NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number n = 6 ? The value of X is ______________.
Your Input ________

Answer

Correct Answer is 15

Explanation

No. of different wavelengths = $${{n(n - 1)} \over 2}$$

$$ = {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$$
2

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
A particle of mass 9.1 $$\times$$ 10$$-$$31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10$$-$$27 kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle.
Your Input ________

Answer

Correct Answer is 910

Explanation

For photon $${\lambda _1} = {h \over P} = {{6.6 \times {{10}^{ - 34}}} \over {{{10}^{ - 27}}}}$$

For particle $${\lambda _2} = {h \over {mv}} = {{6.6 \times {{10}^{ - 34}}} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$$

$$\therefore$$ $${{{\lambda _1}} \over {{\lambda _2}}} = 910$$
3

JEE Main 2021 (Online) 27th July Morning Shift

Numerical
In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 $$\mathop A\limits^o $$. If the speed of electron is 2.2 $$\times$$ 166 m/s, then the current associated with the electron will be _____________ $$\times$$ 10$$-$$2 mA. [Take $$\pi$$ as $${{22} \over 7}$$]
Your Input ________

Answer

Correct Answer is 112

Explanation

$$I = {e \over T} = {{e\omega } \over {2\pi }} = {{eV} \over {2\pi r}}$$

$$I = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 7} \over {2 \times 22 \times 0.5 \times {{10}^{ - 10}}}}$$

= 1.12 mA

= 112 $$\times$$ 10$$-$$2 mA
4

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
The nuclear activity of a radioactive element becomes $${\left( {{1 \over 8}} \right)^{th}}$$ of its initial value in 30 years. The half-life of radioactive element is _____________ years.
Your Input ________

Answer

Correct Answer is 10

Explanation

We know, $$A = {A_0}{e^{ - \lambda t}}$$

For half life

$${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$$

$$ \Rightarrow $$ $${\lambda {t_{1/2}}}$$ = ln 2 .....(1)

And when radioactive element becomes $${\left( {{1 \over 8}} \right)^{th}}$$ of its initial value in 30 years

$${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$$

$$ \Rightarrow $$ 30$$\lambda = 3\ln 2$$

$$ \Rightarrow $$ $$\lambda = {{3\ln 2} \over {30}}$$ .....(2)

Putting value of $$\lambda $$ in (1), we get

$${{3\ln 2} \over {30}} \times {t_{1/2}}$$ = ln 2

$$ \Rightarrow $$ $${t_{1/2}}$$ = 10 years

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12