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.
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**Q1.**What is the dimensional formula of &(planck' s constant)/(linear momentum) ?

Solution

([planck' s constant])/([linear momentum])=([ML

([planck' s constant])/([linear momentum])=([ML

^{1}T^{-1}])/([MLT^{-1}])=[M^{0}LT^{0}]**Q2.**The dimensional formula of coefficient of permittivity for free space (Îµ

_{0}) in the equation F=1/(4Ï€Îµ

_{0}) (q

_{1}q

_{2})/r

^{2}, where symbols have their usual meanings, is

Solution

[Îµ

[Îµ

_{0}]=([A^{2}T^{2}])/([ML^{3}T^{-2}])=[M^{-1}L^{-3}T^{4}A^{2}]**Q3.**The surface tension of a liquid is 70 dyne/cm, in MKS system value is

Solution

1 dyne=10

70 dyne/cm=(70×10

=7×10

1 dyne=10

^{-5}newton,1 cm=10^{-2}m70 dyne/cm=(70×10

^{-5}/10^{-2}N/m=7×10

^{-2}N/m

**Q4.**If unit of length, mass and time each be doubled, the unit of work done is increased by

Solution

n

Given, M

∴n

n

_{2}=n_{1}[M_{2}/M_{1}]^{1}[L_{2}/L_{1}]^{2}[T_{2}/T_{1}]^{-2}Given, M

_{2}=2M_{1},L_{2}=2L_{1},T_{2}=2T_{1}∴n

_{2}=n_{1}[2]^{1}[2]^{2}[2]^{-2}=2n_{1}**Q5.**Unit of impulse is

Solution

Impulse=Force×time=(kg-m/s

Impulse=Force×time=(kg-m/s

^{2})×s=kg-m/s**Q6.**The constant of proportionality 1/(4Ï€Îµ

_{0}) in Coulomb’s law has the following units

Solution

From Coulomb’s law

F=1/(4Ï€Îµ

From Coulomb’s law

F=1/(4Ï€Îµ

_{0}) (q_{1}q_{2}/r^{2}⇒[1/(4Ï€Îµ_{0})]=([F ×r^{2}])/[q]^{2}=([newton] [metre]^{2})/[coulomb]^{2}=Nm^{2}C^{-2}**Q7.**From the dimensional consideration, which of the following equation is correct

Solution

By substituting the dimensions in T=2Ï€√R

By substituting the dimensions in T=2Ï€√R

^{3}/GM we get √L^{3}/(M^{-1}L^{3}T^{-2}×M)=T**Q8.**The length, breadth and thickness of a block are given by l=12cm,b=6 cm and t=2.45cm The volume of block according to the idea of significant figures should be

Solution

Volume V=l×b×t =12×6×2.45=176.4 cm

Volume V=l×b×t =12×6×2.45=176.4 cm

^{3}V=1.764×10^{2}cm^{3}Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V=2×10^{2}cm^{3}**Q9.**A force F is given by F=at+bt

^{2}, where t is time. What are the dimensions of a and b

Solution

From the principle of dimensional homogeneity [a]=[F/t]=[MLT

From the principle of dimensional homogeneity [a]=[F/t]=[MLT

^{-3}] and [b]=[F/t^{2}]=[MLT^{-4}]**Q10.**Which of the following is not a unit of energy

Solution

Kg-m/sec is the unit of linear momentum

Kg-m/sec is the unit of linear momentum