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.**Position of body with acceleration ‘a’ is given by x=Ka

^{m}t

^{n}, here t is time. Find dimensions of m and n

Solution

(b) As x=ka

[M

(b) As x=ka

^{m}×t^{n}[M

^{0}LT^{0}]=[LT^{-2}]^{m}[T]^{n}=[L^{m}T^{(-2m+n)}] ∴m=1 and -2m+n=0 ⇒n=2**Q2.**A pressure of 10

^{6}dyne cm

^{-2}is equivalent to

Solution

(a) 1 Newton = 10

(a) 1 Newton = 10

^{5}dyne and 1m=100 cm**Q3.**The unit of Planck’s constant is

Solution

(d)Joule-s

(d)Joule-s

**Q4.**Candela is the unit of

Solution

(b)Luminous intensity

(b)Luminous intensity

**Q5.**The dimensions of gravitational constant G and the moment of inertia are respectively

Solution

(b)M

(b)M

^{-1}L^{3}T^{-2};ML^{2}T^{0}**Q6.**What will be the unit of time in that system in which the unit of length is metre, unit of mass is kg and unit of force is kg wt?

Solution

(d) We know [F]=[MLT

T

(d) We know [F]=[MLT

^{-2}]T

^{2}=ML/F=(1kg×1m)/(1kg-wt)=(1kg×1m)/9.8N T=1/√9.8 sec**Q7.**If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formula of force is

Solution

(a) Let F ∝P

(a) Let F ∝P

^{x}V^{y}T^{z}By substituting the following dimensions: [P]=[ML^{-1}T^{-2}][V]=[LT^{-1}],[T]=[T] and comparing the dimension of both sides x=1,y=2,z=2, so F=PV^{2}T^{2}**Q8.**If I is the moment of inertia and Ï‰ the angular velocity, what is the dimensional formula of rotational kinetic energy 1/2 IÏ‰

^{2}?

Solution

(c) Do not think in terms of I and Ï‰. Remember; kinetic energy is fundamentally ‘work’ W=Force × distance =[MLT

(c) Do not think in terms of I and Ï‰. Remember; kinetic energy is fundamentally ‘work’ W=Force × distance =[MLT

^{-2}]×[L] =[ML^{2}T^{-2}]**Q9.**The dimensional formula for Planck’s constant (h) is

Solution

(c) E=hv⇒[ML

(c) E=hv⇒[ML

^{2}T^{-2}]=[h][T^{-1}]⇒[h]=[ML^{2}T^{-1}]**Q10.**The physical quantities not having same dimensions are

Solution

(c) Momentum [MLT

(c) Momentum [MLT

^{-1}], Plank’s constant [ML^{2}T^{-1}]