spectrum only when excited, not simply because they are stable — Statement II is incorrect.
The maximum elongation of a steel wire of 1 m length, if the elastic limit of steel and its Young's modulus are respectively 8×108 Nm−2 and 2×1011 Nm−2, is:
Solution
At the elastic limit, strain =Ystress=2×10118×108=4×10−3. Elongation =strain×L=4×10−3×1 m =4 mm.
Consider the following statements A and B and identify the correct answer (refer to the given I–V graph):
A. For a solar cell, the I–V characteristic lies in the IV quadrant of the given graph.
B. In a reverse-biased p-n junction diode, the current measured (in μA) is due to majority charge carriers.

Solution
A solar cell's I–V curve lies in the fourth quadrant (it delivers power, so I and V have opposite signs) — A is correct. Reverse current in a p-n junction is carried by minority carriers — B is incorrect.
A particle moving with uniform speed in a circular path maintains:
Solution
In uniform circular motion the speed is constant but the direction of velocity keeps changing, so velocity varies; the centripetal acceleration has constant magnitude but continuously changing direction, so acceleration varies too.
If c is the velocity of light in free space, the correct statements about a photon among the following are:
A. The energy of a photon is E=hν.
B. The velocity of a photon is c.
C. The momentum of a photon is p=chν.
D. In a photon–electron collision, both total energy and total momentum are conserved.
E. A photon possesses positive charge.
Choose the correct answer from the options given below:
Solution
A photon has energy hν, moves at c, carries momentum p=hν/c, and photon–electron collisions conserve both energy and momentum. A photon is uncharged, so E is false. Hence A, B, C and D only.
Two bodies A and B of the same mass undergo a completely inelastic one-dimensional collision. Body A moves with velocity v1 while body B is at rest before the collision. The velocity of the system after the collision is v2. The ratio v1:v2 is:
Solution
Momentum conservation for a perfectly inelastic collision: mv1=2mv2⇒v1:v2=2:1.
The graph which shows the variation of λ21 with the kinetic energy E of a free particle (where λ is its de Broglie wavelength) is:

Solution
λ=2mEh⇒λ21=h22mE, which is directly proportional to E: a straight line through the origin.
An unpolarised light beam strikes a glass surface at Brewster's angle. Then:
Solution
At Brewster's angle the reflected light is completely plane-polarised (electric vector perpendicular to the plane of incidence), while the refracted beam contains both components and is only partially polarised.
At any instant of time t, the displacement of a particle is given by 2t−1 (SI units) under the influence of a force of 5 N. The value of the instantaneous power is (in SI units):
Solution
Velocity v=dtdx=2 m/s (constant). Instantaneous power P=Fv=5×2=10 W.
A tightly wound 100-turn coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (take the permeability of free space as 4π×10−7 SI units):
Solution
B=2rμ0NI=2×0.14π×10−7×100×7=4.4×10−3 T =4.4 mT.
The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is 2400 gcm2. The length of the 400 g rod is nearly:
Solution
I=12ML2⇒L2=40012×2400=72⇒L≈8.5 cm.
A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If the speed becomes 2ω while keeping the same radius, the tension in the string becomes:
Solution
Tension supplies the centripetal force, T=mω2r. Doubling the angular speed at the same radius quadruples it: 4T.
Match List I with List II.
| List I (Material) | List II (Susceptibility χ) |
|---|---|
| A. Diamagnetic | I. χ=0 |
| B. Ferromagnetic | II. −1≤χ<0 |
| C. Paramagnetic | III. χ≫1 |
| D. Non-magnetic | IV. 0<χ<ε (a small positive number) |
Choose the correct answer from the options given below:
Solution
Diamagnetic: small negative susceptibility, −1≤χ<0 (II). Ferromagnetic: χ≫1 (III). Paramagnetic: small positive, 0<χ<ε (IV). Non-magnetic: χ=0 (I).
In the following circuit, the equivalent capacitance between terminal A and terminal B is:

Solution
The upper and lower branches each have two 2μF capacitors in series, so the midpoints of both sit at the same potential; the central 2μF carries no charge and can be removed. Each branch is 1μF, and the two in parallel give 2μF.
A horizontal force of 10 N is applied to a block A, which is in contact with a block B on the side away from the force. The masses of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:
Solution
Both blocks share acceleration a=2+310=2 m/s2. The only horizontal force on B is the push from A: F=mBa=3×2=6 N.
82290X α Y e+ Z e− P e− Q
In the nuclear emission stated above, the mass number and atomic number of the product Q, respectively, are:
Solution
An α−emission lowers A by 4 and Z by 2: (286,80). Each β−emission leaves A unchanged; a positron lowers Z by 1 and an electron raises it by 1. Net over e+,e−,e−: Z=80−1+1+1=81. So Q is (286,81).
In a vernier callipers, (N+1) divisions of the vernier scale coincide with N divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
Solution
(N+1) VSD =N MSD, so 1 VSD =N+1N MSD. Vernier constant = 1 MSD − 1 VSD =N+11 MSD =N+10.1 mm =100(N+1)1 cm.
If x=5sin(πt+3π) m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of the motion, respectively, are:
Solution
Comparing with x=Asin(ωt+ϕ): amplitude A=5 m and ω=π rad/s, so T=ω2π=2 s.
In the given diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of the induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

Solution
By Lenz's law each solenoid opposes the motion. Solenoid-1 sees the magnet's N pole receding, so its near end (B) becomes an S pole to attract it; solenoid-2 sees the S pole approaching, so its near end (C) becomes an S pole to repel it. Tracing the winding sense that produces those poles gives current directions AB in solenoid-1 and DC in solenoid-2.
A logic circuit provides the output Y as per the following truth table:
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The expression for the output Y is:
Solution
The output is 1 exactly when B=0, regardless of A — the table is that of Bˉ.
A wire of length l and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
Solution
Each part is 10Ω. Five in series give 50Ω; five in parallel give 2Ω; the two groups in series give 50+2=52 Ω.
The output (Y) of the given logic gate circuit is similar to the output of a/an:

Solution
The upper gate has both inputs tied to A, so it outputs Aˉ; the lower NOR with both inputs B outputs Bˉ. The final NOR gives Y=Aˉ+Bˉ=A⋅B — an AND gate.
A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 4πε01=9×109 SI units.)

Solution
Inside a charged spherical shell the field is zero, so the potential is the same at every interior point (and equal to the surface value). The potential difference between C and P is zero.
The mass of a planet is 101th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:
Solution
g=R2GM. With M′=M/10 and R′=R/2: g′=g×101×4=0.4g=3.92 m/s2.
The minimum energy required to launch a satellite of mass m from the surface of the earth (mass M, radius R) into a circular orbit at an altitude of 2R from the surface of the earth is:
Solution
Orbit radius r=3R. Energy in orbit =−2rGMm=−6RGMm; on the surface =−RGMm. Minimum energy =−6RGMm+RGMm=6R5GMm.
A small telescope has an objective of focal length 140 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope for viewing a distant object is:
Solution
For a distant object, m=fefo=5140=28.
The velocity (v)–time (t) plot of the motion of a body is shown below. The acceleration (a)–time (t) graph that best suits this motion is:

Solution
The velocity rises linearly (constant positive acceleration), stays constant (zero acceleration), then falls linearly (constant negative acceleration). The a–t graph is therefore a positive rectangle, zero, then a negative rectangle — graph (2).
Two heaters A and B have power ratings of 1 kW and 2 kW, respectively. The two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
Solution
Series: Ps=P1+P2P1P2=32 kW. Parallel: Pp=P1+P2=3 kW. Ratio =32/3=92.
A force defined by F=αt2+βt acts on a particle at a given time t. The factor which is dimensionless, if α and β are constants, is:
Solution
αt2 and βt both have the dimensions of force, so [αt]=[β]. Hence βαt is dimensionless.
A10μF capacitor is connected to a 210 V, 50 Hz source. The peak current in the circuit is nearly (take π=3.14):
Solution
XC=2πfC1=2×3.14×50×10−51≈318 Ω. Peak current =XCV0=3182102≈0.93 A.
A metallic bar of Young's modulus 0.5×1011 Nm−2 and coefficient of linear thermal expansion 10−5 ∘C−1, length 1 m and area of cross-section 10−3 m2, is heated from 0∘C to 100∘C without expansion or bending. The compressive force developed in it is:
Solution
Thermal strain prevented =αΔT=10−5×100=10−3. Force =YAαΔT=0.5×1011×10−3×10−3=5×104 N =50×103 N.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:
Solution
Between the plates the changing electric field constitutes a displacement current Id=ε0dtdΦE, equal in magnitude to the conduction current and in the same direction, so the current is continuous around the circuit.
Choose the correct circuit (from the four given) which can achieve the bridge balance.

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both non-conducting and non-polar.
Choose the correct statement(s) from the options given below:
Solution
A magnetic sheet is attracted and needs a force to hold it (A). Moving a conductor away from a pole induces eddy currents that oppose the motion, so a force is needed to keep it moving uniformly (C). A non-magnetic sheet feels no force (B false); a non-conducting, non-polar sheet has no induced currents (D false).
If the plates of a parallel plate capacitor connected to a battery are moved closer to each other, then:
A. the charge stored in it increases.
B. the energy stored in it decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
Solution
At constant V (battery connected), C=dε0A rises as d falls, so Q=CV rises and U=21CV2 rises; Q/V=C increases; QV increases. A, C and E are correct; B and D are false.
The following graph represents the T–V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P1, P2 and P3, compared with those of Charles's law represented as dotted lines. Then the correct relation is:

Solution
For an ideal gas at fixed pressure, V=PnRT: the T–V line through the origin has slope ∝P1, so the steepest line corresponds to the lowest pressure. Reading the slopes in the figure gives P1>P2>P3.
The property which is not of an electromagnetic wave travelling in free space is that:
Solution
Electromagnetic waves are produced by accelerating charges; a charge moving with uniform velocity does not radiate. The other three statements are standard properties of EM waves in free space.
An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle of 60∘ with each other. The magnetic moment of this new magnet is:
Solution
Bending does not change the pole strength m; it changes the distance between the poles. The two end poles are now separated by the chord of two arms of length 2L at 60∘: 2×2Lsin30∘=2L. So M′=m×2L=2M.
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is 2x times its original time period. Then the value of x is:
Solution
T=2πgl is independent of mass. Halving l gives T′=2T=22T, so x=2.