JEE Main Practice — Physics Set 6

JEE Main · 2026 · with answers and explanations

questions
25
Total marks
100
Duration
60 minutes
Negative marking
−1

Subjects: Physics

Question 1

Physics · 4 marks · −1

A Zener diode of breakdown voltage 8 V and maximum power rating 0.4 W is used with a series resistor RR across a 20 V supply to regulate the voltage across a 400 Ω400\,\Omega load. The minimum RR that keeps the Zener safe even if the load is disconnected, and the Zener current with that RR when the load is connected, are

  1. A.240 Ω240\,\Omega and 50 mA
  2. B.160 Ω160\,\Omega and 30 mA
  3. C.240 Ω240\,\Omega and 30 mA
  4. D.300 Ω300\,\Omega and 20 mA

Correct answer: C

Explanation: IZ,max=0.48=50I_{Z,max}=\frac{0.4}{8}=50 mA. With no load, all series current flows through the Zener: 20−8R≤0.05⇒Rmin=240 Ω\frac{20-8}{R}\le0.05\Rightarrow R_{min}=240\,\Omega. With the load: series current =12240=50=\frac{12}{240}=50 mA, load current =8400=20=\frac{8}{400}=20 mA, Zener current =30=30 mA.

Question 2

Physics · 4 marks · −1

Silicon with intrinsic carrier concentration 1.5×1016 m−31.5\times10^{16}\ \mathrm{m^{-3}} is doped with 4.5×10224.5\times10^{22} donor atoms per m3^3 (all ionised). The hole concentration in the doped sample is x×109 m−3x\times10^{9}\ \mathrm{m^{-3}}. The value of xx is ______.

Correct answer: 5

Explanation: n≈ND=4.5×1022n\approx N_D=4.5\times10^{22} (since ND≫niN_D\gg n_i). Mass-action law: p=ni2n=2.25×10324.5×1022=5×109 m−3p=\frac{n_i^2}{n}=\frac{2.25\times10^{32}}{4.5\times10^{22}}=5\times10^{9}\ \mathrm{m^{-3}}.

Question 3

Physics · 4 marks · −1

An LED is made from a semiconductor of band gap 2.48 eV. The wavelength of the light it emits (in nm) is ______. (Take hc=1240hc=1240 eV nm.)

Correct answer: 500

Explanation: λ=hcEg=12402.48=500\lambda=\frac{hc}{E_g}=\frac{1240}{2.48}=500 nm.

Question 4

Physics · 4 marks · −1

500 J of heat is supplied to a gas, and the gas does 200 J of work on its surroundings. The change in its internal energy is

  1. A.700 J
  2. B.300 J
  3. C.200 J
  4. D.−300-300 J

Correct answer: B

Explanation: First law: ΔU=Q−W=500−200=300\Delta U=Q-W=500-200=300 J.

Question 5

Physics · 4 marks · −1

The efficiency of a Carnot engine operating between reservoirs at 527∘527^\circC and 127∘127^\circC is

  1. A.76%
  2. B.24%
  3. C.50%
  4. D.33%

Correct answer: C

Explanation: T1=800T_1=800 K, T2=400T_2=400 K. η=1−T2T1=1−400800=50%\eta=1-\frac{T_2}{T_1}=1-\frac{400}{800}=50\%. (Using Celsius temperatures would wrongly give 76%.)

Question 6

Physics · 4 marks · −1

Two moles of an ideal gas expand isothermally and reversibly at 300 K from 10 L to 20 L. The work done by the gas is approximately (R=8.314R=8.314 J mol−1^{-1}K−1^{-1}, ln⁡2=0.693\ln2=0.693)

  1. A.1.73 kJ
  2. B.4.99 kJ
  3. C.3.46 kJ
  4. D.6.92 kJ

Correct answer: C

Explanation: W=nRTln⁡V2V1=2×8.314×300×0.693≈3458W=nRT\ln\frac{V_2}{V_1}=2\times8.314\times300\times0.693\approx3458 J ≈3.46\approx3.46 kJ.

Question 7

Physics · 4 marks · −1

A monatomic ideal gas is compressed adiabatically and reversibly to 18\frac18 of its initial volume. The ratio of final to initial pressure is

  1. A.8
  2. B.4
  3. C.16
  4. D.32

Correct answer: D

Explanation: PVγ=PV^\gamma= const with γ=53\gamma=\frac53: P2P1=85/3=(23)5/3=25=32\frac{P_2}{P_1}=8^{5/3}=(2^3)^{5/3}=2^5=32. (Temperature rises by 82/3=48^{2/3}=4.)

Question 8

Physics · 4 marks · −1

A Carnot refrigerator works between −3∘-3^\circC and 27∘27^\circC. The heat extracted from the cold reservoir for every 1 kJ of work supplied is

  1. A.9 kJ
  2. B.10 kJ
  3. C.0.1 kJ
  4. D.11 kJ

Correct answer: A

Explanation: COP =T2T1−T2=270300−270=9=\frac{T_2}{T_1-T_2}=\frac{270}{300-270}=9. Heat extracted =9×1=9=9\times1=9 kJ. (10 kJ is the heat rejected to the hot reservoir.)

Question 9

Physics · 4 marks · −1

One mole of a monatomic ideal gas is taken round the cycle A→\toB→\toC→\toD→\toA, where A =(P0,V0)=(P_0,V_0), B =(2P0,V0)=(2P_0,V_0), C =(2P0,2V0)=(2P_0,2V_0) and D =(P0,2V0)=(P_0,2V_0), each step being a straight line on the PP–VV diagram (two isochoric and two isobaric steps). The efficiency of this cycle is

  1. A.213\frac{2}{13}
  2. B.15\frac15
  3. C.211\frac{2}{11}
  4. D.14\frac14

Correct answer: A

Explanation: Net work = area of rectangle =P0V0=P_0V_0. Heat is absorbed only on A→\toB (isochoric, TT rises) and B→\toC (isobaric, TT rises): QAB=32(2P0V0−P0V0)=1.5P0V0Q_{AB}=\frac32(2P_0V_0-P_0V_0)=1.5P_0V_0; QBC=52(4P0V0−2P0V0)=5P0V0Q_{BC}=\frac52(4P_0V_0-2P_0V_0)=5P_0V_0. η=P0V06.5P0V0=213\eta=\frac{P_0V_0}{6.5P_0V_0}=\frac{2}{13}.

Question 10

Physics · 4 marks · −1

Two moles of a rigid diatomic ideal gas are heated at constant pressure through 50 K. Taking R=8.3R=8.3 J mol−1^{-1}K−1^{-1}, the heat supplied (in joule) is ______.

Correct answer: 2905

Explanation: Cp=72RC_p=\frac72R. Q=nCpΔT=2×72×8.3×50=2905Q=nC_p\Delta T=2\times\frac72\times8.3\times50=2905 J.

Question 11

Physics · 4 marks · −1

A Carnot engine has efficiency 40% with its sink at 300 K. Keeping the sink temperature fixed, the source temperature must be raised by ______ K to increase the efficiency to 50%.

Correct answer: 100

Explanation: 1−300T1=0.4⇒T1=5001-\frac{300}{T_1}=0.4\Rightarrow T_1=500 K. 1−300T1′=0.5⇒T1′=6001-\frac{300}{T_1'}=0.5\Rightarrow T_1'=600 K. Increase =100=100 K.

Question 12

Physics · 4 marks · −1

The dimensional formula of Planck's constant is the same as that of

  1. A.energy
  2. B.linear momentum
  3. C.angular momentum
  4. D.power

Correct answer: C

Explanation: E=hν⇒[h]=[ML2T−2][T−1]=[ML2T−1]E=h\nu\Rightarrow[h]=\frac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}], the same as angular momentum [mvr]=[ML2T−1][mvr]=[ML^2T^{-1}].

Question 13

Physics · 4 marks · −1

The number of significant figures in 0.004050 is

  1. A.3
  2. B.4
  3. C.6
  4. D.7

Correct answer: B

Explanation: Leading zeros are not significant; the zero between 4 and 5 and the trailing zero after the decimal point are significant: 4, 0, 5, 0 ⇒\Rightarrow 4 significant figures.

Question 14

Physics · 4 marks · −1

In a simple pendulum experiment, gg is determined from g=4π2LT2g=\frac{4\pi^2L}{T^2}. If the length is measured with a maximum error of 1% and the time period with a maximum error of 2%, the maximum percentage error in gg is

  1. A.3%
  2. B.5%
  3. C.4%
  4. D.1%

Correct answer: B

Explanation: Δgg=ΔLL+2ΔTT=1%+4%=5%\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}=1\%+4\%=5\%.

Question 15

Physics · 4 marks · −1

In the van der Waals equation (P+aV2)(V−b)=RT\left(P+\frac{a}{V^2}\right)(V-b)=RT (with VV the molar volume), the dimensions of ab\frac ab are

  1. A.[ML2T−2][ML^2T^{-2}]
  2. B.[ML−1T−2][ML^{-1}T^{-2}]
  3. C.[L3][L^3]
  4. D.[ML5T−2][ML^5T^{-2}]

Correct answer: A

Explanation: [a]=[PV2][a]=[PV^2] and [b]=[V][b]=[V], so [ab]=[PV]=[ML−1T−2][L3]=[ML2T−2]\left[\frac ab\right]=[PV]=[ML^{-1}T^{-2}][L^3]=[ML^2T^{-2}] (energy).

Question 16

Physics · 4 marks · −1

A screw gauge has pitch 0.5 mm and 50 divisions on its circular scale. With the jaws closed, the zero of the circular scale lies 3 divisions below the reference line. While measuring a wire, the main scale reads 2.5 mm and the 20th circular division coincides with the reference line. The corrected diameter of the wire is

  1. A.2.73 mm
  2. B.2.70 mm
  3. C.2.53 mm
  4. D.2.67 mm

Correct answer: D

Explanation: Least count =0.550=0.01=\frac{0.5}{50}=0.01 mm. Zero of circular scale below the reference line ⇒\Rightarrow positive zero error =+3×0.01=+0.03=+3\times0.01=+0.03 mm. Observed reading =2.5+20×0.01=2.70=2.5+20\times0.01=2.70 mm. Corrected =2.70−0.03=2.67=2.70-0.03=2.67 mm.

Question 17

Physics · 4 marks · −1

A system of units takes Planck's constant hh, the gravitational constant GG and the speed of light cc as fundamental quantities. The unit of time in this system is proportional to

  1. A.hcG\sqrt{\frac{hc}{G}}
  2. B.hGc5\sqrt{\frac{hG}{c^5}}
  3. C.hGc3\sqrt{\frac{hG}{c^3}}
  4. D.hc5G\sqrt{\frac{hc^5}{G}}

Correct answer: B

Explanation: Let t=hαGβcγt=h^\alpha G^\beta c^\gamma with [h]=ML2T−1[h]=ML^2T^{-1}, [G]=M−1L3T−2[G]=M^{-1}L^3T^{-2}, [c]=LT−1[c]=LT^{-1}. M: α−β=0\alpha-\beta=0; L: 2α+3β+γ=02\alpha+3\beta+\gamma=0; T: −α−2β−γ=1-\alpha-2\beta-\gamma=1. With α=β\alpha=\beta: 5α+γ=05\alpha+\gamma=0 and −3α−γ=1-3\alpha-\gamma=1, so 2α=12\alpha=1, α=β=12\alpha=\beta=\frac12, γ=−52\gamma=-\frac52. t∝hG/c5t\propto\sqrt{hG/c^5}. (Option A is a mass, C a length, D an energy.)

Question 18

Physics · 4 marks · −1

In a new system of units, the unit of mass is 2 kg, the unit of length is 4 m and the unit of time is 2 s. An energy of 400 J expressed in this system has the numerical value ______.

Correct answer: 50

Explanation: New unit of energy =[ML2T−2]=2×42/22=8=[ML^2T^{-2}]=2\times4^2/2^2=8 J. Numerical value =4008=50=\frac{400}{8}=50.

Question 19

Physics · 4 marks · −1

Five successive measurements of a time period gave 3.12 s, 3.08 s, 3.15 s, 3.05 s and 3.10 s. The percentage error of the result, based on the mean absolute error and rounded to the nearest integer, is ______.

Correct answer: 1

Explanation: Mean =15.505=3.10=\frac{15.50}{5}=3.10 s. Absolute errors: 0.02, 0.02, 0.05, 0.05, 0.00; mean absolute error =0.028≈0.03=0.028\approx0.03 s. Percentage error =0.033.10×100≈0.97%≈1%=\frac{0.03}{3.10}\times100\approx0.97\%\approx1\% (0.90% if 0.028 is used unrounded; either way 1).

Question 20

Physics · 4 marks · −1

In a Young's double-slit experiment, light of wavelength 600 nm is used, the slit separation is 0.2 mm and the screen is 1 m away. The fringe width is

  1. A.0.3 mm
  2. B.3 mm
  3. C.30 mm
  4. D.1.2 mm

Correct answer: B

Explanation: β=λDd=600×10−9×10.2×10−3=3×10−3\beta=\frac{\lambda D}{d}=\frac{600\times10^{-9}\times1}{0.2\times10^{-3}}=3\times10^{-3} m =3=3 mm.

Question 21

Physics · 4 marks · −1

Unpolarised light is incident from air on a glass surface of refractive index 3\sqrt3. The reflected light is completely plane polarised when the angle of incidence is

  1. A.30∘30^\circ
  2. B.45∘45^\circ
  3. C.60∘60^\circ
  4. D.90∘90^\circ

Correct answer: C

Explanation: Brewster's law: tan⁡iB=μ=3⇒iB=60∘\tan i_B=\mu=\sqrt3\Rightarrow i_B=60^\circ.

Question 22

Physics · 4 marks · −1

In a Young's double-slit set-up the fringe width in air is 3 mm. If the entire apparatus is immersed in water of refractive index 4/34/3, the fringe width becomes

  1. A.4 mm
  2. B.3 mm
  3. C.1.69 mm
  4. D.2.25 mm

Correct answer: D

Explanation: Wavelength in water λ′=λ/μ\lambda'=\lambda/\mu, so β′=β/μ=3×34=2.25\beta'=\beta/\mu=3\times\frac34=2.25 mm.

Question 23

Physics · 4 marks · −1

Light of wavelength 500 nm falls normally on a single slit of width 0.1 mm. The width of the central maximum on a screen 2 m away is

  1. A.20 mm
  2. B.10 mm
  3. C.5 mm
  4. D.40 mm

Correct answer: A

Explanation: First minima at sin⁡θ=λa\sin\theta=\frac\lambda a; central maximum width =2λDa=2×5×10−7×210−4=0.02=\frac{2\lambda D}{a}=\frac{2\times5\times10^{-7}\times2}{10^{-4}}=0.02 m =20=20 mm.

Question 24

Physics · 4 marks · −1

Two coherent sources whose intensities are in the ratio 9:19:1 produce an interference pattern. The ratio of maximum to minimum intensity in the pattern is

  1. A.9
  2. B.3
  3. C.4
  4. D.16

Correct answer: C

Explanation: Amplitude ratio =9:1=3:1=\sqrt9:1=3:1. ImaxImin=(3+1)2(3−1)2=164=4\frac{I_{max}}{I_{min}}=\frac{(3+1)^2}{(3-1)^2}=\frac{16}{4}=4.

Question 25

Physics · 4 marks · −1

In a Young's double-slit experiment with light of wavelength 600 nm, one slit is covered by a thin transparent film of refractive index 1.5. The central bright fringe shifts to the position previously occupied by the 5th bright fringe. The thickness of the film and the direction of shift are

  1. A.6 μ6\ \mum, towards the covered slit
  2. B.6 μ6\ \mum, away from the covered slit
  3. C.3 μ3\ \mum, towards the covered slit
  4. D.2 μ2\ \mum, towards the covered slit

Correct answer: A

Explanation: The film adds optical path (μ−1)t(\mu-1)t to that slit, so the zero-path-difference point moves towards the covered slit (its geometric path must be shorter). (μ−1)t=5λ⇒t=5×600×10−90.5=6×10−6(\mu-1)t=5\lambda\Rightarrow t=\frac{5\times600\times10^{-9}}{0.5}=6\times10^{-6} m.

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