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JEE Mains Previous Years Questions with Solutions

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JEE Main 2021 (Online) 27th July Evening Shift

Numerical
The K$$\alpha$$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be __________ keV. (Round off to the nearest integer)

[h = 4.14 $$\times$$ 10$$-$$15 eVs, c = 3 $$\times$$ 108 ms$$-$$1]
Your Input ________

Answer

Correct Answer is 10

Explanation

$${E_{{k_\alpha }}} = {E_k} - {E_L}$$

$${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$$

$${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$$

= 27.5 KeV $$ - {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}$$

EL = (27.5 $$-$$ 17.5) keV

= 10 keV
2

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 $$\Omega$$ in the collector circuit is 0.6 V. If the current gain factor ($$\beta$$) is 24, then the base current is _____________ $$\mu$$A. (Round off to the Nearest Integer)
Your Input ________

Answer

Correct Answer is 25

Explanation

$$\beta = {{{I_C}} \over {{I_B}}} = 24;$$

RC = 1000

$$\Delta$$V = 0.6

$${I_C} = {{0.6} \over {1000}}$$

IC = 6 $$\times$$ 10$$-$$4

$${I_B} = {{{I_C}} \over \beta } = {{6 \times {{10}^{ - 4}}} \over {24}} = 25\mu A$$
3

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $$\mu$$F is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be _____________ $$\Omega$$.
Your Input ________

Answer

Correct Answer is 150

Explanation

Tm =30 ms

C = 200 $$\mu$$F

$${q \over N} = {{{Q_0}{e^{ - t/RC}}} \over {{N_0}{e^{ - \lambda t}}}} = {{{Q_0}} \over {{N_0}}}{e^{t\left( {\lambda - {1 \over {RC}}} \right)}}$$

Since q/N is constant hence

$$\lambda ={1 \over {RC}}$$

$$R = {1 \over {\lambda C}} = {{{T_m}} \over C} = {{30 \times {{10}^{ - 3}}} \over {200 \times {{10}^{ - 6}}}} = 150\Omega $$
4

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ___________ $$\times$$ 10$$-$$7 T.

Given hc = 20 $$\times$$ 10$$-$$26 J-m, mass of electron = 9 $$\times$$ 10$$-$$31 kg
Your Input ________

Answer

Correct Answer is 125

Explanation

By photoelectric equation

$${{hc} \over \lambda } - \phi = {k_{\max }}$$

$${k_{\max }} = {{1240} \over {500}} - 1.25 \approx 1.25$$

$$r = {{\sqrt {2mk} } \over {eB}}$$

$$B = {{\sqrt {2mk} } \over {er}}$$

$$ = 125 \times {10^{ - 7}}T$$

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