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.**Which of the following is not represented in correct unit

Solution

(c) Unit of energy will be kg- m

(c) Unit of energy will be kg- m

_{2}/sec_{2}**Q2.**If 1 g cm s

^{-1}= x newton-second, then the number x is equal to

Solution

(c) x=(1gcms

(c) x=(1gcms

_{-1}/T_{2}=(1gcms_{-1}/(1kg×1ms_{-1}×1s) =(1gcms_{-1}/(10_{3}g×〖10_{2}cms_{2}×1s)=10_{-5}**Q3.**The time taken by an electron to go from ground state to excited state is one

shake (one shake = 10

^{-8}s). this time in nanosecond will be

Solution

(a) n

(a) n

_{1}u_{1}=n_{2}u_{2}n_{2}=(1 shake)/(1 ns) =(10^{-8}s)/(10^{-9}s) ∴ n_{2}=10

**Q4.**The dimensional formula of angular velocity is

Solution

(a) Angular velocity = Î¸/t,[Ï‰]=([M

(a) Angular velocity = Î¸/t,[Ï‰]=([M

^{0}L^{0}T^{0}])/([T])=[T^{-1}]**Q5.**The physical quantity which has the dimensional formula M

^{1}T

^{-3}is

Solution

(b) Solar constant is energy received per unit area per unit time i.e.

([ML

(b) Solar constant is energy received per unit area per unit time i.e.

([ML

^{2}T^{-2}])/([L^{2}][T])=[M^{1}T^{-3}]**Q6.**The dimensions of e

^{2}/4Ï€Îµ

_{0}hc, where e,Îµ

_{0},h and c are electronic charge, electric permittivity,

Planck’s constant and velocity of light in vacuum respectively

Solution

(a) [e]=[AT],∈

And [c]=[LT

∴[e

=[M

(a) [e]=[AT],∈

_{0}=[M^{-1}L^{-3}T^{4}A^{2}],[h]=[ML^{2}T^{-1}]And [c]=[LT

^{-1}]∴[e

^{2}/(4Ï€Ïµ_{0}hc)]=[(A^{2}T^{2})/(M^{-1}L^{-3}T^{4}A^{2}×ML^{2}T^{-1}×LT^{-1})]=[M

^{0}L^{0}T^{0}]**Q7.**Farad is not equivalent to

Solution

(b) Capacitance C=Charge/potential=q/V Also potential =work/charge

(∵V=W/q) ∴ C=q

Thus, (a), (c), (d) are equivalent to farad but (b) is not equivalent to farad.

(b) Capacitance C=Charge/potential=q/V Also potential =work/charge

(∵V=W/q) ∴ C=q

^{2}/J as well as C=J/V^{2}.Thus, (a), (c), (d) are equivalent to farad but (b) is not equivalent to farad.

**Q8.**Given X=(Gh/c

^{2})

^{(1/2)}, where G,h and c are gravitational constant, Planck’s constant and the velocity of light

respectively. Dimensions of X are the same as those of