**Dear Readers,**

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

**Q2. The half life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both them have the same number of atoms. Then**

^{x}k^{y}, where C is a dimensionless constant. The values of x and y are, respectively**Q3. Consider the following reaction ^1 H_2+ ^1 H_2→ _( 1) He^4+Q If m(_1 H^2)=2.0141 u;m(_2 He^4)=4.0024 u, the energy Q released (in MeV) in this fusion reaction is**

**Q4. Which of the following is a correct statement?**

Solution

(a) Both the beta rays and the cathode rays are made up of electrons. So, only option (a) is correct Gamma rays are electromagnetic waves Alpha particles are doubly ionized helium atoms Protons and neutrons have approximately the same mass Therefore, (b), (c) and (d) are wrong options

(a) Both the beta rays and the cathode rays are made up of electrons. So, only option (a) is correct Gamma rays are electromagnetic waves Alpha particles are doubly ionized helium atoms Protons and neutrons have approximately the same mass Therefore, (b), (c) and (d) are wrong options

**Q5. 90% of a radioactive sample is left undecayed after time t has elapsed. What percentage of the initial sample will decay in a total time 2t?;">**

Solution

5 (b) 90% of the sample is left undecayed after time t ∴9/10 N_0=N_0 e^(-Î»t) Î»=1/t ln(10/9) (i) After time 2t, N_c=N_0 e^(-Î»(2t) )=N_0 e^(-1/t [ln(10/9) ]2t) (ii) N=N_0 e^〖-ln〗〖(10/9)^2 〗 =N_0 (9/10)^2≈81% of N_0 (iii) Therefore, 19% of initial value will decay in time 2t

5 (b) 90% of the sample is left undecayed after time t ∴9/10 N_0=N_0 e^(-Î»t) Î»=1/t ln(10/9) (i) After time 2t, N_c=N_0 e^(-Î»(2t) )=N_0 e^(-1/t [ln(10/9) ]2t) (ii) N=N_0 e^〖-ln〗〖(10/9)^2 〗 =N_0 (9/10)^2≈81% of N_0 (iii) Therefore, 19% of initial value will decay in time 2t

**Q6. A physical quantity X is represented by X=(M**

^{x}L^{-y}T^{-z}). Binding energy per nucleon verses mass number curve for nuclei is shown in the figure. W,X,Y and Z are four nuclei indicated on the curve. The process that would release energy is
Solution

| 6 (c) Energy is released in a process when total binding energy (B.E.) of the nucleus is increased or we can say when total B.E. of products is more than the reactants. By calculation we can see that only in case of option (c), this happens Given W→2Y B.E. of reactants =120×7.5=900 MeV and B.E. of products =2×(60×8.5)=1020 MeV i.e.,B.E. of products > B.E. of reactants |

**Q7. The binding energies of nuclei X and Y are E_1 and E_2, respectively. Two atoms of X fuse to give one atom of Y and an energy Q is released. Then,**

Solution

7(b) During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence, Q=E2-〖2E〗_1

7(b) During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is released. Hence, Q=E2-〖2E〗_1

**Q8. A radioactive sample of U^238 decays to Pb through a process for which half life is 4.5×10^9 years. The ratio of number of nuclei of Pb to U^238 after a time of 1.5×10^9 years (given 2^(1/3)=1.26)**

**Q10.In hydrogen spectrum, the wavelength of HÎ± line is 656 nm, whereas in the spectrum of a distant galaxy, HÎ± line wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is**