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.**The figure shows the path of a positively charged particle 1 through a rectangular region of uniform electric field as shown in the figure. What is the direction of electric field and the direction of deflection of particles 2,3 and 4?

Solution

The figure shows the path of a +ve charged particle (1) through a rectangular region of uniform electric field.

The figure shows the path of a +ve charged particle (1) through a rectangular region of uniform electric field.

Since, +ve charged particle moves as a parabolic path in electric field. It means the direction of electric field is upward. The direction of deflection of particle (2) which is –ve is downward. The direction of deflection of particle (3) which is +ve is upward and direction of deflection of particle (4) is downward.

**Q2.**The momentum of a photon in an X-ray beam 10

^{-10}metre wavelength is

Solution

Momentum of photon p=h/λ=(6.6×10

Momentum of photon p=h/λ=(6.6×10

^{-34})/10^{-10}=6.6×10^{-24}kg-m/s**Q3.**According to de-Broglie, the de-Broglie wavelength for electron in an orbit of (radius 5.3×10

^{-11}m) hydrogen atom is 10

^{-10}m. The principle quantum number for this electron is

Solution

2πr=nλ⇒n=2πr/λ=(2×3.14×5.3×10

2πr=nλ⇒n=2πr/λ=(2×3.14×5.3×10

^{-11})/10^{-10}) =3

**Q4.**Bragg’s equation will have no solution is

Solution

Bragg’s law states that 2d sin θ=nλ where =1,2,3….. . If λ>2d, then sinθ will be greater than 1 for n=1, which is not possible

Bragg’s law states that 2d sin θ=nλ where =1,2,3….. . If λ>2d, then sinθ will be greater than 1 for n=1, which is not possible

**Q5.**A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6×10

^{15}Hz, then the maximum kinetic energy of photoelectrons emitted is (h=6.6×10

^{-34}Js)

Solution

Work function W

Work function W

_{0}=hv_{0}=6.6×10^{-34}×1.6×10^{15}=1.056×10^{-18}J=6.6 eV From E=W_{0}+K_{max}⇒K_{max}=E-W_{0}=1.4 eV**Q6.**A tiny spherical oil drop carrying a net charge q is balanced in still air with vertical uniform electric field of strength 81π/7×10

_{5}Vm

^{-1}.When the field is switched off, the drop is observed to fall with terminal velocity 2×10

^{-3}ms

^{-1}. Given g=9.8 ms

^{-2}, viscosity of the air =1.8×10

^{-5}Ns m

^{-2}and the density of oil=900 kgm

^{-3}, the magnitude of q is

Solution

qE =mg …(i)6πηrv=mg 4/3 πr

qE =mg …(i)6πηrv=mg 4/3 πr

^{3}ρg=mg …(ii) ∴ r=(3mg/4πρg)^{(1/3)}…(iii) Substituting the value of r in Eq. (ii), we get 6πηv(3mg/4πpg)^{(1/3)}=mg or (6πηv)^{3}(3mg/4πρg) =(mg)^{3}Again substituting mg=qE, we get (qE)^{2}=(3/4πρg) (6πηv)^{3}Or qE=(3/4πρg)^{(1/2)}(6πηg)^{(3/2)}∴ q=1/E (3/4πρg)^{(1/2)}(6πηv)^{(3/2)}Substituting the values, we get q=7/(81π×10^{5}) √(3/(4π×900×9.8)×216π^{3}) ×√((1.8×10^{-5}×2×10^{-3})^{3})=8.0×10^{-19}C**Q7.**The de-Broglie wavelength of a neutron at 27℃ is λ. What will be its wavelength at 927℃?

Solution

Kinetic energy of a particle at temperature TKis E=3/2 kT. The de-Broglie wavelength associated with it is λ=h/√2mE=h/√(2m×3/2 kT) ie,λ∝1/√T ∴ λ

Kinetic energy of a particle at temperature TKis E=3/2 kT. The de-Broglie wavelength associated with it is λ=h/√2mE=h/√(2m×3/2 kT) ie,λ∝1/√T ∴ λ

_{927}/λ_{27}=√((27+373)/(927+273)) =√(300/1200)=1/2 or λ_{927}=λ_{27}/2=λ/2**Q8.**If in a photoelectric experiment, the wavelength of incident radiation is reduced from 6000 Å to 4000 Å then

Solution

Stopping potential V

Stopping potential V

_{0}=hc/e [1/λ-1/λ_{0}]. As λ decreases so V_{0}increases**Q9.**The energy of a photon of light of wavelength 450 nm is

Solution

E=hc/λ=(6.62×10

E=hc/λ=(6.62×10

^{-34}×3×10^{8})/(450×10^{-9}) )=4.4×10^{-19}J**Q10.**What is the momentum of a photon having frequency 1.5×10

^{13}Hz

Solution

p=hv/c=(6.6×10

p=hv/c=(6.6×10

^{-34}×1.5×10^{13})/(3×10^{8})=3.3×10^{-29}kg-m/s