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Question 45 of 61

Q.(a)

(i) Write the applications of the Dimensional Analysis.
(ii) Check the correctness of the equation (1/2) m v^2 = mgh using dimensional analysis method. OR
(b) Obtain an expression for the surface tension of a liquid by capillary rise method.
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022Subjective· 5mImportance★★★★★
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(i) Dimensional analysis has three main uses: verifying equations, deriving formulas, and converting units. (ii) Checking (1/2)mv² = mgh by comparing dimensions on both sides shows they are equal (ML²T^-2 = ML²T^-2), confirming the equation is dimensionally consistent.

(i) Applications of Dimensional Analysis:

  1. To check the correctness of a physical equation (Principle of Homogeneity of Dimensions): an equation can only be physically correct if the dimensions of every term on both sides are identical; dimensional analysis is a quick way to catch obviously wrong formulas (though it cannot catch errors in pure numbers/dimensionless constants).

  2. To derive a relationship between physical quantities: if we know (or can guess) which quantities a physical quantity depends on, dimensional analysis (by comparing powers of M, L, T on both sides) can be used to derive the form of the relating equation, up to an unknown dimensionless constant.

  3. To convert a physical quantity's value from one system of units to another (e.g., from CGS to SI), using the fact that the numerical value × unit must remain the same physical quantity, and dimensions tell us exactly how each unit scales.

(ii) Checking (1/2)mv² = mgh by dimensional analysis:

LHS = (1/2) m v²

Dimensions: [M] × [L T^-1]² = [M] × [L² T^-2] = [M L² T^-2]

(Note: the pure number 1/2 is dimensionless and does not affect the dimensional formula.)

RHS = m g h

Dimensions: [M] × [L T^-2] × [L] = [M L² T^-2]

Comparing:

LHS dimensions = M L² T^-2 …

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