JEE Main Practice — Physics Set 1

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

In the hydrogen spectrum, λL\lambda_L is the wavelength of the first (longest-wavelength) line of the Lyman series and λB\lambda_B that of the first line of the Balmer series. The ratio λL/λB\lambda_L/\lambda_B is

  1. A.27/527/5
  2. B.5/275/27
  3. C.3/203/20
  4. D.4/94/9

Correct answer: B

Explanation: 1λL=R(1−14)=3R4\frac{1}{\lambda_L}=R\left(1-\frac14\right)=\frac{3R}{4} and 1λB=R(14−19)=5R36\frac{1}{\lambda_B}=R\left(\frac14-\frac19\right)=\frac{5R}{36}. Hence λLλB=5R/363R/4=527\frac{\lambda_L}{\lambda_B}=\frac{5R/36}{3R/4}=\frac{5}{27}.

Question 2

Physics · 4 marks · −1

The activity of a radioactive sample falls to 116\frac{1}{16} of its initial value in 20 days. The half-life of the sample is

  1. A.2.5 days
  2. B.10 days
  3. C.5 days
  4. D.4 days

Correct answer: C

Explanation: AA0=(12)n=116⇒n=4\frac{A}{A_0}=\left(\frac12\right)^{n}=\frac1{16}\Rightarrow n=4 half-lives in 20 days, so T1/2=20/4=5T_{1/2}=20/4=5 days.

Question 3

Physics · 4 marks · −1

A nucleus of mass number 240 with binding energy per nucleon 7.6 MeV splits into two identical nuclei of mass number 120, each with binding energy per nucleon 8.5 MeV. The energy released in the process is

  1. A.216 MeV
  2. B.108 MeV
  3. C.204 MeV
  4. D.432 MeV

Correct answer: A

Explanation: Energy released = (total BE of products) −- (BE of parent) =2×120×8.5−240×7.6=2040−1824=216=2\times120\times8.5-240\times7.6=2040-1824=216 MeV.

Question 4

Physics · 4 marks · −1

The ionisation energy of a hydrogen-like ion (from its ground state) is 122.4 eV. The wavelength of the photon emitted when its electron makes a transition from n=3n=3 to n=2n=2 is (take hc=1240hc=1240 eV nm)

  1. A.656.5 nm
  2. B.218.8 nm
  3. C.91.2 nm
  4. D.72.9 nm

Correct answer: D

Explanation: 13.6Z2=122.4⇒Z=313.6Z^2=122.4\Rightarrow Z=3. ΔE=122.4(14−19)=122.4×536=17\Delta E=122.4\left(\frac14-\frac19\right)=122.4\times\frac{5}{36}=17 eV. λ=124017≈72.9\lambda=\frac{1240}{17}\approx72.9 nm. (656.5 nm is the hydrogen value; 218.8 nm would be for Z=3Z=\sqrt3.)

Question 5

Physics · 4 marks · −1

The nucleus 90234Th^{234}_{90}\mathrm{Th} undergoes a series of α\alpha and β−\beta^- decays to become 83214Bi^{214}_{83}\mathrm{Bi}. The numbers of α\alpha and β−\beta^- particles emitted are respectively

  1. A.4 and 1
  2. B.5 and 2
  3. C.5 and 3
  4. D.10 and 3

Correct answer: C

Explanation: Mass number changes only through α\alpha decay: nα=234−2144=5n_\alpha=\frac{234-214}{4}=5. Five α\alpha decays lower ZZ by 10 (to 80); the final ZZ is 83, so nβ=83−80=3n_{\beta}=83-80=3.

Question 6

Physics · 4 marks · −1

A hydrogen atom in an excited state returns to the ground state by emitting two successive photons of energies 2.55 eV and 10.2 eV. The ratio of the de Broglie wavelength of the electron in the initial excited state to that in the ground state is

  1. A.16
  2. B.2
  3. C.4
  4. D.1/41/4

Correct answer: C

Explanation: 10.210.2 eV corresponds to 2→12\to1. Then 2.552.55 eV must end at n=2n=2: 13.6(14−1n2)=2.55⇒1n2=14−316=11613.6\left(\frac14-\frac1{n^2}\right)=2.55\Rightarrow \frac1{n^2}=\frac14-\frac{3}{16}=\frac1{16}, so n=4n=4. In Bohr's model 2πrn=nλn2\pi r_n=n\lambda_n with rn∝n2r_n\propto n^2, hence λn∝n\lambda_n\propto n. Ratio =4/1=4=4/1=4.

Question 7

Physics · 4 marks · −1

The radius of an 27Al^{27}\mathrm{Al} nucleus is 3.6 fm. Using R=R0A1/3R=R_0A^{1/3}, the radius of a 125Te^{125}\mathrm{Te} nucleus in fm is ______.

Correct answer: 6

Explanation: RTeRAl=(12527)1/3=53\frac{R_{Te}}{R_{Al}}=\left(\frac{125}{27}\right)^{1/3}=\frac53, so RTe=3.6×53=6R_{Te}=3.6\times\frac53=6 fm.

Question 8

Physics · 4 marks · −1

Two radioactive samples A and B have half-lives 2 h and 3 h respectively and initially contain equal numbers of nuclei. After 12 h, the ratio of the activity of B to that of A is x3\frac{x}{3}. The value of xx is ______.

Correct answer: 8

Explanation: After 12 h: NA=N02−6N_A=N_0 2^{-6}, NB=N02−4N_B=N_0 2^{-4}. Activity =λN=\lambda N with λ=ln⁡2/T1/2\lambda=\ln2/T_{1/2}. ABAA=(1/3) 2−4(1/2) 2−6=23×4=83\frac{A_B}{A_A}=\frac{(1/3)\,2^{-4}}{(1/2)\,2^{-6}}=\frac{2}{3}\times4=\frac83. So x=8x=8.

Question 9

Physics · 4 marks · −1

A wire of resistance 5 Ω5\,\Omega is uniformly stretched so that its length doubles (volume unchanged). Its new resistance is

  1. A.10 Ω10\,\Omega
  2. B.20 Ω20\,\Omega
  3. C.2.5 Ω2.5\,\Omega
  4. D.40 Ω40\,\Omega

Correct answer: B

Explanation: With volume constant, A∝1/lA\propto 1/l, so R=ρl/A∝l2R=\rho l/A\propto l^2. Doubling ll gives R′=4R=20 ΩR'=4R=20\,\Omega.

Question 10

Physics · 4 marks · −1

A battery of emf 12 V and internal resistance 0.5 Ω0.5\,\Omega is connected across an external resistor of 5.5 Ω5.5\,\Omega. The terminal potential difference of the battery is

  1. A.10.5 V
  2. B.12 V
  3. C.6 V
  4. D.11 V

Correct answer: D

Explanation: I=125.5+0.5=2I=\frac{12}{5.5+0.5}=2 A. Terminal voltage V=E−Ir=12−2(0.5)=11V=E-Ir=12-2(0.5)=11 V.

Question 11

Physics · 4 marks · −1

In a metre bridge, an unknown resistance XX in the left gap and 12 Ω12\,\Omega in the right gap give a balance point at 40 cm from the left end. What resistance must be connected in parallel with the 12 Ω12\,\Omega resistor so that the balance point shifts to 50 cm?

  1. A.24 Ω24\,\Omega
  2. B.8 Ω8\,\Omega
  3. C.12 Ω12\,\Omega
  4. D.36 Ω36\,\Omega

Correct answer: A

Explanation: X12=4060⇒X=8 Ω\frac{X}{12}=\frac{40}{60}\Rightarrow X=8\,\Omega. Balance at 50 cm needs the right gap to be 8 Ω8\,\Omega: 12R12+R=8⇒12R=96+8R⇒R=24 Ω\frac{12R}{12+R}=8\Rightarrow 12R=96+8R\Rightarrow R=24\,\Omega.

Question 12

Physics · 4 marks · −1

Two bulbs rated 100 W, 220 V and 60 W, 220 V are connected in series across a 220 V supply. Assuming constant resistances, the total power drawn and the brighter bulb are

  1. A.37.5 W; the 60 W bulb
  2. B.37.5 W; the 100 W bulb
  3. C.160 W; the 100 W bulb
  4. D.62.5 W; the 60 W bulb

Correct answer: A

Explanation: R=V2/PR=V^2/P, so R100=484 ΩR_{100}=484\,\Omega and R60=806.7 ΩR_{60}=806.7\,\Omega. Series power =V2R1+R2=P1P2P1+P2=6000160=37.5=\frac{V^2}{R_1+R_2}=\frac{P_1P_2}{P_1+P_2}=\frac{6000}{160}=37.5 W. Same current flows, so the larger resistance (60 W bulb) dissipates more (I2RI^2R) and glows brighter.

Question 13

Physics · 4 marks · −1

A copper wire of cross-sectional area 2 mm22\ \mathrm{mm^2} carries a current of 3.2 A. If the free-electron density of copper is 8×1028 m−38\times10^{28}\ \mathrm{m^{-3}}, the drift speed of electrons is (e=1.6×10−19e=1.6\times10^{-19} C)

  1. A.1.25 mm/s
  2. B.0.0125 mm/s
  3. C.0.125 mm/s
  4. D.12.5 mm/s

Correct answer: C

Explanation: vd=IneA=3.28×1028×1.6×10−19×2×10−6=3.22.56×104=1.25×10−4v_d=\frac{I}{neA}=\frac{3.2}{8\times10^{28}\times1.6\times10^{-19}\times2\times10^{-6}}=\frac{3.2}{2.56\times10^4}=1.25\times10^{-4} m/s =0.125=0.125 mm/s.

Question 14

Physics · 4 marks · −1

Two cells, one of emf 6 V and internal resistance 1 Ω1\,\Omega and the other of emf 4 V and internal resistance 2 Ω2\,\Omega, are connected in parallel with like terminals joined. This combination is connected across a 4 Ω4\,\Omega resistor. The current through the 4 Ω4\,\Omega resistor is

  1. A.10/710/7 A
  2. B.8/78/7 A
  3. C.1 A
  4. D.5/35/3 A

Correct answer: B

Explanation: Let the common terminal voltage be VV. KCL at the junction: 6−V1+4−V2=V4\frac{6-V}{1}+\frac{4-V}{2}=\frac{V}{4}. Multiply by 4: 24−4V+8−2V=V⇒V=32724-4V+8-2V=V\Rightarrow V=\frac{32}{7} V. Current in 4 Ω4\,\Omega: V4=87\frac{V}{4}=\frac{8}{7} A. (Equivalently Eeq=6/1+4/21/1+1/2=163E_{eq}=\frac{6/1+4/2}{1/1+1/2}=\frac{16}{3} V, req=23 Ωr_{eq}=\frac23\,\Omega, I=16/34+2/3=87I=\frac{16/3}{4+2/3}=\frac87 A.)

Question 15

Physics · 4 marks · −1

On a potentiometer, a cell gives a balancing length of 360 cm. When the cell is shunted by a 15 Ω15\,\Omega resistor, the balancing length becomes 300 cm. The internal resistance of the cell (in Ω\Omega) is ______.

Correct answer: 3

Explanation: r=R(l1l2−1)=15(360300−1)=15×0.2=3 Ωr=R\left(\frac{l_1}{l_2}-1\right)=15\left(\frac{360}{300}-1\right)=15\times0.2=3\,\Omega.

Question 16

Physics · 4 marks · −1

A uniform wire of resistance 36 Ω36\,\Omega is bent into a circle. The equivalent resistance (in Ω\Omega) between two points on the circle that subtend an angle of 120∘120^\circ at the centre is ______.

Correct answer: 8

Explanation: The two arcs are 120360×36=12 Ω\frac{120}{360}\times36=12\,\Omega and 24 Ω24\,\Omega, which are in parallel: 12×2436=8 Ω\frac{12\times24}{36}=8\,\Omega.

Question 17

Physics · 4 marks · −1

Photons of energy 4.1 eV fall on a metal surface of work function 2.3 eV. The stopping potential for the emitted photoelectrons is

  1. A.2.3 V
  2. B.6.4 V
  3. C.1.8 V
  4. D.4.1 V

Correct answer: C

Explanation: eV0=hν−ϕ=4.1−2.3=1.8eV_0=h\nu-\phi=4.1-2.3=1.8 eV, so V0=1.8V_0=1.8 V.

Question 18

Physics · 4 marks · −1

An electron accelerated from rest through a potential difference VV has de Broglie wavelength λ\lambda. If the accelerating potential is made 9V9V, the de Broglie wavelength becomes

  1. A.λ/3\lambda/3
  2. B.λ/9\lambda/9
  3. C.3λ3\lambda
  4. D.λ/3\lambda/\sqrt3

Correct answer: A

Explanation: λ=h2meV∝V−1/2\lambda=\frac{h}{\sqrt{2meV}}\propto V^{-1/2}. With 9V9V, λ′=λ/9=λ/3\lambda'=\lambda/\sqrt9=\lambda/3.

Question 19

Physics · 4 marks · −1

When light of wavelength λ\lambda falls on a metal, the stopping potential is VV. When light of wavelength λ/2\lambda/2 is used, the stopping potential becomes 3V3V. The threshold wavelength of the metal is

  1. A.3λ3\lambda
  2. B.4λ4\lambda
  3. C.2λ2\lambda
  4. D.3λ/23\lambda/2

Correct answer: C

Explanation: eV=hcλ−ϕeV=\frac{hc}{\lambda}-\phi and 3eV=2hcλ−ϕ3eV=\frac{2hc}{\lambda}-\phi. Subtracting: 2eV=hcλ2eV=\frac{hc}{\lambda}, so eV=hc2λeV=\frac{hc}{2\lambda} and ϕ=hcλ−hc2λ=hc2λ\phi=\frac{hc}{\lambda}-\frac{hc}{2\lambda}=\frac{hc}{2\lambda}. Threshold wavelength λ0=hcϕ=2λ\lambda_0=\frac{hc}{\phi}=2\lambda.

Question 20

Physics · 4 marks · −1

A proton and an α\alpha-particle are each accelerated from rest through the same potential difference. The ratio of the de Broglie wavelength of the proton to that of the α\alpha-particle is (take mα=4mpm_\alpha=4m_p)

  1. A.2\sqrt2
  2. B.2
  3. C.122\frac{1}{2\sqrt2}
  4. D.222\sqrt2

Correct answer: D

Explanation: λ=h2mqV\lambda=\frac{h}{\sqrt{2mqV}}. λpλα=mαqαmpqp=4×2=22\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}=\sqrt{4\times2}=2\sqrt2.

Question 21

Physics · 4 marks · −1

A 60 W lamp converts 40% of its electrical power into light of wavelength 600 nm. The number of photons emitted per second is approximately (take hc=1240hc=1240 eV nm, 1 eV=1.6×10−191\ \mathrm{eV}=1.6\times10^{-19} J)

  1. A.7.3×10197.3\times10^{19}
  2. B.1.8×10201.8\times10^{20}
  3. C.2.9×10192.9\times10^{19}
  4. D.7.3×10187.3\times10^{18}

Correct answer: A

Explanation: Light power =0.4×60=24=0.4\times60=24 W. Photon energy =1240600=2.067=\frac{1240}{600}=2.067 eV =3.31×10−19=3.31\times10^{-19} J. N=243.31×10−19≈7.3×1019 s−1N=\frac{24}{3.31\times10^{-19}}\approx7.3\times10^{19}\ \mathrm{s^{-1}}. (Using all 60 W would give 1.8×10201.8\times10^{20}.)

Question 22

Physics · 4 marks · −1

Light of wavelength 248 nm falls on a metal of work function 3.0 eV. The most energetic photoelectrons enter a uniform magnetic field perpendicular to their velocity. The field needed to make them move in a circle of radius 5 mm is approximately (hc=1240hc=1240 eV nm, me=9.1×10−31m_e=9.1\times10^{-31} kg, e=1.6×10−19e=1.6\times10^{-19} C)

  1. A.1.9 mT
  2. B.0.95 mT
  3. C.0.48 mT
  4. D.9.5 mT

Correct answer: B

Explanation: Photon energy =1240/248=5=1240/248=5 eV, so Kmax=5−3=2K_{max}=5-3=2 eV =3.2×10−19=3.2\times10^{-19} J. p=2mK=2×9.1×10−31×3.2×10−19=7.63×10−25p=\sqrt{2mK}=\sqrt{2\times9.1\times10^{-31}\times3.2\times10^{-19}}=7.63\times10^{-25} kg m/s. r=peB⇒B=7.63×10−251.6×10−19×5×10−3=9.5×10−4r=\frac{p}{eB}\Rightarrow B=\frac{7.63\times10^{-25}}{1.6\times10^{-19}\times5\times10^{-3}}=9.5\times10^{-4} T ≈0.95\approx0.95 mT.

Question 23

Physics · 4 marks · −1

Light of wavelength 200 nm is incident on a metal surface of work function 2.2 eV. The stopping potential (in volt) is ______. (Take hc=1240hc=1240 eV nm.)

Correct answer: 4

Explanation: E=1240200=6.2E=\frac{1240}{200}=6.2 eV; eV0=6.2−2.2=4.0eV_0=6.2-2.2=4.0 eV, so V0=4V_0=4 V.

Question 24

Physics · 4 marks · −1

The threshold frequency of a metal is ν0\nu_0. Light of frequency 2ν02\nu_0 ejects photoelectrons with maximum speed v1v_1, and light of frequency 10ν010\nu_0 ejects them with maximum speed v2v_2. The value of v2/v1v_2/v_1 is ______.

Correct answer: 3

Explanation: 12mv2=h(ν−ν0)\frac12mv^2=h(\nu-\nu_0). So v12∝hν0v_1^2\propto h\nu_0 and v22∝9hν0v_2^2\propto 9h\nu_0. v2v1=9=3\frac{v_2}{v_1}=\sqrt9=3.

Question 25

Physics · 4 marks · −1

A coil of 200 turns and area 0.05 m20.05\ \mathrm{m^2} is placed with its plane perpendicular to a magnetic field. The field changes uniformly from 0.2 T to 0.8 T in 0.3 s. The magnitude of the average induced emf is

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

Correct answer: D

Explanation: ∣ε∣=NAΔBΔt=200×0.05×0.60.3=20|\varepsilon|=N A\frac{\Delta B}{\Delta t}=200\times0.05\times\frac{0.6}{0.3}=20 V.

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