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Units and Measurements question

2023 · Shift 2 · Q36
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Units and Measurements question

2023 · Shift 2 · Q36

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −1
Young's modulus of elasticity YYY is expressed in terms of three derived quantities, namely, the gravitational constant GGG, Planck's constant hhh and the speed of light ccc, as Y=cαhβGγY=c^\alpha h^\beta G^\gammaY=cαhβGγ. Which of the following is the correct option?
  1. A
    α=7,β=−1,γ=−2\alpha=7, \beta=-1, \gamma=-2α=7,β=−1,γ=−2
  2. B
    α=−7,β=−1,γ=−2\alpha=-7, \beta=-1, \gamma=-2α=−7,β=−1,γ=−2
  3. C
    α=7,β=−1,γ=2\alpha=7, \beta=-1, \gamma=2α=7,β=−1,γ=2
  4. D
    α=−7,β=1,γ=−2\alpha=-7, \beta=1, \gamma=-2α=−7,β=1,γ=−2
View written solutionFree

Correct answer: A

This problem requires us to use the principle of dimensional homogeneity, which states that for an equation to be physically correct, the dimensions of all its terms must be the same. The given equation is Y=cαhβGγY=c^\alpha h^\beta G^\gammaY=cαhβGγ.

Step 1: Determine the dimensions of each physical quantity.

  1. Young's Modulus (Y): Young's modulus is defined as the ratio of stress to strain. Y=StressStrainY = \frac{\text{Stress}}{\text{Strain}}Y=StrainStress​ Stress is force per unit area, and strain is dimensionless. [Force]=[MLT−2][\text{Force}] = [M L T^{-2}][Force]=[MLT−2] [Area]=[L2][\text{Area}] = [L^2][Area]=[L2] [Stress]=[MLT−2][L2]=[ML−1T−2][\text{Stress}] = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}][Stress]=[L2][MLT−2]​=[ML−1T−2] Therefore, the dimensional formula for Young's modulus is [Y]=[M1L−1T−2][Y] = [M^1 L^{-1} T^{-2}][Y]=[M1L−1T−2].

  2. Speed of light (c): The speed of light has the dimensions of velocity. [c]=[LT−1][c] = [L T^{-1}][c]=[LT−1]

  3. Planck's constant (h): From the energy-frequency relation E=hfE = hfE=hf, where EEE is energy and fff is frequency. [h]=[E][f]=[ML2T−2][T−1]=[ML2T−1][h] = \frac{[E]}{[f]} = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}][h]=[f][E]​=[T−1][ML2T−2]​=[ML2T−1]

  4. Gravitational constant (G): From Newton's law of gravitation, F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}F=Gr2m1​m2​​. [G]=[F][r2][m1][m2]=[MLT−2][L2][M][M]=[M−1L3T−2][G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[M L T^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}][G]=[m1​][m2​][F][r2]​=[M][M][MLT−2][L2]​=[M−1L3T−2]

Step 2: Set up the dimensional equation.

Substitute the dimensional formulas into the given equation Y=cαhβGγY=c^\alpha h^\beta G^\gammaY=cαhβGγ:

[Y]=[c]α[h]β[G]γ[Y] = [c]^\alpha [h]^\beta [G]^\gamma[Y]=[c]α[h]β[G]γ [M1L−1T−2]=([LT−1])α([ML2T−1])β([M−1L3T−2])γ[M^1 L^{-1} T^{-2}] = ([L T^{-1}])^\alpha ([M L^2 T^{-1}])^\beta ([M^{-1} L^3 T^{-2}])^\gamma[M1L−1T−2]=([LT−1])α([ML2T−1])β([M−1L3T−2])γ

Step 3: Simplify and equate the powers of M, L, and T.

Combine the terms on the right-hand side:

[M1L−1T−2]=(LαT−α)(MβL2βT−β)(M−γL3γT−2γ)[M^1 L^{-1} T^{-2}] = (L^\alpha T^{-\alpha}) (M^\beta L^{2\beta} T^{-\beta}) (M^{-\gamma} L^{3\gamma} T^{-2\gamma})[M1L−1T−2]=(LαT−α)(MβL2βT−β)(M−γL3γT−2γ) [M1L−1T−2]=Mβ−γLα+2β+3γT−α−β−2γ[M^1 L^{-1} T^{-2}] = M^{\beta-\gamma} L^{\alpha+2\beta+3\gamma} T^{-\alpha-\beta-2\gamma}[M1L−1T−2]=Mβ−γLα+2β+3γT−α−β−2γ

Now, equate the exponents of the fundamental dimensions M, L, and T on both sides:

  1. For Mass (M): 1=β−γ1 = \beta - \gamma1=β−γ (Equation i)
  2. For Length (L): −1=α+2β+3γ-1 = \alpha + 2\beta + 3\gamma−1=α+2β+3γ (Equation ii)
  3. For Time (T): −2=−α−β−2γ-2 = -\alpha - \beta - 2\gamma−2=−α−β−2γ (Equation iii)

Step 4: Solve the system of linear equations.

Add Equation (ii) and Equation (iii): (−1)+(−2)=(α+2β+3γ)+(−α−β−2γ)(-1) + (-2) = (\alpha + 2\beta + 3\gamma) + (-\alpha - \beta - 2\gamma)(−1)+(−2)=(α+2β+3γ)+(−α−β−2γ) −3=β+γ-3 = \beta + \gamma−3=β+γ (Equation iv)

Now we have a simpler system of two equations for β\betaβ and γ\gammaγ:

β−γ=1\beta - \gamma = 1β−γ=1 (from Equation i) β+γ=−3\beta + \gamma = -3β+γ=−3 (from Equation iv)

Adding these two equations: (β−γ)+(β+γ)=1+(−3)( \beta - \gamma ) + ( \beta + \gamma ) = 1 + (-3)(β−γ)+(β+γ)=1+(−3) 2β=−22\beta = -22β=−2 β=−1\beta = -1β=−1

Substitute β=−1\beta = -1β=−1 into Equation (i): −1−γ=1-1 - \gamma = 1−1−γ=1 −γ=2-\gamma = 2−γ=2 γ=−2\gamma = -2γ=−2

Finally, substitute the values of β\betaβ and γ\gammaγ into Equation (iii) to find α\alphaα: −2=−α−(−1)−2(−2)-2 = -\alpha - (-1) - 2(-2)−2=−α−(−1)−2(−2) −2=−α+1+4-2 = -\alpha + 1 + 4−2=−α+1+4 −2=−α+5-2 = -\alpha + 5−2=−α+5 α=5+2\alpha = 5 + 2α=5+2 α=7\alpha = 7α=7

So, the values are α=7,β=−1,γ=−2\alpha=7, \beta=-1, \gamma=-2α=7,β=−1,γ=−2.

Step 5: Compare with the given options.

The calculated values α=7,β=−1,γ=−2\alpha=7, \beta=-1, \gamma=-2α=7,β=−1,γ=−2 match option A.

Option A: α=7,β=−1,γ=−2\alpha=7, \beta=-1, \gamma=-2α=7,β=−1,γ=−2 (Correct) Option B: α=−7,β=−1,γ=−2\alpha=-7, \beta=-1, \gamma=-2α=−7,β=−1,γ=−2 (Incorrect) Option C: α=7,β=−1,γ=2\alpha=7, \beta=-1, \gamma=2α=7,β=−1,γ=2 (Incorrect) Option D: α=−7,β=1,γ=−2\alpha=-7, \beta=1, \gamma=-2α=−7,β=1,γ=−2 (Incorrect)

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