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Atoms and Nuclei question

2017 · 9 Apr · Shift 1 · Q52
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Atoms and Nuclei question

2017 · 9 Apr · Shift 1 · Q52

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Imagine that a reactor converts all given mass into energy and that it operates at a power level of 109 watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is 3×108 m/s)
  1. A
    0.96 gm
  2. B
    0.8 gm
  3. C
    4 ×\times× 10 −-− 2 gm
  4. D
    6.6 ×\times× 10 −-− 5 gm
View written solutionFree

Correct answer: C

  1. Use mass-energy equivalence

    If all mass is converted into energy, then E=mc2E = mc^2E=mc2

    Power is energy produced per unit time: P=EtP = \frac{E}{t}P=tE​

    So for time ttt, the energy produced is E=PtE = PtE=Pt

    Hence the mass consumed is m=Ec2=Ptc2m = \frac{E}{c^2} = \frac{Pt}{c^2}m=c2E​=c2Pt​

  2. Substitute the given values

    Given: P=109 WP = 10^9\ \text{W}P=109 W t=1 hour=3600 st = 1\ \text{hour} = 3600\ \text{s}t=1 hour=3600 s c=3×108 m/sc = 3 \times 10^8\ \text{m/s}c=3×108 m/s

    Therefore, m=109×3600(3×108)2m = \frac{10^9 \times 3600}{(3 \times 10^8)^2}m=(3×108)2109×3600​

  3. Simplify

    First compute c2c^2c2: c2=9×1016c^2 = 9 \times 10^{16}c2=9×1016

    So, m=3.6×10129×1016m = \frac{3.6 \times 10^{12}}{9 \times 10^{16}}m=9×10163.6×1012​

    m=0.4×10−4m = 0.4 \times 10^{-4}m=0.4×10−4

    m=4×10−5 kgm = 4 \times 10^{-5}\ \text{kg}m=4×10−5 kg

  4. Convert into grams

    Since 1 kg=103 g1\ \text{kg} = 10^3\ \text{g}1 kg=103 g, m=4×10−5×103 gm = 4 \times 10^{-5} \times 10^3\ \text{g}m=4×10−5×103 g

    m=4×10−2 gm = 4 \times 10^{-2}\ \text{g}m=4×10−2 g

  5. Match with the options

    4×10−2 g4 \times 10^{-2}\ \text{g}4×10−2 g

    This corresponds to Option C.

  6. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    They match.

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