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 fundamental unit, which has the same power in the dimensional formulae of surface tension and viscosity is

Solution

(a) [surface tension]=[ML

Clearly, mass has the same power in the two physical quantities.

(a) [surface tension]=[ML

^{0}T^{-2}], [viscosity]=[ML^{-1}T^{-1}].Clearly, mass has the same power in the two physical quantities.

**Q2.**Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further,

it is found that the screw gauge has a zero error of -0.03 mm.While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

Solution

(d) Diameter = Main scale reading + Circular scale reading ×LC+Zero error

=3+35×1/(2×50)+0.03=3.38 mm

(d) Diameter = Main scale reading + Circular scale reading ×LC+Zero error

=3+35×1/(2×50)+0.03=3.38 mm

**Q3.**One million electron volt (1 MeV) is equal to

Solution

(b) 1 MeV=10

(b) 1 MeV=10

^{6}eV

**Q4.**Given, Force =Î±/(density+ Î²

^{3}) What are the dimensions of Î±,Î²?

Solution

(d) Dimensions of Î²

Also Î±=force ×density

=[MLT

=[M

(d) Dimensions of Î²

^{3}=dimensions of density=[ML^{-3}] Î²=[〖M^{(1/3)}L^{-1}]Also Î±=force ×density

=[MLT

^{-2}][ML^{-3}]=[M

^{2}L^{-2}T^{-2}]**Q5.**Hertz is the unit for

Solution

(a)Frequency

(a)Frequency

**Q6.**If S=1/3 ft

^{3},f has the dimensions of

Solution

(a) Here, [f]=([S])/[t

(a) Here, [f]=([S])/[t

^{3}] =[M^{0}LT^{-3}].**Q7.**If x=a-b, then the maximum percentage error in the measurement of x will be

Solution

(a) Maximum absolute error is Î”a+Î”b. Now work out the relative error ad finally the percentage error.

(a) Maximum absolute error is Î”a+Î”b. Now work out the relative error ad finally the percentage error.

**Q8.**If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

Solution

(d) As v=4/3 Ï€r

(d) As v=4/3 Ï€r

^{3}dv/v=3(dr/r) ∴ Percentage error in determination of volume =3 (Percentage error in measurement of radius) =3(2%)=6%**Q9.**Which pair has the same dimensions

Solution

(c) Impulse = change in momentum so dimensions of both quantities will be same and equal to MLT

(c) Impulse = change in momentum so dimensions of both quantities will be same and equal to MLT

^{-1}**Q10.**The value of Planck’s constant is

Solution

(a) 6.63 ×10

(a) 6.63 ×10

^{-34}J-sec