Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) The density of the copper (binom6429mathrmCu) nucleus is greater than that of the carbon (binom126mathrmC) nucleus. [cite: 20] Reason (R): The nucleus of mass number A has a radius proportional to mathrmA^1/3. [cite: 21] In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 22]

Solution & Explanation

### Related Formula R = R_0 A^1/3 [cite: 667] rho = fractextMasstextVolume = fracm_n Afrac43pi R^3 [cite: 664] ### Core Logic Substituting the expression for radius R into the density equation: [cite: 664] rho = fracm_n Afrac43pi (R_0 A^1/3)^3 = fracm_n Afrac43pi R_0^3 A = fracm_nfrac43pi R_0^3 [cite: 664] As observed, the mass number A cancels out perfectly, implying that the density of all nuclei is roughly identical and constant regardless of their mass numbers[cite: 664, 666]. Thus, the nuclear density of copper is equal to that of carbon, meaning Assertion (A) is incorrect[cite: 20, 663]. Reason (R) is correct since R propto A^1/3 is a foundational empirical law of nuclear physics[cite: 21, 668]. ### Pattern Recognition Nuclear mass scales with A, while volume scales with R^3 propto (A^1/3)^3 = A[cite: 664]. Therefore, textDensity propto fracAA = textconstant[cite: 664, 666]. Always look out for options asserting varying nuclear densities across heavy vs light elements—it is a common trap. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

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Q60 jee_main_2024_31_jan_evening Nuclear Size and Density
A nucleus has mass number A_1 and volume V_1. Another nucleus has mass number A_2 and volume V_2. If relation between mass number is A_2 = 4A_1, then fracV_2V_1 = ________.
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula R = R_0 A^1/3 V = frac43pi R^3 ### Core Logic Since radius R is proportional to A^1/3, the volume V (which depends on R^3) will be directly proportional to the mass number A. ### Step 1: Show Proportionality V = frac43pi (R_0 A^1/3)^3 = frac43pi R_0^3 A This proves that V propto A. ### Step 2: Calculate Ratio fracV_2V_1 = fracA_2A_1 Given that A_2 = 4A_1: fracV_2V_1 = frac4A_1A_1 = 4 ### Pattern Recognition Nuclear density is constant for all nuclei. Therefore, Mass propto Volume. Since Mass number (A) represents mass, Volume is strictly linearly proportional to Mass number. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei
Q60 jee_main_2024_31_jan_morning Mass Defect And Energy
The mass defect in a particular reaction is 0.4mathrm\ g. The amount of energy liberated is n times 10^7mathrm\ kWh where n = _______. (speed of light = 3 times 10^8mathrm\ m/s)
Numerical Answer. Answer: 1 to 1

Solution

### Related Formula E = Delta m c^2 1 text kWh = 3.6 times 10^6 text J ### Core Logic Given the mass defect: Delta m = 0.4mathrm\,g = 0.4 times 10^-3mathrm\,kg The total energy liberated in Joules is: E = (0.4 times 10^-3) times (3 times 10^8)^2 E = 0.4 times 10^-3 times 9 times 10^16 E = 3.6 times 10^13mathrm\,J ### Step 2: Conversion to kWh We need the answer in mathrmkWh. Since 1mathrm\,kWh = 1000mathrm\,W times 3600mathrm\,s = 3.6 times 10^6mathrm\,J: E = frac3.6 times 10^133.6 times 10^6mathrm\,kWh E = 10^7mathrm\,kWh Comparing this to n times 10^7mathrm\,kWh, we get: n = 1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

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