AIPMT 2014 · Physics

AIPMT 2014 Physics Questions with Solutions

The AIPMT 2014 paper had 49 Physics questions from 24 chapters.

Dual Nature of Radiation and Matter and Laws of Motion had the most questions (4 each), followed by Current Electricity, Moving Charges and Magnetism and 3 other chapters with 3 each.

8 questions below have the answer and explanation free; the other 41 are in Premium.

Physics questions
49
Chapters covered
24
Solved free here
8 of 49
Easy / Medium / Hard
17 / 27 / 5

Difficulty is NEET MIND's own tag for each question.

Chapter-wise: AIPMT 2014 Physics

How many questions each chapter had in AIPMT 2014. Open a chapter for its questions from every year.

  1. Dual Nature of Radiation and Matter4 Qs
  2. Laws of Motion4 Qs
  3. Current Electricity3 Qs
  4. Moving Charges and Magnetism3 Qs
  5. Nuclei3 Qs
  6. System of Particles and Rotational Motion3 Qs
  7. Waves3 Qs
  8. Electric Charges and Fields2 Qs
  9. Gravitation2 Qs
  10. Motion in a Plane2 Qs
  11. Ray Optics and Optical Instruments2 Qs
  12. Semiconductor Electronics2 Qs
  13. Thermal Properties of Matter2 Qs
  14. Thermodynamics2 Qs
  15. Units and Measurements2 Qs
  16. Wave Optics2 Qs
  17. Alternating Current1 Q
  18. Atoms1 Q
  19. Electromagnetic Induction1 Q
  20. Electrostatic Potential and Capacitance1 Q
  21. Kinetic Theory1 Q
  22. Mechanical Properties of Fluids1 Q
  23. Mechanical Properties of Solids1 Q
  24. Oscillations1 Q

All 49 AIPMT 2014 Physics questions

In paper order. Try each one, then open the answer where it is shown.

  1. Question 1 (AIPMT 2014, Q1)

    Units and MeasurementsEasy
    If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are:
    1. Option A: [F V T⁻¹]
    2. Option B: [F V T⁻²]
    3. Option C: [F V⁻¹ T⁻¹]
    4. Option D: [F V⁻¹ T]

    The answer and explanation for this question are in NEET MIND Premium.

  2. Question 2 (AIPMT 2014, Q2)

    Motion in a PlaneEasy
    A projectile is fired from the surface of the earth with a velocity of 5 m s−15\,\text{m s}^{-1} and angle θ\theta with the horizontal. Another projectile fired from another planet with a velocity of 3 m s−13\,\text{m s}^{-1} at the same angle follows a trajectory which is identical to the trajectory of the projectile fired from earth. The value of the acceleration due to gravity on the planet is (in m s−2\text{m s}^{-2}) given g=9.8 m s−2g = 9.8\,\text{m s}^{-2}.
    1. Option A: 3.5
    2. Option B: 5.9
    3. Option C: 16.3
    4. Option D: 110.8
    Show answer & explanation

    Correct answer: (A) 3.5

    Explanation

    For projectile motion, the range is given by: R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g} For identical trajectories at the same angle: u12g1=u22g2\frac{u_1^2}{g_1} = \frac{u_2^2}{g_2} Substituting values: 529.8=32gplanet\frac{5^2}{9.8} = \frac{3^2}{g_{\text{planet}}} gplanet=(35)2×9.8g_{\text{planet}} = \left(\frac{3}{5}\right)^2 \times 9.8 gplanet≈3.5 m s−2g_{\text{planet}} \approx 3.5\,\text{m s}^{-2} Hence, the correct answer is option (A).

  3. Question 3 (AIPMT 2014, Q3)

    Motion in a PlaneEasy
    A particle is moving such that its position coordinates (x,y)(x, y) are (2 m,3 m)(2\,\text{m}, 3\,\text{m}) at time t=0t = 0, (6 m,7 m)(6\,\text{m}, 7\,\text{m}) at time t=2 st = 2\,\text{s} and (13 m,14 m)(13\,\text{m}, 14\,\text{m}) at time t=5 st = 5\,\text{s}. Average velocity vector (v⃗av)\left(\vec{v}_{\text{av}}\right) from t=0t = 0 to t=5 st = 5\,\text{s} is:
    1. Option A: 15(13i^+14j^)\dfrac{1}{5}(13\hat{i} + 14\hat{j})
    2. Option B: 73(i^+j^)\dfrac{7}{3}(\hat{i} + \hat{j})
    3. Option C: 2(i^+j^)2(\hat{i} + \hat{j})
    4. Option D: 115(i^+j^)\dfrac{11}{5}(\hat{i} + \hat{j})
    Show answer & explanation

    Correct answer: (D) 115(i^+j^)\dfrac{11}{5}(\hat{i} + \hat{j})

    Explanation

    Average velocity is given by: v⃗av=Δr⃗Δt\vec{v}_{\text{av}} = \frac{\Delta \vec{r}}{\Delta t} Initial position: r⃗1=2i^+3j^\vec{r}_1 = 2\hat{i} + 3\hat{j} Final position: r⃗2=13i^+14j^\vec{r}_2 = 13\hat{i} + 14\hat{j} Displacement: Δr⃗=(13−2)i^+(14−3)j^\Delta \vec{r} = (13 - 2)\hat{i} + (14 - 3)\hat{j} Δr⃗=11i^+11j^\Delta \vec{r} = 11\hat{i} + 11\hat{j} Time interval: Δt=5−0=5 s\Delta t = 5 - 0 = 5\,\text{s} Therefore, v⃗av=11i^+11j^5\vec{v}_{\text{av}} = \frac{11\hat{i} + 11\hat{j}}{5} v⃗av=115(i^+j^)\vec{v}_{\text{av}} = \frac{11}{5}(\hat{i} + \hat{j}) Hence, the correct answer is option (D).

  4. Question 4 (AIPMT 2014, Q4)

    Laws of MotionMedium
    A system consists of three masses m1m_1, m2m_2 and m3m_3 connected by a string passing over a pulley PP. The mass m1m_1 hangs freely and m2m_2 and m3m_3 are on a rough horizontal table (the coefficient of friction = μ\mu). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1m_1 is: Assume m1=m2=m3=mm_1 = m_2 = m_3 = m.
    1. Option A: g(1−gμ)9\dfrac{g(1-g\mu)}{9}
    2. Option B: 2gμ3\dfrac{2g\mu}{3}
    3. Option C: g(1−2μ)3\dfrac{g(1-2\mu)}{3}
    4. Option D: g(1−2μ)2\dfrac{g(1-2\mu)}{2}

    The answer and explanation for this question are in NEET MIND Premium.

  5. Question 5 (AIPMT 2014, Q5)

    Laws of MotionEasy
    The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is :
    1. Option A: 24 Ns24\,\text{Ns}
    2. Option B: 20 Ns20\,\text{Ns}
    3. Option C: 12 Ns12\,\text{Ns}
    4. Option D: 6 Ns6\,\text{Ns}

    The answer and explanation for this question are in NEET MIND Premium.

  6. Question 6 (AIPMT 2014, Q6)

    Laws of MotionMedium
    A balloon with mass 'm' is descending down with an acceleration 'a' (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration 'a'?
    1. Option A: 2mag+a\dfrac{2ma}{g+a}
    2. Option B: 2mag−a\dfrac{2ma}{g-a}
    3. Option C: mag+a\dfrac{ma}{g+a}
    4. Option D: mag−a\dfrac{ma}{g-a}

    The answer and explanation for this question are in NEET MIND Premium.

  7. Question 7 (AIPMT 2014, Q7)

    Laws of MotionMedium
    A body of mass (4m) is lying in the x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m), move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is:
    1. Option A: mv2mv^2
    2. Option B: 32mv2\dfrac{3}{2}mv^2
    3. Option C: 2mv22mv^2
    4. Option D: 4mv24mv^2

    The answer and explanation for this question are in NEET MIND Premium.

  8. Question 8 (AIPMT 2014, Q8)

    OscillationsMedium
    The oscillation of a body on a smooth horizontal surface is represented by the equation: x=Acos⁡(ωt)x = A\cos(\omega t) where: x=displacement at time tx = \text{displacement at time } t ω=angular frequency of oscillation\omega = \text{angular frequency of oscillation} Which one of the following graphs shows correctly the variation of acceleration 'a' with time 't'?
    1. Option A: Option (1)
    2. Option B: Option (2)
    3. Option C: Option (3)
    4. Option D: Option (4)

    The answer and explanation for this question are in NEET MIND Premium.

  9. Question 9 (AIPMT 2014, Q9)

    System of Particles and Rotational MotionMedium
    A solid cylinder of mass 50 kg50\,\text{kg} and radius 0.5 m0.5\,\text{m} is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and the other hanging freely. The tension in the string required to produce an angular acceleration of 22 revolutions s−2\text{s}^{-2} is:
    1. Option A: 25 N25\,\text{N}
    2. Option B: 50 N50\,\text{N}
    3. Option C: 78.5 N78.5\,\text{N}
    4. Option D: 157 N157\,\text{N}

    The answer and explanation for this question are in NEET MIND Premium.

  10. Question 10 (AIPMT 2014, Q10)

    System of Particles and Rotational MotionMedium
    The ratio of the accelerations for a solid sphere (mass 'm' and radius 'R') rolling down an incline of angle θ\theta without slipping and slipping down the incline without rolling is:
    1. Option A: 5:75:7
    2. Option B: 2:32:3
    3. Option C: 2:52:5
    4. Option D: 7:57:5

    The answer and explanation for this question are in NEET MIND Premium.

  11. Question 11 (AIPMT 2014, Q11)

    GravitationMedium
    A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass =5.98×1024 kg= 5.98 \times 10^{24}\,\text{kg}) have to be compressed to be a black hole?
    1. Option A: 10−9 m10^{-9}\,\text{m}
    2. Option B: 10−6 m10^{-6}\,\text{m}
    3. Option C: 10−2 m10^{-2}\,\text{m}
    4. Option D: 100 m100\,\text{m}

    The answer and explanation for this question are in NEET MIND Premium.

  12. Question 12 (AIPMT 2014, Q12)

    GravitationMedium
    Dependence of intensity of gravitational field (E)(E) of earth with distance (r)(r) from the centre of earth is correctly represented by:
    1. Option A: Option (1)
    2. Option B: Option (2)
    3. Option C: Option (3)
    4. Option D: Option (4)

    The answer and explanation for this question are in NEET MIND Premium.

  13. Question 13 (AIPMT 2014, Q13)

    Mechanical Properties of SolidsMedium
    Copper of fixed volume 'V; is drawn into a wire of length 'l'. When this wire is subjected to a constant force 'F', the extension produced in the wire is 'Δ\Deltal'. Which of the following graphs is a straight line?
    1. Option A: Δl\Delta l versus 1l\dfrac{1}{l}
    2. Option B: Δl\Delta l versus l2l^2
    3. Option C: Δl\Delta l versus 1l2\dfrac{1}{l^2}
    4. Option D: Δl\Delta l versus ll
    Show answer & explanation

    Correct answer: (B) Δl\Delta l versus l2l^2

    Explanation

    Young's modulus is given by: Y=FAΔllY = \frac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}} Rearranging: Δl=FlAY\Delta l = \frac{Fl}{AY} Since the volume of the wire is constant: V=AlV = Al Therefore: A=VlA = \frac{V}{l} Substituting into the expression for extension: Δl=Fl(Vl)Y\Delta l = \frac{Fl}{\left(\dfrac{V}{l}\right)Y} Δl=Fl2VY\Delta l = \frac{Fl^2}{VY} Thus, Δl∝l2\Delta l \propto l^2 Hence, the graph between $\Delta landandl^2$ is a straight line. Therefore, option (B) is correct.

  14. Question 14 (AIPMT 2014, Q14)

    Mechanical Properties of FluidsMedium
    A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:
    1. Option A: energy =4VT(1r−1R)= 4VT\left(\dfrac{1}{r} - \dfrac{1}{R}\right) is released
    2. Option B: energy =3VT(1r+1R)= 3VT\left(\dfrac{1}{r} + \dfrac{1}{R}\right) is absorbed
    3. Option C: energy =3VT(1r−1R)= 3VT\left(\dfrac{1}{r} - \dfrac{1}{R}\right) is released
    4. Option D: Energy is neither released nor absorbed

    The answer and explanation for this question are in NEET MIND Premium.

  15. Question 15 (AIPMT 2014, Q15)

    Thermal Properties of MatterMedium
    Steam at 100∘C100^{\circ}\mathrm{C} is passed into 20 g20\,\mathrm{g} of water at 10∘C10^{\circ}\mathrm{C}. When the water acquires a temperature of 80∘C80^{\circ}\mathrm{C}, the mass of water present will be: (Take specific heat of water =1 cal g−1 ∘C−1= 1\,\mathrm{cal\,g^{-1}\,^{\circ}C^{-1}} and latent heat of steam =540 cal g−1= 540\,\mathrm{cal\,g^{-1}})
    1. Option A: 24 g24\,\mathrm{g}
    2. Option B: 31.5 g31.5\,\mathrm{g}
    3. Option C: 42.5 g42.5\,\mathrm{g}
    4. Option D: 22.5 g22.5\,\mathrm{g}
    Show answer & explanation

    Correct answer: (D) 22.5 g22.5\,\mathrm{g}

    Explanation

    Let mm grams of steam condense. Heat released by steam: mL+mc(100−80)mL + mc(100-80) Heat gained by water: 20×1×(80−10)20 \times 1 \times (80-10) Substituting values: m(540)+m(20)=20×70m(540) + m(20) = 20 \times 70 560m=1400560m = 1400 m=2.5 gm = 2.5\,\mathrm{g} Therefore, total mass of water present: 20+2.5=22.5 g20 + 2.5 = 22.5\,\mathrm{g} Hence, the correct answer is option (D).

  16. Question 16 (AIPMT 2014, Q16)

    Thermal Properties of MatterHard
    A quantity of water cools from 70∘C70^{\circ}\mathrm{C} to 60∘C60^{\circ}\mathrm{C} in the first 55 minutes and to 54∘C54^{\circ}\mathrm{C} in the next 55 minutes. The temperature of the surroundings is:
    1. Option A: 45∘C45^{\circ}\mathrm{C}
    2. Option B: 20∘C20^{\circ}\mathrm{C}
    3. Option C: 42∘C42^{\circ}\mathrm{C}
    4. Option D: 10∘C10^{\circ}\mathrm{C}
    Show answer & explanation

    Correct answer: (A) 45∘C45^{\circ}\mathrm{C}

    Explanation

    According to Newton's law of cooling: θ1−θ2t=k(θ1+θ22−θ0)\frac{\theta_1-\theta_2}{t}=k\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right) For the first $5minutes:minutes:70−605=k(65−θ0)\frac{70-60}{5}=k(65-\theta_0)2=k(65−θ0)(1)2=k(65-\theta_0) \qquad (1)ForthenextFor the next5minutes:minutes:60−545=k(57−θ0)\frac{60-54}{5}=k(57-\theta_0)65=k(57−θ0)(2)\frac{6}{5}=k(57-\theta_0) \qquad (2)Dividing(1)by(2):Dividing (1) by (2):26/5=65−θ057−θ0\frac{2}{6/5}=\frac{65-\theta_0}{57-\theta_0}Solving,Solving,θ0=45∘C\theta_0=45^{\circ}\mathrm{C}$ Hence, the correct answer is option (A).

  17. Question 17 (AIPMT 2014, Q17)

    ThermodynamicsHard
    A monoatomic gas at a pressure PP, having a volume VV expands isothermally to a volume 2V2V and then adiabatically to a volume 16V16V. The final pressure of the gas is: (Take γ=53\gamma = \dfrac{5}{3})
    1. Option A: 64P64P
    2. Option B: 32P32P
    3. Option C: P64\dfrac{P}{64}
    4. Option D: 16P16P

    The answer and explanation for this question are in NEET MIND Premium.

  18. Question 18 (AIPMT 2014, Q18)

    ThermodynamicsHard
    A thermodynamic system undergoes the cyclic process ABCDAABCDA as shown in the figure. The work done by the system in the cycle is:
    1. Option A: P0V0P_0V_0
    2. Option B: 2P0V02P_0V_0
    3. Option C: P0V02\dfrac{P_0V_0}{2}
    4. Option D: Zero

    The answer and explanation for this question are in NEET MIND Premium.

  19. Question 19 (AIPMT 2014, Q19)

    Kinetic TheoryEasy
    The mean free path of molecules of a gas, (radius 'r') is inversely proportional to:
    1. Option A: r3r^3
    2. Option B: r2r^2
    3. Option C: rr
    4. Option D: r\sqrt{r}
    Show answer & explanation

    Correct answer: (B) r2r^2

    Explanation

    The mean free path of gas molecules is: λ=12πd2n\lambda = \frac{1}{\sqrt{2}\pi d^2 n} where: - $d=diameterofamolecule−= diameter of a molecule -n=numberdensityofmoleculesSince:= number density of molecules Since:d=2rd = 2rtherefore,therefore,λ∝1d2∝1r2\lambda \propto \frac{1}{d^2} \propto \frac{1}{r^2}Hence,themeanfreepathisinverselyproportionaltoHence, the mean free path is inversely proportional tor^2$. Therefore, option (B) is correct.

  20. Question 20 (AIPMT 2014, Q20)

    WavesMedium
    If n1n_1, n2n_2 and n3n_3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency nn of the string is given by:
    1. Option A: 1n=1n1+1n2+1n3\dfrac{1}{n}=\dfrac{1}{n_1}+\dfrac{1}{n_2}+\dfrac{1}{n_3}
    2. Option B: 1n=1n1+1n2+1n3\dfrac{1}{\sqrt{n}}=\dfrac{1}{\sqrt{n_1}}+\dfrac{1}{\sqrt{n_2}}+\dfrac{1}{\sqrt{n_3}}
    3. Option C: n=n1+n2+n3\sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}
    4. Option D: n=n1+n2+n3n=n_1+n_2+n_3
    Show answer & explanation

    Correct answer: (A) 1n=1n1+1n2+1n3\dfrac{1}{n}=\dfrac{1}{n_1}+\dfrac{1}{n_2}+\dfrac{1}{n_3}

    Explanation

    For a stretched string, n=12lTμn=\frac{1}{2l}\sqrt{\frac{T}{\mu}} For the same string, tension $Tandlineardensityand linear density\muremainconstant.Therefore,remain constant. Therefore,n∝1ln\propto \frac{1}{l}Lettheoriginalstringlengthbe:Let the original string length be:l=l1+l2+l3l=l_1+l_2+l_3Usingtheproportionality:Using the proportionality:1n∝l\frac{1}{n}\propto lHence,Hence,1n=1n1+1n2+1n3\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$ Therefore, option (A) is correct.

  21. Question 21 (AIPMT 2014, Q21)

    WavesMedium
    The number of possible natural oscillations of an air column in a pipe closed at one end of length 85 cm85\,\text{cm} whose frequencies lie below 1250 Hz1250\,\text{Hz} are: (velocity of sound = 340 ms⁻¹)
    1. Option A: 4
    2. Option B: 5
    3. Option C: 7
    4. Option D: 6

    The answer and explanation for this question are in NEET MIND Premium.

  22. Question 22 (AIPMT 2014, Q22)

    WavesMedium
    A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the horn as heard by him will be:
    1. Option A: 1332 Hz1332\,\text{Hz}
    2. Option B: 1372 Hz1372\,\text{Hz}
    3. Option C: 1412 Hz1412\,\text{Hz}
    4. Option D: 1464 Hz1464\,\text{Hz}

    The answer and explanation for this question are in NEET MIND Premium.

  23. Question 23 (AIPMT 2014, Q23)

    Electrostatic Potential and CapacitanceHard
    Two thin dielectric slabs of dielectric constants K1K_1 and K2K_2 (K1<K2K_1 < K_2) are inserted between the plates of a parallel plate capacitor as shown in the figure. The variation of electric field 'E' between the plates with distance 'd' as measured from plate PP is correctly shown by:
    1. Option A: Option (1)
    2. Option B: Option (2)
    3. Option C: Option (3)
    4. Option D: Option (4)

    The answer and explanation for this question are in NEET MIND Premium.

  24. Question 24 (AIPMT 2014, Q24)

    Electric Charges and FieldsEasy
    A conducting sphere of radius RR is given a charge QQ. The electric potential and the electric field at the centre of the sphere respectively are:
    1. Option A: Zero and Q4πε0R2\dfrac{Q}{4\pi\varepsilon_0R^2}
    2. Option B: Q4πε0R\dfrac{Q}{4\pi\varepsilon_0R} and Zero
    3. Option C: Q4πε0R\dfrac{Q}{4\pi\varepsilon_0R} and Q4πε0R2\dfrac{Q}{4\pi\varepsilon_0R^2}
    4. Option D: Both are zero

    The answer and explanation for this question are in NEET MIND Premium.

  25. Question 25 (AIPMT 2014, Q25)

    Electric Charges and FieldsMedium
    In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz where V is in volts and xx, yy, zz are in metres. The electric force experienced by a charge of 2 coulomb situated at the point (1,1,1)(1, 1, 1) is:
    1. Option A: 65 N6\sqrt{5}\,\mathrm{N}
    2. Option B: 30 N30\,\mathrm{N}
    3. Option C: 24 N24\,\mathrm{N}
    4. Option D: 435 N4\sqrt{35}\,\mathrm{N}

    The answer and explanation for this question are in NEET MIND Premium.

  26. Question 26 (AIPMT 2014, Q26)

    Current ElectricityEasy
    Two cities are 150 km150\,\mathrm{km} apart. Electric power is sent from one city to another through copper wires. The fall of potential per kilometre is 8 V8\,\mathrm{V} and the average resistance per kilometre is 0.5 Ω0.5\,\Omega. The power loss in the wires is:
    1. Option A: 19.2 W19.2\,\mathrm{W}
    2. Option B: 19.2 kW19.2\,\mathrm{kW}
    3. Option C: 19.2 J19.2\,\mathrm{J}
    4. Option D: 12.2 kW12.2\,\mathrm{kW}

    The answer and explanation for this question are in NEET MIND Premium.

  27. Question 27 (AIPMT 2014, Q27)

    Current ElectricityMedium
    The resistance in the two arms of the meter bridge are 5Ω and RΩ, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 ℓ₁. The resistance R is:
    1. Option A: 10 Ω
    2. Option B: 15 Ω
    3. Option C: 20 Ω
    4. Option D: 25 Ω

    The answer and explanation for this question are in NEET MIND Premium.

  28. Question 28 (AIPMT 2014, Q28)

    Current ElectricityMedium
    A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire has an emf of 2.0 V and negligible internal resistance. The potentiometer wire itself is 4m long. When the resistance R connected across the given cell has values of (i) infinity (ii) 9.5 Ω, the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of the internal resistance of the cell is:
    1. Option A: 0.25 Ω
    2. Option B: 0.95 Ω
    3. Option C: 0.50 Ω
    4. Option D: 0.75 Ω

    The answer and explanation for this question are in NEET MIND Premium.

  29. Question 29 (AIPMT 2014, Q29)

    Moving Charges and MagnetismMedium
    Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment m⃗\vec{m}. Which configuration has highest net magnetic dipole moment?
    1. Option A: Option (a)
    2. Option B: Option (b)
    3. Option C: Option (c)
    4. Option D: Option (d)

    The answer and explanation for this question are in NEET MIND Premium.

  30. Question 30 (AIPMT 2014, Q30)

    Moving Charges and MagnetismMedium
    In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be:
    1. Option A: 1/499 G
    2. Option B: 499/500 G
    3. Option C: 1/500 G
    4. Option D: 500/499 G

    The answer and explanation for this question are in NEET MIND Premium.

  31. Question 31 (AIPMT 2014, Q31)

    Moving Charges and MagnetismMedium
    Two identical long conducting wires AOB and COD are placed at right angles to each other, with one above the other such that O is their common point. The wires carry currents I₁ and I₂ respectively. Point P lies at a distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at point P will be:
    1. Option A: μ₀ / 2πd (I₁ / I₂)
    2. Option B: μ₀ / 2πd (I₁ + I₂)
    3. Option C: μ₀ / 2πd (I₁² − I₂²)
    4. Option D: μ₀ / 2πd (I₁² + I₂²)¹ᐟ²

    The answer and explanation for this question are in NEET MIND Premium.

  32. Question 32 (AIPMT 2014, Q32)

    Electromagnetic InductionMedium
    A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is v, is:
    1. Option A: Zero
    2. Option B: Bvπr²/2 and P is at higher potential
    3. Option C: πrBv and R is at higher potential
    4. Option D: 2rBv and R is at higher potential

    The answer and explanation for this question are in NEET MIND Premium.

  33. Question 33 (AIPMT 2014, Q33)

    Alternating CurrentEasy
    A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil are respectively:
    1. Option A: 300 V, 15A
    2. Option B: 450 V, 15A
    3. Option C: 450 V, 13.5A
    4. Option D: 600 V, 15A

    The answer and explanation for this question are in NEET MIND Premium.

  34. Question 34 (AIPMT 2014, Q34)

    Dual Nature of Radiation and MatterEasy
    Light with an energy flux of 25 × 10⁴ W m⁻² falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm², the average force exerted on the surface is:
    1. Option A: 1.25 × 10⁻⁶ N
    2. Option B: 2.50 × 10⁻⁶ N
    3. Option C: 1.20 × 10⁻⁶ N
    4. Option D: 3.0 × 10⁻⁶ N

    The answer and explanation for this question are in NEET MIND Premium.

  35. Question 35 (AIPMT 2014, Q35)

    Wave OpticsMedium
    A beam of light of λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:
    1. Option A: 1.2 cm
    2. Option B: 1.2 mm
    3. Option C: 2.4 cm
    4. Option D: 2.4 mm

    The answer and explanation for this question are in NEET MIND Premium.

  36. Question 36 (AIPMT 2014, Q36)

    Wave OpticsEasy
    In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, λ being the wavelength of light used. The intensity at a point where the path difference is λ/4 will be:
    1. Option A: K
    2. Option B: K/4
    3. Option C: K/2
    4. Option D: Zero

    The answer and explanation for this question are in NEET MIND Premium.

  37. Question 37 (AIPMT 2014, Q37)

    Ray Optics and Optical InstrumentsEasy
    If the focal length of objective lens is increased, then magnifying power of:
    1. Option A: Microscope will increase but that of telescope decrease.
    2. Option B: Microscope and telescope both will increase.
    3. Option C: Microscope and telescope both will decrease.
    4. Option D: Microscope will decrease but that of telescope increase.

    The answer and explanation for this question are in NEET MIND Premium.

  38. Question 38 (AIPMT 2014, Q38)

    Ray Optics and Optical InstrumentsHard
    The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface return back through the same path after suffering reflection at the silvered surface. The refractive index μ of the prism is:
    1. Option A: 2sinA
    2. Option B: 2cosA
    3. Option C: 1/2 cosA
    4. Option D: tanA

    The answer and explanation for this question are in NEET MIND Premium.

  39. Question 39 (AIPMT 2014, Q39)

    Dual Nature of Radiation and MatterMedium
    When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increases from 0.5 eV to 0.8 eV. The work function of the metal is:
    1. Option A: 0.65 eV
    2. Option B: 1.0 eV
    3. Option C: 1.3 eV
    4. Option D: 1.5 eV

    The answer and explanation for this question are in NEET MIND Premium.

  40. Question 40 (AIPMT 2014, Q40)

    Dual Nature of Radiation and MatterEasy
    If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
    1. Option A: 25
    2. Option B: 75
    3. Option C: 60
    4. Option D: 50

    The answer and explanation for this question are in NEET MIND Premium.

  41. Question 41 (AIPMT 2014, Q41)

    AtomsMedium
    Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be:
    1. Option A: 3
    2. Option B: 2
    3. Option C: 6
    4. Option D: 10
    Show answer & explanation

    Correct answer: (C) 6

    Explanation

    The energy of the incident photon is E = 12400/975 ≈ 12.72 eV. For hydrogen, excitation from n = 1 to n = 4 requires 13.6(1 − 1/16) = 12.75 eV, which closely matches the photon energy. Thus the atom is excited to n = 4. The number of possible spectral lines emitted during de-excitation is n(n − 1)/2 = 4×3/2 = 6.

  42. Question 42 (AIPMT 2014, Q42)

    NucleiMedium
    The binding energy per nucleon of ⁷₃Li and ⁴₂He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ⁷₃Li + ¹₁H → ⁴₂He + Q, the value of energy Q released is:
    1. Option A: 19.6 MeV
    2. Option B: −2.4 MeV
    3. Option C: 8.4 MeV
    4. Option D: 17.3 MeV

    The answer and explanation for this question are in NEET MIND Premium.

  43. Question 43 (AIPMT 2014, Q43)

    NucleiMedium
    A radio isotope X with a half-life 1.4 × 10⁹ years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is:
    1. Option A: 1.96 × 10⁹ years
    2. Option B: 3.92 × 10⁹ years
    3. Option C: 4.20 × 10⁹ years
    4. Option D: 8.40 × 10⁹ years

    The answer and explanation for this question are in NEET MIND Premium.

  44. Question 44 (AIPMT 2014, Q44)

    Semiconductor ElectronicsMedium
    The given graph represents V-I characteristic for a semiconductor device. Which of the following statement is correct?
    1. Option A: It is V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current.
    2. Option B: It is for a solar cell and point A and B represent open circuit voltage and current, respectively.
    3. Option C: It is a photodiode and points A and B represent open circuit voltage and current, respectively.
    4. Option D: It is for a LED and points A and B represent open circuit voltage and short circuit current, respectively.

    The answer and explanation for this question are in NEET MIND Premium.

  45. Question 45 (AIPMT 2014, Q45)

    Semiconductor ElectronicsEasy
    The barrier potential of a p-n junction depends on: (a) type of semiconductor material (b) amount of doping (c) temperature. Which one of the following is correct?
    1. Option A: (a) and (b) only
    2. Option B: (b) only
    3. Option C: (b) and (c) only
    4. Option D: (a), (b) and (c)

    The answer and explanation for this question are in NEET MIND Premium.

  46. Question 46 (AIPMT 2014, Q46)

    NucleiEasy
    What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, ℓ = 1, mℓ_ℓ = 0
    1. Option A: 1
    2. Option B: 2
    3. Option C: 3
    4. Option D: 4

    The answer and explanation for this question are in NEET MIND Premium.

  47. Question 47 (AIPMT 2014, Q47)

    Dual Nature of Radiation and MatterEasy
    Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck's constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹)
    1. Option A: 6.67 × 10¹⁵
    2. Option B: 6.67 × 10¹¹
    3. Option C: 4.42 × 10⁻¹⁵
    4. Option D: 4.42 × 10⁻¹⁸

    The answer and explanation for this question are in NEET MIND Premium.

  48. Question 48 (AIPMT 2014, Q48)

    Units and MeasurementsEasy
    Equal masses of H₂, O₂ and methane have been taken in container of volume V at temperature 27°C under identical conditions. The ratio of the volumes of gases H₂ : O₂ : methane would be:
    1. Option A: 8 : 16 : 1
    2. Option B: 16 : 8 : 1
    3. Option C: 16 : 1 : 2
    4. Option D: 8 : 1 : 2

    The answer and explanation for this question are in NEET MIND Premium.

  49. Question 49 (AIPMT 2014, Q49)

    System of Particles and Rotational MotionEasy
    If a is the length of the side of a cube, the distance between the body centered atom and one corner of the cube will be:
    1. Option A: 2/√3 a
    2. Option B: 4/√3 a
    3. Option C: √3/4 a
    4. Option D: √3/2 a

    The answer and explanation for this question are in NEET MIND Premium.

NEET Physics questions in other years

All NEET Physics PYQs, chapter-wise

Practise AIPMT 2014 Physics with your progress saved

A free account gives you 15 PYQs in every chapter with explanations, and keeps every mistake for revision.