NEET 2017 · Physics

NEET 2017 Physics Questions with Solutions

The NEET 2017 paper had 45 Physics questions from 25 chapters.

Current Electricity had the most questions (4), followed by Gravitation, Ray Optics and Optical Instruments and System of Particles and Rotational Motion with 3 each.

14 questions below have the answer and explanation free; the other 31 are in Premium.

Physics questions
45
Chapters covered
25
Solved free here
14 of 45
Easy / Medium / Hard
6 / 25 / 14

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

Chapter-wise: NEET 2017 Physics

How many questions each chapter had in NEET 2017. Open a chapter for its questions from every year.

  1. Current Electricity4 Qs
  2. Gravitation3 Qs
  3. Ray Optics and Optical Instruments3 Qs
  4. System of Particles and Rotational Motion3 Qs
  5. Dual Nature of Radiation and Matter2 Qs
  6. Electromagnetic Waves2 Qs
  7. Electrostatic Potential and Capacitance2 Qs
  8. Laws of Motion2 Qs
  9. Motion in a Plane2 Qs
  10. Moving Charges and Magnetism2 Qs
  11. Oscillations2 Qs
  12. Semiconductor Electronics2 Qs
  13. Thermodynamics2 Qs
  14. Wave Optics2 Qs
  15. Waves2 Qs
  16. Atoms1 Q
  17. Electric Charges and Fields1 Q
  18. Electromagnetic Induction1 Q
  19. Kinetic Theory1 Q
  20. Mechanical Properties of Fluids1 Q
  21. Mechanical Properties of Solids1 Q
  22. Nuclei1 Q
  23. Thermal Properties of Matter1 Q
  24. Units and Measurements1 Q
  25. Work, Energy and Power1 Q

All 45 NEET 2017 Physics questions

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

  1. Question 1 (NEET 2017, Q91)

    OscillationsMedium
    A spring of force constant kk is cut into lengths of ratio 1:2:31:2:3. They are connected in series and the new force constant is k′k'. Then they are connected in parallel and force constant is k′′k''. Then k′:k′′k' : k'' is
    1. Option A: 1:141 : 14
    2. Option B: 1:61 : 6
    3. Option C: 1:91 : 9
    4. Option D: 1:111 : 11

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  2. Question 2 (NEET 2017, Q92)

    ThermodynamicsHard
    Thermodynamic processes are indicated in the following diagram.
    1. Option A: P → d, Q → b, R → a, S → c
    2. Option B: P → a, Q → c, R → d, S → b
    3. Option C: P → c, Q → a, R → d, S → b
    4. Option D: P → c, Q → d, R → b, S → a

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  3. Question 3 (NEET 2017, Q93)

    Electrostatic Potential and CapacitanceMedium
    A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system
    1. Option A: Increases by a factor of 22
    2. Option B: Increases by a factor of 44
    3. Option C: Decreases by a factor of 22
    4. Option D: Remains the same

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  4. Question 4 (NEET 2017, Q94)

    Mechanical Properties of FluidsHard
    A U-tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm10\ \text{mm} above the water level on the other side. Meanwhile the water rises by 65 mm65\ \text{mm} from its original level (see diagram). The density of the oil is
    1. Option A: 928 kg m−3928\ \text{kg m}^{-3}
    2. Option B: 650 kg m−3650\ \text{kg m}^{-3}
    3. Option C: 425 kg m−3425\ \text{kg m}^{-3}
    4. Option D: 800 kg m−3800\ \text{kg m}^{-3}
    Show answer & explanation

    Correct answer: (A) 928 kg m−3928\ \text{kg m}^{-3}

    Explanation

    Let the density of oil be ρo\rho_o. The oil column height above the oil-water interface is: 65 mm+65 mm+10 mm=140 mm65\ \text{mm} + 65\ \text{mm} + 10\ \text{mm} = 140\ \text{mm} The difference in water levels is: 65 mm+65 mm=130 mm65\ \text{mm} + 65\ \text{mm} = 130\ \text{mm} Equating pressures at the same horizontal level: ρog(140)=ρwg(130)\rho_o g (140) = \rho_w g (130) ρo=130140×1000\rho_o = \frac{130}{140} \times 1000 ρo≈928 kg m−3\rho_o \approx 928\ \text{kg m}^{-3}

  5. Question 5 (NEET 2017, Q95)

    Dual Nature of Radiation and MatterHard
    The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature TT Kelvin and mass mm, is
    1. Option A: 2hmkT\dfrac{2h}{\sqrt{mkT}}
    2. Option B: hmkT\dfrac{h}{\sqrt{mkT}}
    3. Option C: h3mkT\dfrac{h}{\sqrt{3mkT}}
    4. Option D: 2h3mkT\dfrac{2h}{\sqrt{3mkT}}

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  6. Question 6 (NEET 2017, Q96)

    GravitationMedium
    The acceleration due to gravity at a height 1 km1\ \text{km} above the earth is the same as at a depth dd below the surface of earth. Then
    1. Option A: d=2 kmd = 2\ \text{km}
    2. Option B: d=12 kmd = \dfrac{1}{2}\ \text{km}
    3. Option C: d=1 kmd = 1\ \text{km}
    4. Option D: d=32 kmd = \dfrac{3}{2}\ \text{km}

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  7. Question 7 (NEET 2017, Q97)

    Motion in a PlaneMedium
    The xx and yy coordinates of the particle at any time are x=5t−2t2x = 5t - 2t^2 and y=10ty = 10t respectively, where xx and yy are in meters and tt in seconds. The acceleration of the particle at t=2 st = 2\ \text{s} is
    1. Option A: −8 m/s2-8\ \text{m/s}^{2}
    2. Option B: 00
    3. Option C: 5 m/s25\ \text{m/s}^{2}
    4. Option D: −4 m/s2-4\ \text{m/s}^{2}
    Show answer & explanation

    Correct answer: (D) −4 m/s2-4\ \text{m/s}^{2}

    Explanation

    Given: x=5t−2t2,y=10tx = 5t - 2t^2, \qquad y = 10t Velocity components are: vx=dxdt=5−4tv_x = \frac{dx}{dt} = 5 - 4t vy=dydt=10v_y = \frac{dy}{dt} = 10 Acceleration components are: ax=dvxdt=−4a_x = \frac{dv_x}{dt} = -4 ay=dvydt=0a_y = \frac{dv_y}{dt} = 0 Hence, acceleration of the particle at $t = 2\ \text{s}is:is:−4 m s−2-4\ \text{m s}^{-2}$

  8. Question 8 (NEET 2017, Q98)

    Current ElectricityMedium
    In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V3\ \text{V}. The resistance of collector is 3 kΩ3\ \text{k}\Omega. If current gain is 100100 and the base resistance is 2 kΩ2\ \text{k}\Omega, the power gain of the amplifier is
    1. Option A: 2020 and 20002000
    2. Option B: 200200 and 10001000
    3. Option C: 1515 and 200200
    4. Option D: 150150 and 1500015000

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  9. Question 9 (NEET 2017, Q99)

    Moving Charges and MagnetismHard
    An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying same current II along the same direction is shown in figure. Magnitude of force per unit length on the middle wire BB is given by
    1. Option A: Option (1)
    2. Option B: Option (2)
    3. Option C: Option (3)
    4. Option D: Option (4)

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  10. Question 10 (NEET 2017, Q100)

    GravitationEasy
    Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
    1. Option A: Will become stationary
    2. Option B: Keep floating at the same distance between them
    3. Option C: Move towards each other
    4. Option D: Move away from each other

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  11. Question 11 (NEET 2017, Q101)

    ThermodynamicsHard
    A Carnot engine having an efficiency of 110\dfrac{1}{10} as heat engine, is used as a refrigerator. If the work done on the system is 10 J10\ \text{J}, the amount of energy absorbed from the reservoir at lower temperature is
    1. Option A: 100 J100\ \text{J}
    2. Option B: 1 J1\ \text{J}
    3. Option C: 90 J90\ \text{J}
    4. Option D: 99 J99\ \text{J}

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  12. Question 12 (NEET 2017, Q102)

    Moving Charges and MagnetismHard
    A 250250-turn rectangular coil of length 2.1 cm2.1\ \text{cm} and width 1.25 cm1.25\ \text{cm} carries a current of 85 μA85\ \mu\text{A} and is subjected to a magnetic field of 0.85 T0.85\ \text{T}. Work done for rotating the coil by 180∘180^{\circ} against the torque is
    1. Option A: 1.15 μJ1.15\ \mu\text{J}
    2. Option B: 9.1 μJ9.1\ \mu\text{J}
    3. Option C: 4.55 μJ4.55\ \mu\text{J}
    4. Option D: 2.3 μJ2.3\ \mu\text{J}

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  13. Question 13 (NEET 2017, Q103)

    Semiconductor ElectronicsMedium
    Which one of the following represents forward bias diode?
    1. Option A: Circuit (1)
    2. Option B: Circuit (2)
    3. Option C: Circuit (3)
    4. Option D: Circuit (4)

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  14. Question 14 (NEET 2017, Q104)

    AtomsMedium
    The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
    1. Option A: 0.50.5
    2. Option B: 22
    3. Option C: 11
    4. Option D: 44
    Show answer & explanation

    Correct answer: (D) 44

    Explanation

    For the last line of Balmer series: 1λB=R(122−1∞2)\frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) λB=4R\lambda_B = \frac{4}{R} For the last line of Lyman series: 1λL=R(112−1∞2)\frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) λL=1R\lambda_L = \frac{1}{R} Hence, λBλL=4\frac{\lambda_B}{\lambda_L} = 4

  15. Question 15 (NEET 2017, Q105)

    Current ElectricityHard
    Figure shows a circuit contains three identical resistors with resistance R=9.0 ΩR = 9.0\ \Omega each, two identical inductors with inductance L=2.0 mHL = 2.0\ \text{mH} each, and an ideal battery with ε=18 V\varepsilon = 18\ \text{V}. The current ii through the battery just after the switch is closed is
    1. Option A: 0 A0\ \text{A}
    2. Option B: 2 mA2\ \text{mA}
    3. Option C: 0.2 A0.2\ \text{A}
    4. Option D: 2 A2\ \text{A}

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  16. Question 16 (NEET 2017, Q106)

    Laws of MotionHard
    Two blocks AA and BB of masses 3m3m and mm respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of AA and BB immediately after the string is cut, are respectively
    1. Option A: g3, g3\dfrac{g}{3},\ \dfrac{g}{3}
    2. Option B: g, g3g,\ \dfrac{g}{3}
    3. Option C: g3, g\dfrac{g}{3},\ g
    4. Option D: g, gg,\ g

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  17. Question 17 (NEET 2017, Q107)

    Electromagnetic InductionMedium
    A long solenoid of diameter 0.1 m0.1\,\text{m} has 2×1042 \times 10^4 turns per meter. At the centre of the solenoid, a coil of 100100 turns and radius 0.01 m0.01\,\text{m} is placed with its axis coinciding with the solenoid axis. The current in the solenoid decreases uniformly from 4 A4\,\text{A} to 0 A0\,\text{A} in 0.05 s0.05\,\text{s}. If the resistance of the coil is 10π2 Ω10\pi^2\,\Omega, then the total charge flowing through the coil during this time is:
    1. Option A: 16π μC16\pi\,\mu\text{C}
    2. Option B: 32π μC32\pi\,\mu\text{C}
    3. Option C: 16 μC16\,\mu\text{C}
    4. Option D: 32 μC32\,\mu\text{C}
    Show answer & explanation

    Correct answer: (D) 32 μC32\,\mu\text{C}

    Explanation

    Magnetic field inside the solenoid is: B=μ0nIB = \mu_0 n I Magnetic flux linked with the coil: Φ=NBA=N(μ0nI)(πr2)\Phi = NBA = N(\mu_0 n I)(\pi r^2) Total induced charge: ΔQ=NΔΦR\Delta Q = \frac{N\Delta \Phi}{R} Substituting the values: ΔQ=100×4π×10−7×2×104×4×π(0.01)210π2\Delta Q = \frac{100 \times 4\pi \times 10^{-7} \times 2 \times 10^4 \times 4 \times \pi (0.01)^2}{10\pi^2} ΔQ=32 μC\Delta Q = 32\,\mu\text{C}

  18. Question 18 (NEET 2017, Q108)

    Units and MeasurementsHard
    A physical quantity having dimensions of length can be formed using the quantities cc, GG and e24πε0\dfrac{e^2}{4\pi\varepsilon_0}. Which of the following combinations represents 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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  19. Question 19 (NEET 2017, Q109)

    Electromagnetic WavesEasy
    In an electromagnetic wave in free space, the root mean square value of electric field is Erms=6 V/mE_{\text{rms}} = 6\,\text{V/m}. The peak value of magnetic field is:
    1. Option A: 4.23×10−8 T4.23 \times 10^{-8}\,\text{T}
    2. Option B: 1.41×10−8 T1.41 \times 10^{-8}\,\text{T}
    3. Option C: 2.83×10−8 T2.83 \times 10^{-8}\,\text{T}
    4. Option D: 0.70×10−8 T0.70 \times 10^{-8}\,\text{T}

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  20. Question 20 (NEET 2017, Q110)

    Current ElectricityEasy
    The resistance of a wire is R ΩR\,\Omega. If it is melted and stretched to nn times its original length, then its new resistance will be:
    1. Option A: Rn2\dfrac{R}{n^2}
    2. Option B: nRnR
    3. Option C: Rn\dfrac{R}{n}
    4. Option D: n2Rn^2R

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  21. Question 21 (NEET 2017, Q111)

    Ray Optics and Optical InstrumentsEasy
    The ratio of resolving powers of an optical microscope for two wavelengths λ1=4000 A˚\lambda_1 = 4000\,\text{\AA} and λ2=6000 A˚\lambda_2 = 6000\,\text{\AA} is:
    1. Option A: 16:8116:81
    2. Option B: 8:278:27
    3. Option C: 9:49:4
    4. Option D: 3:23:2

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  22. Question 22 (NEET 2017, Q112)

    Ray Optics and Optical InstrumentsMedium
    A thin prism having refracting angle 10∘10^\circ is made of glass of refractive index 1.421.42. This prism is combined with another thin prism of glass of refractive index 1.71.7. The combination produces dispersion without deviation. The refracting angle of the second prism should be:
    1. Option A: 10∘10^\circ
    2. Option B: 4∘4^\circ
    3. Option C: 6∘6^\circ
    4. Option D: 8∘8^\circ

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  23. Question 23 (NEET 2017, Q113)

    Wave OpticsMedium
    Two Polaroids P1P_1 and P2P_2 are placed with their transmission axes perpendicular to each other. Unpolarised light of intensity I0I_0 is incident on P1P_1. A third Polaroid P3P_3 is placed between P1P_1 and P2P_2 such that its axis makes an angle 45∘45^\circ with that of P1P_1. The intensity of transmitted light through P2P_2 is:
    1. Option A: I016\dfrac{I_0}{16}
    2. Option B: I02\dfrac{I_0}{2}
    3. Option C: I04\dfrac{I_0}{4}
    4. Option D: I08\dfrac{I_0}{8}
    Show answer & explanation

    Correct answer: (D) I08\dfrac{I_0}{8}

    Explanation

    After passing through the first Polaroid: I1=I02I_1 = \frac{I_0}{2} Using Malus' law through $P_3::I2=I1cos⁡245∘I_2 = I_1 \cos^2 45^\circI2=I02×12=I04I_2 = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}AgainthroughAgain throughP_2::I3=I2cos⁡245∘I_3 = I_2 \cos^2 45^\circI3=I04×12=I08I_3 = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$

  24. Question 24 (NEET 2017, Q114)

    Current ElectricityEasy
    A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
    1. Option A: A combination of cells, galvanometer and resistances
    2. Option B: Cells
    3. Option C: Potential gradients
    4. Option D: A condition of no current flow through the galvanometer

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  25. Question 25 (NEET 2017, Q115)

    WavesMedium
    The two nearest harmonics of a tube closed at one end and open at the other end are 220 Hz220\,\text{Hz} and 260 Hz260\,\text{Hz}. What is the fundamental frequency of the system?
    1. Option A: 40 Hz40\,\text{Hz}
    2. Option B: 10 Hz10\,\text{Hz}
    3. Option C: 20 Hz20\,\text{Hz}
    4. Option D: 30 Hz30\,\text{Hz}
    Show answer & explanation

    Correct answer: (C) 20 Hz20\,\text{Hz}

    Explanation

    For a pipe closed at one end, only odd harmonics are present. Thus, nv4l=220\frac{nv}{4l} = 220 and (n+2)v4l=260\frac{(n+2)v}{4l} = 260 Subtracting the equations: 2v4l=40\frac{2v}{4l} = 40 v4l=20 Hz\frac{v}{4l} = 20\,\text{Hz} Hence, the fundamental frequency is: 20 Hz20\,\text{Hz}

  26. Question 26 (NEET 2017, Q116)

    Ray Optics and Optical InstrumentsMedium
    A beam of light from a source LL is incident normally on a plane mirror fixed at a distance xx from the source. The beam is reflected back as a spot on a scale placed just above the source LL. When the mirror is rotated through a small angle θ\theta, the spot of light is found to move through a distance yy on the scale. The angle θ\theta is given by:
    1. Option A: xy\dfrac{x}{y}
    2. Option B: y2x\dfrac{y}{2x}
    3. Option C: yx\dfrac{y}{x}
    4. Option D: x2y\dfrac{x}{2y}

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  27. Question 27 (NEET 2017, Q117)

    Semiconductor ElectronicsMedium
    The given electrical network is equivalent to:
    1. Option A: NOT gate
    2. Option B: AND gate
    3. Option C: OR gate
    4. Option D: NOR gate

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  28. Question 28 (NEET 2017, Q118)

    OscillationsMedium
    A particle executes linear simple harmonic motion with amplitude 3 cm3\,\text{cm}. When the particle is at a distance 2 cm2\,\text{cm} from the mean position, the magnitude of its velocity is equal to that of its acceleration. The time period of the motion is:
    1. Option A: 2π3 s\dfrac{2\pi}{\sqrt{3}}\,\text{s}
    2. Option B: 5π s\dfrac{\sqrt{5}}{\pi}\,\text{s}
    3. Option C: 52π s\dfrac{\sqrt{5}}{2\pi}\,\text{s}
    4. Option D: 4π5 s\dfrac{4\pi}{\sqrt{5}}\,\text{s}

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  29. Question 29 (NEET 2017, Q119)

    Motion in a PlaneMedium
    Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1t_1. On another day, if she remains stationary on the moving escalator, then the escalator takes her up in time t2t_2. The time taken by her to walk up on the moving escalator will be:
    1. Option A: (t1−t2)(t_1-t_2)
    2. Option B: t1+t22\dfrac{t_1+t_2}{2}
    3. Option C: t1t2t2−t1\dfrac{t_1t_2}{t_2-t_1}
    4. Option D: t1t2t2+t1\dfrac{t_1t_2}{t_2+t_1}
    Show answer & explanation

    Correct answer: (D) t1t2t2+t1\dfrac{t_1t_2}{t_2+t_1}

    Explanation

    Let the length of escalator be dd. Velocity of Preeti relative to escalator: vp=dt1v_p = \frac{d}{t_1} Velocity of escalator: ve=dt2v_e = \frac{d}{t_2} Net velocity while walking on moving escalator: v=vp+vev = v_p + v_e If time taken is $t,,dt=dt1+dt2\frac{d}{t}=\frac{d}{t_1}+\frac{d}{t_2}1t=1t1+1t2\frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}t=t1t2t1+t2t=\frac{t_1t_2}{t_1+t_2}$

  30. Question 30 (NEET 2017, Q120)

    System of Particles and Rotational MotionHard
    Two discs of the same moment of inertia rotating about their regular axis passing through the centre and perpendicular to the plane of the disc with angular velocities ω1\omega_1 and ω2\omega_2 are brought into contact face to face with their axes coinciding. The expression for loss of energy 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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  31. Question 31 (NEET 2017, Q121)

    Kinetic TheoryMedium
    A gas mixture consists of 22 moles of O2O_2 and 44 moles of Ar at temperature TT. Neglecting all vibrational modes, the total internal energy of the system is:
    1. Option A: 11RT11RT
    2. Option B: 4RT4RT
    3. Option C: 15RT15RT
    4. Option D: 9RT9RT
    Show answer & explanation

    Correct answer: (A) 11RT11RT

    Explanation

    For diatomic gas O2O_2 without vibrational modes: UO2=n52RTU_{O_2}=n\frac{5}{2}RT UO2=2×52RT=5RTU_{O_2}=2\times\frac{5}{2}RT=5RT For monoatomic gas Ar: UAr=n32RTU_{Ar}=n\frac{3}{2}RT UAr=4×32RT=6RTU_{Ar}=4\times\frac{3}{2}RT=6RT Total internal energy: U=5RT+6RT=11RTU=5RT+6RT=11RT

  32. Question 32 (NEET 2017, Q122)

    Mechanical Properties of SolidsMedium
    The bulk modulus of a spherical object is BB. If it is subjected to uniform pressure pp, then the fractional decrease in radius is:
    1. Option A: p3B\dfrac{p}{3B}
    2. Option B: pB\dfrac{p}{B}
    3. Option C: B3p\dfrac{B}{3p}
    4. Option D: 3pB\dfrac{3p}{B}
    Show answer & explanation

    Correct answer: (A) p3B\dfrac{p}{3B}

    Explanation

    Bulk modulus is defined as: B=pΔV/VB=\frac{p}{\Delta V/V} For a sphere: V=43πr3V=\frac{4}{3}\pi r^3 Hence, ΔVV=3Δrr\frac{\Delta V}{V}=3\frac{\Delta r}{r} Therefore, B=p3Δr/rB=\frac{p}{3\Delta r/r} So the fractional decrease in radius is: Δrr=p3B\frac{\Delta r}{r}=\frac{p}{3B}

  33. Question 33 (NEET 2017, Q123)

    Laws of MotionEasy
    One end of a string of length ll is connected to a particle of mass mm and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed vv, the net force on the particle directed towards the centre will be (TT represents the tension in the string):
    1. Option A: Zero
    2. Option B: TT
    3. Option C: T+mv2lT + \dfrac{mv^2}{l}
    4. Option D: T−mv2lT - \dfrac{mv^2}{l}

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  34. Question 34 (NEET 2017, Q124)

    System of Particles and Rotational MotionMedium
    A rope is wound around a hollow cylinder of mass 3 kg3\,\text{kg} and radius 40 cm40\,\text{cm}. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N30\,\text{N}?
    1. Option A: 5 m/s25\,\text{m/s}^2
    2. Option B: 25 m/s225\,\text{m/s}^2
    3. Option C: 0.25 rad/s20.25\,\text{rad/s}^2
    4. Option D: 25 rad/s225\,\text{rad/s}^2

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  35. Question 35 (NEET 2017, Q125)

    Wave OpticsHard
    Young's double slit experiment is first performed in air and then repeated in a medium other than air. It is found that the eighth bright fringe in the medium lies where the fifth dark fringe lies in air. The refractive index of the medium is nearly:
    1. Option A: 1.781.78
    2. Option B: 1.251.25
    3. Option C: 1.591.59
    4. Option D: 1.691.69

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  36. Question 36 (NEET 2017, Q126)

    Electric Charges and FieldsHard
    Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe)(e + \Delta e). If the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance dd apart are zero, then the value of Δe\Delta e is of the order of: (Given: mass of hydrogen atom mh=1.67×10−27 kgm_h = 1.67 \times 10^{-27}\,\text{kg})
    1. Option A: 10−47 C10^{-47}\,\text{C}
    2. Option B: 10−20 C10^{-20}\,\text{C}
    3. Option C: 10−23 C10^{-23}\,\text{C}
    4. Option D: 10−37 C10^{-37}\,\text{C}
    Show answer & explanation

    Correct answer: (D) 10−37 C10^{-37}\,\text{C}

    Explanation

    For equilibrium: Fe=FgF_e = F_g 14πε0(Δe)2d2=GmH2d2\frac{1}{4\pi\varepsilon_0}\frac{(\Delta e)^2}{d^2} = \frac{Gm_H^2}{d^2} 9×109(Δe)2=6.67×10−11(1.67×10−27)29 \times 10^9 (\Delta e)^2 = 6.67 \times 10^{-11}(1.67 \times 10^{-27})^2 Solving: Δe≈10−37 C\Delta e \approx 10^{-37}\,\text{C}

  37. Question 37 (NEET 2017, Q127)

    Thermal Properties of MatterMedium
    Two rods AA and BB of different materials are welded together as shown in the figure. Their thermal conductivities are K1K_1 and K2K_2. The thermal conductivity of the composite rod will be:
    1. Option A: 2(K1+K2)2(K_1+K_2)
    2. Option B: K1+K22\dfrac{K_1+K_2}{2}
    3. Option C: 3(K1+K2)2\dfrac{3(K_1+K_2)}{2}
    4. Option D: K1+K2K_1+K_2
    Show answer & explanation

    Correct answer: (B) K1+K22\dfrac{K_1+K_2}{2}

    Explanation

    The rods are connected in parallel for heat flow. Total heat current: H=K1A(T1−T2)d+K2A(T1−T2)dH = \frac{K_1A(T_1-T_2)}{d} + \frac{K_2A(T_1-T_2)}{d} If equivalent conductivity is $K_{eq}fortotalareafor total area2A::H=Keq(2A)(T1−T2)dH = \frac{K_{eq}(2A)(T_1-T_2)}{d}Comparing:Comparing:Keq=K1+K22K_{eq} = \frac{K_1+K_2}{2}$

  38. Question 38 (NEET 2017, Q128)

    Dual Nature of Radiation and MatterMedium
    The photoelectric threshold wavelength of silver is 3250×10−10 m3250 \times 10^{-10}\,\text{m}. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10−10 m2536 \times 10^{-10}\,\text{m} is: (Given: h=4.14×10−15 eV sh = 4.14 \times 10^{-15}\,\text{eV s} and c=3×108 m/sc = 3 \times 10^8\,\text{m/s})
    1. Option A: ≈ 0.3 × 10⁶ ms⁻¹
    2. Option B: ≈ 6 × 10⁵ ms⁻¹
    3. Option C: ≈ 0.6 × 10⁶ ms⁻¹
    4. Option D: ≈ 61 × 10³ ms⁻¹

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  39. Question 39 (NEET 2017, Q129)

    WavesMedium
    Two cars moving in opposite directions approach each other with speeds of 22 m/s22\,\text{m/s} and 16.5 m/s16.5\,\text{m/s} respectively. The driver of the first car blows a horn having frequency 400 Hz400\,\text{Hz}. The frequency heard by the driver of the second car is: (Velocity of sound =340 m/s= 340\,\text{m/s})
    1. Option A: 448 Hz448\,\text{Hz}
    2. Option B: 350 Hz350\,\text{Hz}
    3. Option C: 361 Hz361\,\text{Hz}
    4. Option D: 411 Hz411\,\text{Hz}
    Show answer & explanation

    Correct answer: (A) 448 Hz448\,\text{Hz}

    Explanation

    Using Doppler effect: f′=f(v+vov−vs)f' = f\left(\frac{v+v_o}{v-v_s}\right) f′=400(340+16.5340−22)f' = 400\left(\frac{340+16.5}{340-22}\right) f′≈448 Hzf' \approx 448\,\text{Hz}

  40. Question 40 (NEET 2017, Q130)

    Work, Energy and PowerMedium
    Consider a drop of rain water having mass 1 g1\,\text{g} falling from a height of 1 km1\,\text{km}. It hits the ground with a speed of 50 m/s50\,\text{m/s}. Take gg constant with a value 10 m/s210\,\text{m/s}^2. The work done by (i) gravitational force and (ii) resistive force of air is:
    1. Option A: (i) 10 J10\,\text{J} (ii) −8.75 J-8.75\,\text{J}
    2. Option B: (i) −10 J-10\,\text{J} (ii) −8.25 J-8.25\,\text{J}
    3. Option C: (i) 1.25 J1.25\,\text{J} (ii) −8.25 J-8.25\,\text{J}
    4. Option D: (i) 100 J100\,\text{J} (ii) 8.75 J8.75\,\text{J}
    Show answer & explanation

    Correct answer: (A) (i) 10 J10\,\text{J} (ii) −8.75 J-8.75\,\text{J}

    Explanation

    Mass: m=1 g=10−3 kgm = 1\,\text{g} = 10^{-3}\,\text{kg} Height: h=1000 mh = 1000\,\text{m} Work done by gravity: Wg=mghW_g = mgh Wg=10−3×10×1000=10 JW_g = 10^{-3} \times 10 \times 1000 = 10\,\text{J} Final kinetic energy: Kf=12mv2K_f = \frac{1}{2}mv^2 Kf=12(10−3)(50)2=1.25 JK_f = \frac{1}{2}(10^{-3})(50)^2 = 1.25\,\text{J} Using work-energy theorem: Wg+Wair=KfW_g + W_{air} = K_f 10+Wair=1.2510 + W_{air} = 1.25 Wair=−8.75 JW_{air} = -8.75\,\text{J}

  41. Question 41 (NEET 2017, Q131)

    Electromagnetic WavesMedium
    A spherical black body with a radius of 12 cm12\,\text{cm} radiates 450 W450\,\text{W} power at 500 K500\,\text{K}. If the radius were halved and the temperature doubled, the power radiated in watt would be:
    1. Option A: 18001800
    2. Option B: 225225
    3. Option C: 450450
    4. Option D: 10001000

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  42. Question 42 (NEET 2017, Q132)

    Electrostatic Potential and CapacitanceMedium
    The diagrams below show regions of equipotentials. A positive charge is moved from AA to BB in each diagram. Identify the correct statement regarding the work required.
    1. Option A: Maximum work is required to move qq in figure (b)
    2. Option B: Maximum work is required to move qq in figure (c)
    3. Option C: In all the four cases the work done is the same
    4. Option D: Minimum work is required to move qq in figure (a)

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  43. Question 43 (NEET 2017, Q133)

    System of Particles and Rotational MotionMedium
    Which of the following statements are correct? (a) Centre of mass of a body always coincides with the centre of gravity of the body. (b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero. (c) A couple on a body produces both translational and rotational motion in a body. (d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
    1. Option A: (c) and (d)
    2. Option B: (b) and (d)
    3. Option C: (a) and (b)
    4. Option D: (b) and (c)

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  44. Question 44 (NEET 2017, Q134)

    GravitationHard
    If θ1\theta_1 and θ2\theta_2 are the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ\theta is given by:
    1. Option A: tan⁡2θ=tan⁡2θ1−tan⁡2θ2\tan^2\theta = \tan^2\theta_1 - \tan^2\theta_2
    2. Option B: cot⁡2θ=cot⁡2θ1+cot⁡2θ2\cot^2\theta = \cot^2\theta_1 + \cot^2\theta_2
    3. Option C: tan⁡2θ=tan⁡2θ1+tan⁡2θ2\tan^2\theta = \tan^2\theta_1 + \tan^2\theta_2
    4. Option D: cot⁡2θ=cot⁡2θ1−cot⁡2θ2\cot^2\theta = \cot^2\theta_1 - \cot^2\theta_2

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  45. Question 45 (NEET 2017, Q135)

    NucleiHard
    Radioactive material AA has decay constant 8λ8\lambda and material BB has decay constant λ\lambda. Initially they have the same number of nuclei. After what time will the ratio of number of nuclei of material BB to that of material AA be 1e\dfrac{1}{e}?
    1. Option A: 19λ\dfrac{1}{9\lambda}
    2. Option B: 1λ\dfrac{1}{\lambda}
    3. Option C: 17λ\dfrac{1}{7\lambda}
    4. Option D: 18λ\dfrac{1}{8\lambda}
    Show answer & explanation

    Correct answer: (C) 17λ\dfrac{1}{7\lambda}

    Explanation

    Radioactive decay law: N=N0e−λtN = N_0e^{-\lambda t} For materials $AandandB::NA=N0e−8λtN_A = N_0e^{-8\lambda t}NB=N0e−λtN_B = N_0e^{-\lambda t}Given:Given:NBNA=1e\frac{N_B}{N_A} = \frac{1}{e}e7λt=e−1e^{7\lambda t} = e^{-1}Takinglogarithm:Taking logarithm:7λt=17\lambda t = 1t=17λt = \frac{1}{7\lambda}$

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