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Units and Measurements question

2020 · 9 Jan · Shift 1 · Q67
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Units and Measurements question

2020 · 9 Jan · Shift 1 · Q67

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
A quantity f is given by f=hc5Gf = \sqrt {{{h{c^5}} \over G}}f=Ghc5​​ where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of :
  1. A
    Energy
  2. B
    Momentum
  3. C
    Area
  4. D
    Volume
View written solutionFree

Correct answer: A

  1. Write dimensions of each constant

    We use:

    [c]=LT−1[c] = LT^{-1}[c]=LT−1 [G]=M−1L3T−2[G] = M^{-1}L^3T^{-2}[G]=M−1L3T−2 [h]=Planck constant=energy×time[h] = \text{Planck constant} = \text{energy} \times \text{time}[h]=Planck constant=energy×time

    Since energy has dimension ML2T−2ML^2T^{-2}ML2T−2,

    [h]=ML2T−1[h] = ML^2T^{-1}[h]=ML2T−1

  2. Find dimension of c5c^5c5

    [c5]=(LT−1)5=L5T−5[c^5] = (LT^{-1})^5 = L^5T^{-5}[c5]=(LT−1)5=L5T−5

  3. Find dimension of hc5hc^5hc5

    [hc5]=(ML2T−1)(L5T−5)=ML7T−6[hc^5] = (ML^2T^{-1})(L^5T^{-5}) = ML^7T^{-6}[hc5]=(ML2T−1)(L5T−5)=ML7T−6

  4. Divide by GGG

    [hc5G]=ML7T−6M−1L3T−2\left[\frac{hc^5}{G}\right] = \frac{ML^7T^{-6}}{M^{-1}L^3T^{-2}}[Ghc5​]=M−1L3T−2ML7T−6​

    =M1−(−1)L7−3T−6−(−2)= M^{1-(-1)}L^{7-3}T^{-6-(-2)}=M1−(−1)L7−3T−6−(−2)

    =M2L4T−4= M^2L^4T^{-4}=M2L4T−4

  5. Take square root

    Given f=hc5Gf = \sqrt{\frac{hc^5}{G}}f=Ghc5​​

    So, [f]=M2L4T−4=ML2T−2[f] = \sqrt{M^2L^4T^{-4}} = ML^2T^{-2}[f]=M2L4T−4​=ML2T−2

  6. Identify the physical quantity

    ML2T−2ML^2T^{-2}ML2T−2 is the dimension of energy.

  7. Check options

    • A: Energy →\rightarrow→ correct
    • B: Momentum has dimension MLT−1MLT^{-1}MLT−1
    • C: Area has dimension L2L^2L2
    • D: Volume has dimension L3L^3L3

Therefore, the correct option is A.

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