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. Heaviest particle is:
Solution
Neutron has more mass among all
Neutron has more mass among all
Q2.Beryllium’s fourth electron will have the four quantum numbers:
n l m s
Solution
4th electron of Be is in 2s-subshell.
4th electron of Be is in 2s-subshell.
Q3. Which set of quantum numbers is possible for the last electron of Mg^+ ion?
Solution
Last electron of Mg^+ is 3s^1.
Last electron of Mg^+ is 3s^1.
Q4. For which of the following, the radius will be same as for hydrogen atom having n=1?
Solution
〖_(r_2 )〗Be^(3+)=(r_1 H)/4×2^2 (∵r_2 H=r_(1 H)×2^2 and _(r_n ) Be^(3+)=(r_n H)/n)
〖_(r_2 )〗Be^(3+)=(r_1 H)/4×2^2 (∵r_2 H=r_(1 H)×2^2 and _(r_n ) Be^(3+)=(r_n H)/n)
Q5.Evidence for the existence of different energy levels in atom is supplied by:
Solution
Spectral lines of different λ suggest for different energy levels.
Spectral lines of different λ suggest for different energy levels.
Q6. Potassium ion is isoelectronic with the atom of :
Solution
K^+ and Ar both have 18 electrons.
K^+ and Ar both have 18 electrons.
Q8.The electronic transitions from n=2 to n=1 will produce shortest wavelength in (where n=principle quantum state)
Solution
1/λ=Z^2.R_H [1/(n_1^2 )-1/(n_2^2 )] ⟹1/λ=(Z)^2.R_H {1/1-1/4}=3/4 R_H Z^2 ∴λ∝1/Z^2 Hence for shortest λ,Z must be maximum, which is for Li^(2+).
1/λ=Z^2.R_H [1/(n_1^2 )-1/(n_2^2 )] ⟹1/λ=(Z)^2.R_H {1/1-1/4}=3/4 R_H Z^2 ∴λ∝1/Z^2 Hence for shortest λ,Z must be maximum, which is for Li^(2+).
Q9.X-rays and γ-rays of same energies may be distinguished by:
Solution
Both have different modes of preparation.
Both have different modes of preparation.
Q10. Which one is not true for the cathode rays ?
Solution
Cathode rays are fastly moving electrons.
Cathode rays are fastly moving electrons.