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 ratio of radius of III and IV Bohr's orbits in hydrogen atom is:
Solution
r_n=r_1×n^2
r_n=r_1×n^2
Q2.Which has minimum number of unpaired d-electrons?
Solution
Co^(2+) has 1s^2,2s^2 2p^6,3s^2 3p^6 3d^7 configuration having 3 unpaired electron only,
Co^(2+) has 1s^2,2s^2 2p^6,3s^2 3p^6 3d^7 configuration having 3 unpaired electron only,
Q3. The number of d-electrons retained in Fe^(2+)(At. No. Fe=26) ions is
Solution
〖_26〗Fe=[Ar]3d^6 4s^2 Fe^(2+)(24 electrons)=[Ar]3d^6 4s^0
〖_26〗Fe=[Ar]3d^6 4s^2 Fe^(2+)(24 electrons)=[Ar]3d^6 4s^0
Q4. Positive charge in an atom is :
Solution
All the protons carrying +ve charge are present in nucleus
All the protons carrying +ve charge are present in nucleus
Q5.The difference between ions and atoms is of:
Solution
Ions have charge, different size and configuration than atom.
Ions have charge, different size and configuration than atom.
Q6. The nucleus and an atom can be assumed to be spherical. The radius of the nucleus of mass no. A is given by 1.25×10^(-13)×A^(1/3) cm. The atomic radius of atom is 1 â„«. If the mass no. is 64, the fraction of the atomic volume that is occupied by nucleus is:
Solution
76 (d) V_n/V_a =((43)Ï€r_n^3)/((43)Ï€r_a^3 )=(r_n^3)/(r_a^3 )=[1.25×10^(-13)×(64)^(1/3) ]^3/(10^(-8) )^3
76 (d) V_n/V_a =((43)Ï€r_n^3)/((43)Ï€r_a^3 )=(r_n^3)/(r_a^3 )=[1.25×10^(-13)×(64)^(1/3) ]^3/(10^(-8) )^3
Q7.Which of the following sets of quantum numbers is correct for an electron in 4f-orbital?
Solution
For n=4,l≠4, for n=l=3,m≠4
For n=4,l≠4, for n=l=3,m≠4
Q8.Electronic configuration of niobium (Nb=41) is:
Solution
No doubt in Cr it is 3d^5,4s^1; but in Nb it is 4d^4,5s^1.
No doubt in Cr it is 3d^5,4s^1; but in Nb it is 4d^4,5s^1.
Q9.The filling of 4p-sublevel starts in the element of atomic number:
Solution
At. no. 30 has configuration …3d^10,4s^2 and thus, 31 has …3d^10,4s^2 4p^1
At. no. 30 has configuration …3d^10,4s^2 and thus, 31 has …3d^10,4s^2 4p^1
Q10. The energy of the electron at infinite distance from the nucleus in Bohr's model is taken as :
Solution
The energy level increase with increase in distance from the nucleus and the negative values of electrons energy near to nucleus decrease to zero at infinite distance.
The energy level increase with increase in distance from the nucleus and the negative values of electrons energy near to nucleus decrease to zero at infinite distance.