NEET 2016 · Physics

NEET 2016 Physics Questions with Solutions

The NEET 2016 paper had 45 Physics questions from 26 chapters.

System of Particles and Rotational Motion had the most questions (4), followed by Moving Charges and Magnetism, Ray Optics and Optical Instruments and Semiconductor Electronics with 3 each.

11 questions below have the answer and explanation free; the other 34 are in Premium.

Physics questions
45
Chapters covered
26
Solved free here
11 of 45
Easy / Medium / Hard
10 / 27 / 8

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

Chapter-wise: NEET 2016 Physics

How many questions each chapter had in NEET 2016. Open a chapter for its questions from every year.

  1. System of Particles and Rotational Motion4 Qs
  2. Moving Charges and Magnetism3 Qs
  3. Ray Optics and Optical Instruments3 Qs
  4. Semiconductor Electronics3 Qs
  5. Alternating Current2 Qs
  6. Current Electricity2 Qs
  7. Dual Nature of Radiation and Matter2 Qs
  8. Gravitation2 Qs
  9. Laws of Motion2 Qs
  10. Mechanical Properties of Fluids2 Qs
  11. Thermal Properties of Matter2 Qs
  12. Thermodynamics2 Qs
  13. Wave Optics2 Qs
  14. Waves2 Qs
  15. Atoms1 Q
  16. Electric Charges and Fields1 Q
  17. Electromagnetic Induction1 Q
  18. Electromagnetic Waves1 Q
  19. Electrostatic Potential and Capacitance1 Q
  20. Kinetic Theory1 Q
  21. Motion in a Plane1 Q
  22. Motion in a Straight Line1 Q
  23. Nuclei1 Q
  24. Oscillations1 Q
  25. Units and Measurements1 Q
  26. Work, Energy and Power1 Q

All 45 NEET 2016 Physics questions

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

  1. Question 1 (NEET 2016, Q1)

    Units and MeasurementsMedium
    Planck’s constant (h)(h), speed of light in vacuum (c)(c) and Newton’s gravitational constant (G)(G) are three fundamental constants. Which of the following combinations of these has the dimension of length?
    1. Option A: Option 1
    2. Option B: Option 2
    3. Option C: Option 3
    4. Option D: Option 4

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  2. Question 2 (NEET 2016, Q2)

    Motion in a Straight LineEasy
    Two cars PP and QQ start from a point at the same time in a straight line and their positions are represented by: xP(t)=at+bt2x_P(t)=at+bt^2 and xQ(t)=ft−t2x_Q(t)=ft-t^2 At what time do the cars have the same velocity?
    1. Option A: a−f1+b\dfrac{a-f}{1+b}
    2. Option B: a+f2(b−1)\dfrac{a+f}{2(b-1)}
    3. Option C: a+f2(1+b)\dfrac{a+f}{2(1+b)}
    4. Option D: f−a2(1+b)\dfrac{f-a}{2(1+b)}
    Show answer & explanation

    Correct answer: (D) f−a2(1+b)\dfrac{f-a}{2(1+b)}

    Explanation

    Velocity of car PP: vP=dxPdt=a+2btv_P = \frac{dx_P}{dt} = a + 2bt Velocity of car $Q::vQ=dxQdt=f−2tv_Q = \frac{dx_Q}{dt} = f - 2tForsamevelocity:For same velocity:a+2bt=f−2ta + 2bt = f - 2t2t(b+1)=f−a2t(b+1)=f-at=f−a2(1+b)t=\frac{f-a}{2(1+b)}$ Hence option (D) is correct.

  3. Question 3 (NEET 2016, Q3)

    Motion in a PlaneMedium
    In the given figure, a=15 m/s2a = 15\ \text{m/s}^2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5 mR = 2.5\ \text{m} at a given instant of time. The speed of the particle is:
    1. Option A: 4.5 m/s
    2. Option B: 5.0 m/s
    3. Option C: 5.7 m/s
    4. Option D: 6.2 m/s
    Show answer & explanation

    Correct answer: (C) 5.7 m/s

    Explanation

    The centripetal acceleration is the radial component of the total acceleration. From the figure: ac=acos⁡30∘a_c = a \cos 30^\circ v2r=15×32\frac{v^2}{r} = 15 \times \frac{\sqrt{3}}{2} Given: r=2.5 mr = 2.5\ \text{m} Therefore, v2=15×32×2.5v^2 = 15 \times \frac{\sqrt{3}}{2} \times 2.5 v≈5.7 m/sv \approx 5.7\ \text{m/s} Hence option (C) is correct.

  4. Question 4 (NEET 2016, Q4)

    Laws of MotionMedium
    A rigid ball of mass mm strikes a rigid wall at 60∘60^\circ and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be:
    1. Option A: mvmv
    2. Option B: 2mv2mv
    3. Option C: mv2\dfrac{mv}{2}
    4. Option D: mv3\dfrac{mv}{3}

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  5. Question 5 (NEET 2016, Q5)

    System of Particles and Rotational MotionMedium
    A bullet of mass 10 g10\ \text{g} moving horizontally with a velocity of 400 m s−1400\ \text{m s}^{-1} strikes a wood block of mass 2 kg2\ \text{kg} suspended by a light inextensible string of length 5 m5\ \text{m}. As a result, the centre of gravity of the block rises through a vertical distance of 10 cm10\ \text{cm}. The speed of the bullet after it emerges horizontally from the block will be:
    1. Option A: 100 ms−1100\ \text{ms}^{-1}
    2. Option B: 80 ms−180\ \text{ms}^{-1}
    3. Option C: 120 ms−1120\ \text{ms}^{-1}
    4. Option D: 160 ms−1160\ \text{ms}^{-1}

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  6. Question 6 (NEET 2016, Q6)

    Laws of MotionEasy
    Two identical balls AA and BB having velocities of 0.5 m/s0.5\ \text{m/s} and −0.3 m/s-0.3\ \text{m/s} respectively collide elastically in one dimension. The velocities of BB and AA after the collision respectively will be
    1. Option A: −0.5 m/s-0.5\ \text{m/s} and 0.3 m/s0.3\ \text{m/s}
    2. Option B: 0.5 m/s0.5\ \text{m/s} and −0.3 m/s-0.3\ \text{m/s}
    3. Option C: −0.3 m/s-0.3\ \text{m/s} and 0.5 m/s0.5\ \text{m/s}
    4. Option D: 0.3 m/s0.3\ \text{m/s} and 0.5 m/s0.5\ \text{m/s}

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  7. Question 7 (NEET 2016, Q7)

    Work, Energy and PowerEasy
    A particle moves from a point (−2i^+5j^)(-2\hat{i}+5\hat{j}) to (4j^+3k^)(4\hat{j}+3\hat{k}) when a force (4i^+3j^) N(4\hat{i}+3\hat{j})\ \text{N} is applied. How much work has been done by the force?
    1. Option A: 8 J
    2. Option B: 11 J
    3. Option C: 5 J
    4. Option D: 2 J
    Show answer & explanation

    Correct answer: (C) 5 J

    Explanation

    Displacement vector: s⃗=r2⃗−r1⃗\vec{s} = \vec{r_2}-\vec{r_1} =(4j^+3k^)−(−2i^+5j^)= (4\hat{j}+3\hat{k}) - (-2\hat{i}+5\hat{j}) =2i^−j^+3k^= 2\hat{i}-\hat{j}+3\hat{k} Force: F⃗=4i^+3j^\vec{F} = 4\hat{i}+3\hat{j} Work done: W=F⃗⋅s⃗W = \vec{F} \cdot \vec{s} =(4)(2)+(3)(−1)+(0)(3)= (4)(2) + (3)(-1) + (0)(3) =8−3=5 J= 8 - 3 = 5\ \text{J} Hence option (C) is correct.

  8. Question 8 (NEET 2016, Q8)

    System of Particles and Rotational MotionMedium
    Two rotating bodies AA and BB of masses mm and 2m2m with moments of inertia IAI_A and IBI_B (IBI_B > IAI_A) have equal kinetic energy of rotation. If LAL_A and LBL_B are their angular momenta respectively, then:
    1. Option A: LA=LB2L_A = \dfrac{L_B}{2}
    2. Option B: LA=2LBL_A = 2L_B
    3. Option C: LB>LAL_B > L_A
    4. Option D: LA>LBL_A > L_B

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  9. Question 9 (NEET 2016, Q9)

    System of Particles and Rotational MotionMedium
    A solid sphere of mass mm and radius RR is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (Esphere/Ecylinder)(E_{sphere}/E_{cylinder}) will be:
    1. Option A: 2 : 3
    2. Option B: 1 : 5
    3. Option C: 1 : 4
    4. Option D: 3 : 1

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  10. Question 10 (NEET 2016, Q10)

    System of Particles and Rotational MotionMedium
    A light rod of length ll has two masses m1m_1 and m2m_2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:
    1. Option A: Option 1
    2. Option B: Option 2
    3. Option C: Option 3
    4. Option D: Option 4

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  11. Question 11 (NEET 2016, Q11)

    GravitationMedium
    Starting from the centre of the earth having radius RR, the variation of g (acceleration due to gravity) is shown by:
    1. Option A: Graph (1)
    2. Option B: Graph (2)
    3. Option C: Graph (3)
    4. Option D: Graph (4)

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  12. Question 12 (NEET 2016, Q12)

    GravitationMedium
    A satellite of mass mm is orbiting the earth (of radius RR) at a height hh from its surface. In terms of g0g_0, the value of the total energy of the satellite is:
    1. Option A: mg0R22(R+h)\dfrac{mg_0R^2}{2(R+h)}
    2. Option B: −mg0R2R+h-\dfrac{mg_0R^2}{R+h}
    3. Option C: 2mg0R2R+h\dfrac{2mg_0R^2}{R+h}
    4. Option D: −2mg0R2R+h-\dfrac{2mg_0R^2}{R+h}

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  13. Question 13 (NEET 2016, Q13)

    Mechanical Properties of FluidsMedium
    A rectangular film of liquid is extended from (4 cm×2 cm)(4\ \text{cm} \times 2\ \text{cm}) to (5 cm×4 cm)(5\ \text{cm} \times 4\ \text{cm}). If the work done is 3×10−4 J3 \times 10^{-4}\ \text{J}, the value of surface tension of the liquid is:
    1. Option A: 0.250 N m−10.250\ \text{N m}^{-1}
    2. Option B: 0.125 N m−10.125\ \text{N m}^{-1}
    3. Option C: 0.2 N m−10.2\ \text{N m}^{-1}
    4. Option D: 8.0 N m−18.0\ \text{N m}^{-1}
    Show answer & explanation

    Correct answer: (B) 0.125 N m−10.125\ \text{N m}^{-1}

    Explanation

    Work done in increasing surface area: W=2T(A2−A1)W = 2T(A_2-A_1) Initial area: A1=4×2=8 cm2=8×10−4 m2A_1 = 4 \times 2 = 8\ \text{cm}^2 = 8 \times 10^{-4}\ \text{m}^2 Final area: A2=5×4=20 cm2=20×10−4 m2A_2 = 5 \times 4 = 20\ \text{cm}^2 = 20 \times 10^{-4}\ \text{m}^2 Thus: 3×10−4=2T(12×10−4)3\times10^{-4} = 2T(12\times10^{-4}) T=0.125 N m−1T = 0.125\ \text{N m}^{-1} Hence option (B) is correct.

  14. Question 14 (NEET 2016, Q14)

    Mechanical Properties of FluidsHard
    Three liquids of densities ρ1\rho_1, ρ2\rho_2 and ρ3\rho_3 (with ρ1>ρ2>ρ3\rho_1 > \rho_2 > \rho_3), having the same value of surface tension TT, rise to the same height in three identical capillaries. The angles of contact θ1\theta_1, θ2\theta_2 and θ3\theta_3 obey:
    1. Option A: π2>θ1>θ2>θ3≥0\dfrac{\pi}{2} > \theta_1 > \theta_2 > \theta_3 \geq 0
    2. Option B: 0≤θ1<θ2<θ3<π20 \leq \theta_1 < \theta_2 < \theta_3 < \dfrac{\pi}{2}
    3. Option C: π2<θ1<θ2<θ3<π\dfrac{\pi}{2} < \theta_1 < \theta_2 < \theta_3 < \pi
    4. Option D: π>θ1>θ2>θ3>π2\pi > \theta_1 > \theta_2 > \theta_3 > \dfrac{\pi}{2}

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  15. Question 15 (NEET 2016, Q15)

    Thermal Properties of MatterMedium
    Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100∘C100^\circ C, while the other one is at 0∘C0^\circ C. If the two bodies are brought in contact assuming no heat loss, then the final common temperature is:
    1. Option A: 50∘C50^\circ C
    2. Option B: More than 50∘C50^\circ C
    3. Option C: Less than 50∘C50^\circ C but greater than 0∘C0^\circ C
    4. Option D: 0∘C0^\circ C
    Show answer & explanation

    Correct answer: (B) More than 50∘C50^\circ C

    Explanation

    Since heat capacity increases with temperature, the hotter body has a larger heat capacity. Therefore, for the same temperature fall, the hot body loses more heat than the cold body gains for equal temperature rise. Thus the equilibrium temperature shifts towards the hotter body. Hence the final temperature is greater than 50∘C50^\circ C. Hence option (B) is correct.

  16. Question 16 (NEET 2016, Q16)

    Thermal Properties of MatterHard
    A body cools from a temperature 3T3T to 2T2T in 10 minutes. The room temperature is TT. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of the next 10 minutes will be:
    1. Option A: Option 1
    2. Option B: Option 2
    3. Option C: Option 3
    4. Option D: Option 4
    Show answer & explanation

    Correct answer: (B) Option 2

    Explanation

    Using Newton’s law of cooling: dθdt∝(θ−T)\frac{d\theta}{dt} \propto (\theta - T) Initially: Temperature excess decreases from: 3T−T=2T3T - T = 2T to: 2T−T=T2T - T = T in 10 minutes. Thus the excess temperature halves in 10 minutes. After the next 10 minutes, excess temperature becomes: T2\frac{T}{2} Therefore final temperature: T+T2=3T2T + \frac{T}{2} = \frac{3T}{2} Hence option (B) is correct.

  17. Question 17 (NEET 2016, Q17)

    ThermodynamicsHard
    One mole of an ideal monoatomic gas undergoes a process described by the equation PV3=constantPV^3 = \text{constant}. The heat capacity of the gas during this process is:
    1. Option A: Option 1
    2. Option B: Option 2
    3. Option C: Option 3
    4. Option D: Option 4

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  18. Question 18 (NEET 2016, Q18)

    ThermodynamicsMedium
    The temperature inside a refrigerator is t2∘Ct_2^\circ C and the room temperature is t1∘Ct_1^\circ C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:
    1. Option A: t1t1−t2\dfrac{t_1}{t_1-t_2}
    2. Option B: t1+273t1−t2\dfrac{t_1+273}{t_1-t_2}
    3. Option C: t2+273t1−t2\dfrac{t_2+273}{t_1-t_2}
    4. Option D: t1+t2t1+273\dfrac{t_1+t_2}{t_1+273}

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  19. Question 19 (NEET 2016, Q19)

    Kinetic TheoryEasy
    A given sample of an ideal gas occupies a volume VV at a pressure PP and absolute temperature TT. The mass of each molecule of the gas is mm. Which of the following gives the density of the gas?
    1. Option A: PkT\dfrac{P}{kT}
    2. Option B: PmkT\dfrac{Pm}{kT}
    3. Option C: PkTV\dfrac{P}{kTV}
    4. Option D: mkTmkT
    Show answer & explanation

    Correct answer: (B) PmkT\dfrac{Pm}{kT}

    Explanation

    For an ideal gas: PV=NkTPV = NkT where $Nisthenumberofmolecules.Density:is the number of molecules. Density:ρ=NmV\rho = \frac{Nm}{V}Fromidealgasequation:From ideal gas equation:NV=PkT\frac{N}{V} = \frac{P}{kT}Therefore:Therefore:ρ=mPkT\rho = m\frac{P}{kT}ρ=PmkT\rho = \frac{Pm}{kT}$ Hence option (B) is correct.

  20. Question 20 (NEET 2016, Q20)

    OscillationsMedium
    A body of mass mm is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass is slightly pulled down and released, it oscillates with a time period of 3 s3\ \text{s}. When the mass is increased by 1 kg1\ \text{kg}, the time period of oscillations becomes 5 s5\ \text{s}. The value of mm in kg is:
    1. Option A: 34\dfrac{3}{4}
    2. Option B: 43\dfrac{4}{3}
    3. Option C: 169\dfrac{16}{9}
    4. Option D: 916\dfrac{9}{16}

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  21. Question 21 (NEET 2016, Q21)

    WavesMedium
    The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe of length LL. The length of the open pipe will be:
    1. Option A: LL
    2. Option B: 2L2L
    3. Option C: L2\dfrac{L}{2}
    4. Option D: 4L4L
    Show answer & explanation

    Correct answer: (B) 2L2L

    Explanation

    For a closed organ pipe: First overtone frequency: fc=3v4Lf_c = \frac{3v}{4L} For an open organ pipe: Second overtone frequency: fo=3v2Lof_o = \frac{3v}{2L_o} Given: fo=fcf_o = f_c 3v2Lo=3v4L\frac{3v}{2L_o} = \frac{3v}{4L} Therefore: Lo=2LL_o = 2L Hence option (B) is correct.

  22. Question 22 (NEET 2016, Q22)

    WavesMedium
    Three sound waves of equal amplitudes have frequencies (n−1)(n-1), nn, and (n+1)(n+1). They superimpose to give beats. The number of beats produced per second will be:
    1. Option A: 1
    2. Option B: 4
    3. Option C: 3
    4. Option D: 2
    Show answer & explanation

    Correct answer: (D) 2

    Explanation

    The frequencies are: (n−1), n, (n+1)(n-1),\ n,\ (n+1) Beat frequency between $(n-1)andandn::1 beat/s1\ \text{beat/s}BeatfrequencybetweenBeat frequency betweennandand(n+1)::1 beat/s1\ \text{beat/s}Totalbeatsproduced:Total beats produced:1+1=2 beats/s1+1=2\ \text{beats/s}$ Hence option (D) is correct.

  23. Question 23 (NEET 2016, Q23)

    Electric Charges and FieldsMedium
    An electric dipole is placed at an angle of 30∘30^\circ with an electric field intensity 2×105 N/C2 \times 10^5\ \text{N/C}. It experiences a torque equal to 4 N m4\ \text{N m}. The charge on the dipole, if the dipole length is 2 cm2\ \text{cm}, is:
    1. Option A: 8 mC
    2. Option B: 2 mC
    3. Option C: 5 mC
    4. Option D: 7 μC

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  24. Question 24 (NEET 2016, Q24)

    Electrostatic Potential and CapacitanceHard
    A parallel-plate capacitor of area AA, plate separation dd and capacitance CC is filled with four dielectric materials having dielectric constants k1k_1, k2k_2, k3k_3 and k4k_4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance CC, then its dielectric constant kk is given by:
    1. Option A: k=k1+k2+k3+3k4k = k_1+k_2+k_3+3k_4
    2. Option B: k=23(k1+k2+k3)+2k4k = \dfrac{2}{3}(k_1+k_2+k_3)+2k_4
    3. Option C: 2k=3k1+k2+k3+1k4\dfrac{2}{k} = \dfrac{3}{k_1+k_2+k_3}+\dfrac{1}{k_4}
    4. Option D: 1k=1k1+1k2+1k3+32k4\dfrac{1}{k}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}+\dfrac{3}{2k_4}

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  25. Question 25 (NEET 2016, Q25)

    Current ElectricityEasy
    The potential difference (VA−VB)(V_A - V_B) between the points AA and BB in the given figure is:
    1. Option A: −3 V-3\ \text{V}
    2. Option B: +3 V+3\ \text{V}
    3. Option C: +6 V+6\ \text{V}
    4. Option D: +9 V+9\ \text{V}

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  26. Question 26 (NEET 2016, Q26)

    Current ElectricityEasy
    A filament lamp (500 W, 100 V)(500\ \text{W},\ 100\ \text{V}) is to be used in a 230 V230\ \text{V} main supply. When a resistance RR is connected in series, it works perfectly and consumes 500 W500\ \text{W}. The value of RR is:
    1. Option A: 230 Ω230\ \Omega
    2. Option B: 46 Ω46\ \Omega
    3. Option C: 26 Ω26\ \Omega
    4. Option D: 13 Ω13\ \Omega

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  27. Question 27 (NEET 2016, Q27)

    Moving Charges and MagnetismMedium
    A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is BB. If it is bent into a circular coil of nn turns, the magnetic field at the centre of this coil of nn turns will be:
    1. Option A: nBnB
    2. Option B: n2Bn^2B
    3. Option C: 2nB2nB
    4. Option D: 2n2B2n^2B

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  28. Question 28 (NEET 2016, Q28)

    Moving Charges and MagnetismMedium
    A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60∘60^\circ is WW. Now the torque required to keep the magnet in this new position is:
    1. Option A: W3\dfrac{W}{\sqrt{3}}
    2. Option B: 3W\sqrt{3}W
    3. Option C: 3W2\dfrac{\sqrt{3}W}{2}
    4. Option D: 2W3\dfrac{2W}{\sqrt{3}}

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  29. Question 29 (NEET 2016, Q29)

    Moving Charges and MagnetismMedium
    An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10⁻² T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is:
    1. Option A: 1 GHz
    2. Option B: 100 MHz
    3. Option C: 62.8 MHz
    4. Option D: 6.28 MHz

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  30. Question 30 (NEET 2016, Q30)

    Alternating CurrentMedium
    Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
    1. Option A: R=20 Ω, L=1.5 H, C=35 μFR = 20\ \Omega,\ L = 1.5\ \text{H},\ C = 35\ \mu\text{F}
    2. Option B: R=25 Ω, L=2.5 H, C=45 μFR = 25\ \Omega,\ L = 2.5\ \text{H},\ C = 45\ \mu\text{F}
    3. Option C: R=15 Ω, L=3.5 H, C=30 μFR = 15\ \Omega,\ L = 3.5\ \text{H},\ C = 30\ \mu\text{F}
    4. Option D: R=25 Ω, L=1.5 H, C=45 μFR = 25\ \Omega,\ L = 1.5\ \text{H},\ C = 45\ \mu\text{F}

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  31. Question 31 (NEET 2016, Q31)

    Electromagnetic InductionMedium
    A uniform magnetic field is restricted within a region of radius rr. The magnetic field changes with time at a rate dB⃗\vec{B}/dt. Loop 1 of radius R>rR > r encloses the region and loop 2 of radius RR is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is:
    1. Option A: Option 1
    2. Option B: Option 2
    3. Option C: Option 3
    4. Option D: Option 4
    Show answer & explanation

    Correct answer: (D) Option 4

    Explanation

    Magnetic flux linked with loop 1 is only through the magnetic field region of radius rr. Therefore, Φ=Bπr2\Phi = B\pi r^2 Using Faraday’s law: E=−dΦdt\mathcal{E} = -\frac{d\Phi}{dt} E1=−dBdtπr2\mathcal{E}_1 = -\frac{dB}{dt}\pi r^2 Loop 2 lies completely outside the magnetic field region, so magnetic flux through it is zero. Hence: E2=0\mathcal{E}_2 = 0 Therefore option (D) is correct.

  32. Question 32 (NEET 2016, Q32)

    Alternating CurrentEasy
    The potential differences across the resistance, capacitance and inductance are 80 V80\ \text{V}, 40 V40\ \text{V} and 100 V100\ \text{V} respectively in an L-C-R circuit. The power factor of this circuit is:
    1. Option A: 0.4
    2. Option B: 0.5
    3. Option C: 0.8
    4. Option D: 1.0

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  33. Question 33 (NEET 2016, Q33)

    Electromagnetic WavesMedium
    A 100 Ω100\ \Omega resistance and a capacitor of reactance 100 Ω100\ \Omega are connected in series across 220 V source. When the capacitor is 50% charged, peak value of the displacement current is:
    1. Option A: 2.2 A
    2. Option B: 11 A
    3. Option C: 4.4 A
    4. Option D: 11211\sqrt{2} A

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  34. Question 34 (NEET 2016, Q34)

    Ray Optics and Optical InstrumentsHard
    Two identical glass (μg=3/2)(\mu_g = 3/2) equiconvex lenses of focal length ff are kept in contact. The space between the two lenses is filled with water (μw=4/3)(\mu_w = 4/3). The focal length of the combination is:
    1. Option A: f3\dfrac{f}{3}
    2. Option B: ff
    3. Option C: 4f3\dfrac{4f}{3}
    4. Option D: 3f4\dfrac{3f}{4}

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  35. Question 35 (NEET 2016, Q35)

    Ray Optics and Optical InstrumentsEasy
    An air bubble in a glass slab with refractive index 1.51.5 (near normal incidence) is 5 cm5\ \text{cm} deep when viewed from one surface and 3 cm3\ \text{cm} deep when viewed from the opposite face. The thickness of the slab is:
    1. Option A: 8 cm
    2. Option B: 10 cm
    3. Option C: 12 cm
    4. Option D: 16 cm

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  36. Question 36 (NEET 2016, Q36)

    Wave OpticsHard
    The interference pattern is obtained with two coherent light sources of intensity ratio nn. In the interference pattern, the ratio Imax−IminImax+Imin\frac{I_{max}-I_{min}}{I_{max}+I_{min}} will be:
    1. Option A: nn+1\dfrac{\sqrt{n}}{n+1}
    2. Option B: 2nn+1\dfrac{2\sqrt{n}}{n+1}
    3. Option C: n(n+1)2\dfrac{\sqrt{n}}{(n+1)^2}
    4. Option D: 2n(n+1)2\dfrac{2\sqrt{n}}{(n+1)^2}

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  37. Question 37 (NEET 2016, Q37)

    Ray Optics and Optical InstrumentsEasy
    A person can see clearly objects only when they lie between 50 cm50\ \text{cm} and 400 cm400\ \text{cm} from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be:
    1. Option A: Convex, +2.25 diopter
    2. Option B: Concave, -0.25 diopter
    3. Option C: Concave, -0.2 diopter
    4. Option D: Convex, +0.15 diopter

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  38. Question 38 (NEET 2016, Q38)

    Wave OpticsMedium
    A linear aperture whose width is 0.02 cm0.02\ \text{cm} is placed immediately in front of a lens of focal length 60 cm60\ \text{cm}. The aperture is illuminated normally by a parallel beam of wavelength 5×10−5 cm5 \times 10^{-5}\ \text{cm}. The distance of the first dark band from the centre of the screen is:
    1. Option A: 0.10 cm
    2. Option B: 0.25 cm
    3. Option C: 0.20 cm
    4. Option D: 0.15 cm

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  39. Question 39 (NEET 2016, Q39)

    Dual Nature of Radiation and MatterHard
    Electrons of mass mm with de-Broglie wavelength λ\lambda fall on the target in an X-ray tube. The cutoff wavelength (λ0)(\lambda_0) of the emitted X-rays is:
    1. Option A: λ0=2mcλ2h\lambda_0 = \dfrac{2mc\lambda^2}{h}
    2. Option B: λ0=2hmc\lambda_0 = \dfrac{2h}{mc}
    3. Option C: λ0=2m2c2λ3h2\lambda_0 = \dfrac{2m^2c^2\lambda^3}{h^2}
    4. Option D: λ0=λ\lambda_0 = \lambda

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  40. Question 40 (NEET 2016, Q40)

    Dual Nature of Radiation and MatterEasy
    Photons with energy 5 eV5\ \text{eV} are incident on a cathode CC in a photoelectric cell. The maximum kinetic energy of emitted photoelectrons is 2 eV2\ \text{eV}. When photons of energy 6 eV6\ \text{eV} are incident on CC, no photoelectrons will reach anode AA, if the stopping potential of AA relative to CC is:
    1. Option A: +3 V
    2. Option B: +4 V
    3. Option C: -1 V
    4. Option D: -3 V

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  41. Question 41 (NEET 2016, Q41)

    AtomsHard
    If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ\lambda. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
    1. Option A: 1625λ\dfrac{16}{25}\lambda
    2. Option B: 916λ\dfrac{9}{16}\lambda
    3. Option C: 207λ\dfrac{20}{7}\lambda
    4. Option D: 2013λ\dfrac{20}{13}\lambda
    Show answer & explanation

    Correct answer: (C) 207λ\dfrac{20}{7}\lambda

    Explanation

    Using Rydberg formula: 1λ=R(1n12−1n22)\frac{1}{\lambda} = R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) For transition $3 \to 2::1λ=R(122−132)\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)FortransitionFor transition4 \to 3withwavelengthwith wavelength\lambda'::1λ′=R(132−142)\frac{1}{\lambda'} = R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)Takingratio:Taking ratio:λ′λ=(122−132)(132−142)\frac{\lambda'}{\lambda} = \frac{\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\left(\frac{1}{3^2}-\frac{1}{4^2}\right)}=5367144= \frac{\frac{5}{36}}{\frac{7}{144}}=207= \frac{20}{7}Thus:Thus:λ′=207λ\lambda' = \frac{20}{7}\lambda$ Hence option (C) is correct.

  42. Question 42 (NEET 2016, Q42)

    NucleiMedium
    The half-life of a radioactive substance is 3030 minutes. The time (in minutes) taken between 40%40\% decay and 85%85\% decay of the same radioactive substance is:
    1. Option A: 15
    2. Option B: 30
    3. Option C: 45
    4. Option D: 60

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  43. Question 43 (NEET 2016, Q43)

    Semiconductor ElectronicsMedium
    For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ2\ \text{k}\Omega is 4 V4\ \text{V}. If the current amplification factor of the transistor is 100100 and the base resistance is 1 kΩ1\ \text{k}\Omega, then the input signal voltage is:
    1. Option A: 10 mV
    2. Option B: 20 mV
    3. Option C: 30 mV
    4. Option D: 15 mV

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  44. Question 44 (NEET 2016, Q44)

    Semiconductor ElectronicsMedium
    The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R1R_1 will be:
    1. Option A: 2.5 A
    2. Option B: 10.0 A
    3. Option C: 1.43 A
    4. Option D: 3.13 A

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  45. Question 45 (NEET 2016, Q45)

    Semiconductor ElectronicsMedium
    What is the output YY in the following circuit, when all the three inputs AA, BB, CC are first 0 and then 1?
    1. Option A: 0, 1
    2. Option B: 0, 0
    3. Option C: 1, 0
    4. Option D: 1, 1

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