As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

**Q1.**An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is

Solution

Here, u=0;a=eE/m;v=?;t=t ∴ v=u+at=0+eE/m t de-Broglie wavelength, Î»=h/mv=h/(m(eEt/m))=h/eEt Rate of change of de-Broglie wavelength dÎ»/dt=h/eE (-1/t

Here, u=0;a=eE/m;v=?;t=t ∴ v=u+at=0+eE/m t de-Broglie wavelength, Î»=h/mv=h/(m(eEt/m))=h/eEt Rate of change of de-Broglie wavelength dÎ»/dt=h/eE (-1/t

^{2})=-h)/eEt^{2}**Q2.**One electron and one proton is accelerated by equal potential. Ratio in their de-Broglie wavelength is

Solution

If a charge particle of mass m and charge q is accelerated through a potential difference V and E is the energy acquired by the particle , then E=qV If v is velocity of particle, then E=1/2 mv

If a charge particle of mass m and charge q is accelerated through a potential difference V and E is the energy acquired by the particle , then E=qV If v is velocity of particle, then E=1/2 mv

^{2}Or v=√(2E/m) Now, de-Broglie wavelength of particle is Î»=h/mv=h/(m√(2E /m) Substituting the value of E, we get Î»=h/√2mqV For electron, Î»_{e}=h/√(2m_{e}eV) For proton, Î»_{p}=h/√(2m_{p}eV) ∴ Î»_{e}/Î»_{p}=√(m_{p}/m_{e})**Q3.**A 1Î¼A beam of protons with a cross-sectional area of 0.5 sq.mm is moving with a velocity of 3×10

^{4}ms

^{-1}. Then charge density of beam is

Solution

For one second, distance = Velocity =3×10

For one second, distance = Velocity =3×10

^{4}m/sec and Q=i×1=10^{-6}C.Charge density =Charge/Volume =10^{-6}/(3×10^{4}×0.5×10^{-6})=6.6×10^{-5}C/m^{3}

**Q4.**The dependence of the short wavelength limit Î»

_{min}on the accelerating potential V is represented by the curve of figure

Solution

Î»

Î»

_{min}=hc/eV⇒log.Î»_{min}=log (hc/e)-logV ⇒log Î»_{min}=-log V+log (hc/e) This is the equation of straight line having slope (-1) and intercept log(hc/e) on+log_{e}Î»_{min}axis**Q5.**X-rays are produced due to

Solution

Changing in atomic energy level

Changing in atomic energy level

**Q6.**In Millikan’s oil drop experiment, a charged drop of mass 1.8×10

^{-14}kg is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is

Solution

In equilibrium, qE=mg Also, E=V/d and q=ne ∴ ne×V/d=mg ⇒ n=mgd/eV Given, m=1.8×10

In equilibrium, qE=mg Also, E=V/d and q=ne ∴ ne×V/d=mg ⇒ n=mgd/eV Given, m=1.8×10

^{-14}kg, g=10ms^{-2}, d=0.9 cm=0.9×10^{-2}m, e=1.6×10^{-19}C, V=2000 V ∴ n=(1.8×10^{-14}×10×0.9×10^{-2})/(1.6×10^{-19}×2000) n=81/16≈5 electrons**Q7.**In Millikan’s oil drop experiment, a charged drop falls with terminal velocity V. if an electric field E is applied in vertically upward direction then it starts moving in upward direction with terminal velocity 2V. if magnitude of electric field is decreased to E/2, then terminal velocity will becomes

Solution

In the absence of electric field (i.e.E=0) mg=6Ï€Î·rv

In the absence of electric field (i.e.E=0) mg=6Ï€Î·rv

…(i)
In the presence of Electric field
mg+QE=6Ï€Î·r(2v)

…(ii)
When electric field to reduced to E/2
mg+Q(E/2)=6Ï€Î·r(v

^{'}) …(iii)
After solving (i), (ii) and (iii)
We get v

^{'}=3/2 v**Q8.**The wavelength of de-Broglie wave is 2Î¼m, then its momentum is (h=6.63× 10

^{-34}Js)

Solution

According to de-Broglie hypothesis Î»=h/p Where, h is Planck’s constant. ⇒ p=h/Î» Given, h=6.63×10

According to de-Broglie hypothesis Î»=h/p Where, h is Planck’s constant. ⇒ p=h/Î» Given, h=6.63×10

^{-34}J-s Î»=2Î¼m=2×10^{-6}m∴ ∴ p=(6.63×10^{-34})/(2×10^{-6}) =3.315×10^{-28}kg-ms^{-1}**Q9.**When a monochromatic point source of light is at a distance 0.2 m from a photoelectric cell, the saturation current and cut-off voltage are 12.0 mA and 0.5 V. If the same source is placed 0.4 m away from the photoelectric cell, then the saturation current and the stopping potential respectively are

Solution

The cut-off voltage or stopping potential measure maximum kinetic energy of the electron. It depends on the frequency of incident light whereas the current depends on the number of photons incident. Hence, cut-off voltage will be 0.5 V. Now by inverse square law, 12∝ 1/(0.2)

The cut-off voltage or stopping potential measure maximum kinetic energy of the electron. It depends on the frequency of incident light whereas the current depends on the number of photons incident. Hence, cut-off voltage will be 0.5 V. Now by inverse square law, 12∝ 1/(0.2)

^{2}or I∝1/(0.4)^{2}∴ I/12=(0.2)^{2}/(0.4)^{2}=1/4 or I=12/4=3mA**Q10.**The momentum of a charged particle moving in a perpendicular magnetic field depends on

Solution

q v B=mv

q v B=mv

^{2}/r or mv=q r B.