JEE Advanced Past Questions: Units and Measurements

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Q1. Plank’s constant has dimension ………….. (1985 2M)

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Q2. In the formula X = 3 YZ², X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are ……….. (1988 2M)

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Q3. The dimensions of Electrical conductivity are ………… (1997 2M)

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Q4. The equation of state for real gas is given by (P + a/V²)(V – b) = RT. The dimensions of the constant ‘a’ is ……….. (1997 2M)

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Solutions:

Solution 01:

We have E = hν

∴ h = E ν^-1

∴ [h] = [M¹L²T^-2] [T^-1]^-1

∴ [h] = [M¹L²T^-2] [T¹]

∴ [h] = [M¹L²T^-1]

Ans: Plank’s constant has dimension [M¹L²T^-1]

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Solution 02:

Given [X] = [C] = [M^-1L^-2T²Q²]

Given [Z] = [B] = [MT^-1Q^-1]

Given X = 3 YZ²

Y = (1/3) X Z^-2

[Y] = [M^-1L^-2T²Q²] [MT^-1Q^-1]^-2

[Y] = [M^-1L^-2T²Q²] [M^-2T²Q²]

[Y] = [M^-3L^-2T⁴Q⁴]

Ans: In the formula X = 3 YZ², X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are [M^-3L^-2T⁴Q⁴]

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Solution 03:

We have

[ρ] = [R] [A] [l]^-1

[ρ] = [M L² T^-3 A^-2] [ L² ] [ L]^-1

[ρ] = [M L³ T^-3 A^-2]

Ans: The dimensions of Electrical conductivity are [M L³ T^-3 A^-2]

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Solution 04:

Due to addition between P and a/V², their dimensions should be same

[P] = [a/V²]

[a] = [P] [V]² = [ML^-1T^-2] [L³]²

[a] = [ML^-1T^-2] [L⁶] = [ML⁵T^-2]

Ans: The equation of state for real gas is given by (P + a/V²)(V – b) = RT. The dimensions of the constant ‘a’ is [ML⁵T^-2]

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