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JEE Main 2023 Question Paper & Answers

The official JEE Main 2023 question paper, with the official answer key. Attempt as a real timed exam (with negative marking) or switch to practice mode for instant per-question answers and explanations.

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Official PYQ

JEE Main 2023 - February 1 Shift 1

180 Mins 300 Marks
75Questions
300Max Marks

Sample questions from JEE Main 2023 - February 1 Shift 1

1A child stands on the edge of a cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 msβˆ’1^{-1}. Neglecting air resistance, the speed with which the stone hits the ground will be ___ msβˆ’1^{-1} (given, g = 10 msβˆ’2^{-2}).
  • A) 15
  • B) 20
  • C) 30
  • D) 25
Answer: A
2Let Οƒ\sigma be the uniform surface charge density of two infinite thin plane sheets shown in the figure (one carrying +Οƒ+\sigma, the other βˆ’Οƒ-\sigma, placed parallel to each other). Then the electric fields in the three different regions E1E_1 (left of both sheets), E2E_2 (between the sheets) and E3E_3 (right of both sheets) are:
  • A) E1=E2=E3=σΡ0E_1=E_2=E_3=\dfrac{\sigma}{\varepsilon_0}
  • B) E1=E3=0,Β E2=σΡ0E_1=E_3=0,\ E_2=\dfrac{\sigma}{\varepsilon_0}
  • C) E1=E3=Οƒ2Ξ΅0,Β E2=0E_1=E_3=\dfrac{\sigma}{2\varepsilon_0},\ E_2=0
  • D) E1=E2=0,Β E3=σΡ0E_1=E_2=0,\ E_3=\dfrac{\sigma}{\varepsilon_0}
Answer: B
3A mercury drop of radius 10βˆ’210^{-2} m is broken into 125 equal-sized droplets. Surface tension of mercury is 0.45 Nmβˆ’1^{-1}. The gain in surface energy is:
  • A) 28Γ—10βˆ’528\times10^{-5} J
  • B) 17.5Γ—10βˆ’517.5\times10^{-5} J
  • C) 5Γ—10βˆ’55\times10^{-5} J
  • D) 2.26Γ—10βˆ’52.26\times10^{-5} J
Answer: D
4If earth has a mass nine times and radius twice that of a planet P, then the minimum velocity required by a rocket to pull out of the gravitational field of P is x3ve\dfrac{x}{3}v_e, where vev_e is the escape velocity on earth. The value of xx is:
  • A) 8
  • B) 3
  • C) 18
  • D) 2
Answer: D
5One mole of an ideal gas at initial temperature TT undergoes a quasi-static, reversible adiabatic expansion in which its volume becomes twice its initial value. If Ξ³\gamma is the ratio of specific heats, the work done by the gas in the process is (a form of):
  • A) W=RTΞ³βˆ’1[2βˆ’2Ξ³βˆ’1]W=\dfrac{RT}{\gamma-1}\left[2-2^{\gamma-1}\right]
  • B) W=RT[2βˆ’2]W=RT\left[2-\sqrt2\right]
  • C) W=RTΞ³βˆ’1[2Ξ³βˆ’1βˆ’2]W=\dfrac{RT}{\gamma-1}\left[2^{\gamma-1}-2\right]
  • D) W=RTΞ³[2βˆ’2]W=\dfrac{RT}{\gamma}\left[2-\sqrt2\right]
Answer: B

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JEE Main 2023 Previous Year Question Paper with Answers (PYQ) | ClassScribe