JEE Mains is a PAN India level exam, and if you dream of qualifying for it, you have to be the best with your preparations. It comprises three sections: Physics, Chemistry, and Mathematics. And, it is necessary to attempt all three sections if you want to be enlisted as one of the toppers.

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To ensure that you are preparing on the right track, check your preparations for each section separately. This article aims to help you out with your practices. Here are some important Physics questions with solutions from the previous years’ JEE Mains exam papers. Have a look.

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**Important Physics Questions with Solutions from ****JEE Mains Previous Years’ Papers**

1. A Plano-convex lens with a refractive index of one and a focal length f1 and a Plano concave lens with a refractive index of two and focal length f2 are in proximity. If the radius of curvature of spherical faces of both lenses is R each and f1 = 2f2, what is the relation between 1 and 2?

1f2 – 1f1 = (2- 1)(1 – 1-R)

1f2 = (2– 1)(1-R – 1)

(1– 1)R = (2– 1)2R

21– 2= 1

Therefore, the required relation is 21– 2= 1.

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2. A cable of mass 5g and the length of 1 m is fixed at both ends. The tension in the cable is 8.0 N. The cable is set into vibration using an extrinsic vibrator of frequency 100 Hz. How close to the separation between sequential nodes on the cable?

Velocity of wave on cable, V = T = 85 x 1000

V = 40 m/s

Now, wavelength of wave = vn = 40100 m

Therefore, separation between successive nodes, 2 = 20100 m = 20 cm

3. Two guns, X and Y, can fire bullets at 2km/s and 4 km/s. They are ejected in all feasible directions from a point parallel to the ground. What will be the ratio of utmost areas engulfed by the bullets ejected by two guns on the ground?

R = u² sin 2g and A = R2

A ∝ R2 ∝ u4

A₁A₂ = u₁⁴u₂⁴ = [24]4 = 116

Therefore, the required ratio is 1:16.

4. The TV transmission tower has a height of 140m. The height of the receiving aerial is 40m. What is the maximal distance upto that can broadcast this tower’s signals in the LOS (Line of Sight) system? [Given: Radius of Earth = 6.4 x 106 m]

Maximum distance upto that can broadcast signals is dmax = 2RhT + 2Rhg

dmax = 2 x 6.4 x 106 [104 + 40

dmax = 65 km

Therefore, the required maximum distance is 65 km.

5. If the magnetic field of a horizontal electromagnetic wave is given by

B = 100 x 10-6 sin [2 x 2 x 1015(t – xc)],

what will be the maximum electric field associated with it? [Given: Speed of light = 3 x 108 m/s]

E0 = B0 x C

E0 = 100 x 10-6 x 3 x 108 = 3 x 104

Therefore, the maximum electric field associated is 3 x 104 N/C.

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6. A hydrogen atom initially at the ground state is excited by absorbing a photon of wavelength 980Å. What will be the atom’s radius in the excited state in the form of Bohr’s radius a0?

Energy of photon = 12500980 = 12.75 eV

Electrons will be excited to the n = 4 level.

As R ∝ n2

Therefore, the radius of the atom will be 16a0.

7. The unbending diatomic ideal gas via an adiabatic process at room temperature. The connection between volume and temperature of this process is TVx = constant. What will be the value of x?

For adiabatic process,

TV-1 = constant

For diatomic process, – 1 = 75 – 1

x = 75 – 1 = 25

Therefore, the value of x is 25.

8. Two equal resistances are connected in series to a battery and consume 60 W of electric power. If these resistances are now fastened in parallel combinations to the same battery, what will be the amount of electric power they consume?

In series combination, equivalent resistance = 2R

60 W power consumed = 22R

In parallel combination, equivalent resistance = R2

New power, P0 = 2(R2)

P0 = 4P = 240 W

Therefore, the amount of electric power consumed by them = 240 W.

9. Suppose a circular coil of wire containing current, I, forms a magnetic dipole. The magnetic flux across a boundless plane that contains the circular coil and excludes the circular coil area is given by 1. The magnetic flux throughout the area is given by 0. What will be the relation between 1 and 0?

As magnetic field lines always form a closed loop, every magnetic field line creating magnetic flux in the inner region must pass all over the outer region. The magnetic flux across the area is given by 0. Since the flux in two regions are in opposite directions,

1 = – 0

Therefore, the required relation is 1 + 0 = 0.

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10. A 60 HP electric motor lifts a lift having a maximal total load range of 2000 kg. If the frictional force on the lift is 4000 N, find the elevator’s speed at full load.

The speed of elevator at full load = (4000 x v) + (mg x v)

(4000 x v) + (mg x v) = P

v x (4000 + mg) = P

v = P(4000 + mg) = 60 x 7464000+20000

v = 1.86 m/s

Therefore, the required speed is 1.9 m/s.

11. Two moles of an absolute gas with CpCv = 53 are mixed with 3 moles of another absolute gas with CpCv = 43. What will be the value of CpCv for the mixture?

The value of CpCv for the mixture = mixture

mixture= n1Cp1+n2Cp2n1Cv1+n2Cv2

mixture= n11R1-1 + n22R2-1n1R1-1 + n2R2-1

On rearranging,

n1+n2mix-1 = n11-1 + n22-1

5mix-1 = 313 + 223

mixture= 1712 = 1.42

Therefore, the value of CpCv for the mixture is 1.42.

12. In the shown circuit of the following, the battery is of EMF 4 and internal resistance 8 . Find the potential difference between points x and y as shown in the circuit. [Given: R = 12 .]

I = 420 = 210

Vx– Vy = IR = I x 12

Vx– Vy = 210 x 12

Vx– Vy = 2.4 V

Therefore, the potential difference between points x and y is 2.4 V.

13. A body cools from 100℃ to 90℃ in 20 minutes. Find the time to cool from 110℃ to 100℃.

By Newton’s law of cooling,

dQdt ∝ (T – T0)

Here, (T – T0) is the temperature difference between the body and the surroundings. If this difference is more, then the rate of cooling is higher. Or cooling is fast.

Therefore, to cool from 110℃ to 100℃ takes less than 20 minutes compared to the time taken to cool from 100℃ to 90℃.

14. For a transistor, the emitter current is 40 mA, and the collector current is 35 mA. What will be the of this transistor?

= ICIb = IcIe- Ic

= 3540-35 = 355 = 7

Therefore, the of this transistor is 7.

15. Choose the correct option (s) regarding the Hydrogen spectrum

- A is the series limit of Lyman.
- B is the third line of Balmer.
- C is the second line of Paschen.
- All the above are correct.

- The correct option is (d).

A → Series limit of Lyman.

B → 3rd line of Balmer.

C → 2nd line of Paschen.

16. Two satellites revolve around a planet in orbit. The ratio of their periods is 1:8. Find the ratio of their angular velocities.

= 2T

12 = T2T1 = 81

Therefore, the ratio of their angular velocities is 8:1.

17. A 220 volts AC supply is given to the main circuit of the transformer, and the output of 12 volts DC is abstracted using a rectifier. If the subsidiary number of turns was 24, find the number of turns in the main coil.

- NPNS = VPVS

NP24 = 22012

NP= 440

Hence, the number of turns in the primary coil is 440.

18. An electromagnetic wave propagates in a medium, where r = r = 2. If the speed of light in such a medium is x x 107 m/s, find the value of x.

n = rr = 2

v = cn = 3 x 10⁸2 = 15 x 107 m/s [c = speed of light = 3 x 10⁸ m/s]

Therefore, the value of x = 15.

### Conclusion

Although Physics is a typical subject and the toughest among the three JEE Mains subjects, practice can help you qualify it with an excellent score. It is a golden word that can solve every problem. While preparing, please focus on the concepts behind the questions and strike the answer or formulae according to them. Understanding the concepts is equally important as knowing the formula to crack JEE Mains. Therefore, before jumping on mock papers, go through your syllabus and notes and revise them properly and carefully.

Be calm, positive, and concentrate on your aim. And in case you get stuck with any JEE Mains question or problem, contact your peers, teachers, or experts. Otherwise, find Aakash near you!

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