BrainIt

Physics · chapter PYQs

Nuclei, on the exam screen.

5 questions from NEET 2026, Re-NEET 2026, NEET 2020, 5 minutes, marked +4 / −1. See them with solutions.

Nuclei · PYQs

Candidate: You

Time Left

Q1/5

  1. Question 1 · NEET 2026 · Q31 · +4 / −1

    Four statements are given (AA is mass number). Choose the correct answer from the options given below:

    Statement I: The volume of a nucleus is proportional to A1/3A^{1/3}.
    Statement II: The volume of a nucleus is proportional to AA.
    Statement III: The difference in mass of an atom and its nucleus is called the mass defect.
    Statement IV: The difference in mass of a nucleus and its constituents is called the mass defect.

    Answer and explanation

    Answer: (4)

    Nuclear radius r=r0A1/3r = r_0 A^{1/3}, so volume ∝r3∝A\propto r^3 \propto A (statement II correct, I wrong). The mass defect is the difference between the total mass of the constituent nucleons and the actual mass of the nucleus (statement IV correct, III wrong). Hence only II and IV are true.

  2. Question 2 · NEET 2026 · Q32 · +4 / −1

    An unknown nucleus has a nuclear density of 2.29×10172.29 \times 10^{17} kg/m3^3 and mass of 19.926×10−2719.926 \times 10^{-27} kg. Its mass number AA is approximately: (Take R0=1.2×10−15R_0 = 1.2 \times 10^{-15} m, 4π=12.564\pi = 12.56)

    Answer and explanation

    Answer: (1)

    Using density = mass/volume with volume =43πR3= \frac{4}{3}\pi R^3 and R=R0A1/3R = R_0 A^{1/3}, we get 43πR03A=mρ\frac{4}{3}\pi R_0^3 A = \frac{m}{\rho}. Solving, A=mρ⋅34πR03=19.926×10−27×32.29×1017×12.56×(1.2×10−15)3≈12A = \frac{m}{\rho}\cdot\frac{3}{4\pi R_0^3} = \frac{19.926\times10^{-27}\times3}{2.29\times10^{17}\times12.56\times(1.2\times10^{-15})^3} \approx 12.

  3. Question 3 · Re-NEET 2026 · Q7 · +4 / −1

    Consider the following nuclear reaction 238U→234Th+4He^{238}\mathrm{U} \rightarrow ^{234}\mathrm{Th} + ^{4}\mathrm{He}. Take masses of 238U^{238}\mathrm{U}, 234Th^{234}\mathrm{Th} and 4He^{4}\mathrm{He} as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: 1 u = 931.5 MeV c−2c^{-2}]

    Answer and explanation

    Answer: (1)

    The Q value is the mass defect converted to energy: Δm=238.050−(234.043+4.003)=0.004 u\Delta m = 238.050 - (234.043 + 4.003) = 0.004\ \text{u}. Multiplying by the conversion factor, Q=0.004×931.5=3.726 MeV=3726 keVQ = 0.004 \times 931.5 = 3.726\ \text{MeV} = 3726\ \text{keV}. The positive value confirms the decay is spontaneous. NCERT Class 12, Nuclei (mass-energy and nuclear reactions).

  4. Question 4 · NEET 2020 · Q151 · +4 / −1

    When a uranium isotope 92235^{235}_{92}U is bombarded with a neutron, it generates 3689^{89}_{36}Kr, three neutrons and

    Answer and explanation

    Answer: (4)

    The nuclear reaction is 92235U+01n→3689Kr+3 01n+56144Ba^{235}_{92}U + ^1_0n → ^{89}_{36}Kr + 3\ ^1_0n + ^{144}_{56}Ba. The mass and charge numbers must be conserved, so the remaining product is 56144Ba^{144}_{56}Ba. Chapter: Nuclei.

  5. Question 5 · NEET 2020 · Q174 · +4 / −1

    The energy equivalent of 0.5 g of a substance is

    Answer and explanation

    Answer: (1)

    E=mc2=0.5×10−3×(3×108)2=4.5×1013E = mc^{2} = 0.5 \times 10^{-3} \times (3 \times 10^{8})^{2} = 4.5 \times 10^{13} J. Chapter: Nuclei.