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Question

The speed of sound in a medium depends on the modulus of elasticity $E$ and density $\rho$ of the medium.  Using dimensional analysis, determine the relationship between these quantities.

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option C
## CONCEPT FRAMEWORK
	**Primary Concept**: Dimensional Analysis
	**Related Topics**: Units and Dimensions, Physical Quantities
	**Prerequisites**: Understanding of fundamental dimensions (length, mass, time), and the concept of dimensional homogeneity.
## SOLUTION PATHWAY
**Given Information**:
	* Speed of sound, $v$
	* Modulus of elasticity, $E$
	* Density, $ho$
	*  We want to find a relation of the form: $v \propto E^a ho^b$
**Method & Steps**:
	1.  Write down the dimensions of each quantity:
		Reasoning: We need to express each physical quantity in terms of fundamental dimensions (M, L, T).
		Formula/Rule: Speed: $[v] = LT^{-1}$, Modulus of Elasticity: $[E] = ML^{-1}T^{-2}$, Density: $[ho] = ML^{-3}$
	2.  Substitute the dimensions into the proportionality relation:
		Working: $LT^{-1} \propto (ML^{-1}T^{-2})^a (ML^{-3})^b$. This gives us: $LT^{-1} \propto M^{a+b}L^{-a-3b}T^{-2a}$
		Key Point: For dimensional homogeneity, the powers of M, L, and T on both sides must be equal.
	3.  Equate the exponents of M, L, and T:
		Working: From the exponents of M: $0 = a+b$ ; From the exponents of L: $1 = -a - 3b$; From the exponents of T: $-1 = -2a$
	4.  Solve the system of equations:
		Working: From -1 = -2a, we get a = 1/2. Substituting into 0 = a + b, we get 0 = 1/2 + b => b = -1/2. Let's verify if the exponents of L holds true: 1 = -1/2 - 3(-1/2) = -1/2 + 3/2 = 2/2 = 1. This is correct.
	5.  Write the relationship:
		Result: $v \propto E^{1/2} ho^{-1/2}$ which simplifies to $v \propto \sqrt{\frac{E}{ho}}$
		Verification: The dimensions of the right hand side are $\sqrt{\frac{ML^{-1}T^{-2}}{ML^{-3}}} = \sqrt{L^2T^{-2}} = LT^{-1}$, which matches the dimensions of speed.
 ## OPTION ANALYSIS
[A] :x:  The dimensions of $\sqrt{Eho}$ are $\sqrt{(ML^{-1}T^{-2})(ML^{-3})} = \sqrt{M^2L^{-4}T^{-2}}=ML^{-2}T^{-1}$, which does not match the dimensions of speed.
[B] :x:  The dimensions of $\frac{E}{ho}$ are $\frac{ML^{-1}T^{-2}}{ML^{-3}} = L^2T^{-2}$, which does not match the dimensions of speed.
[C] ✓: The dimensions of $\sqrt{\frac{E}{ho}}$ are $\sqrt{\frac{ML^{-1}T^{-2}}{ML^{-3}}} = \sqrt{L^2T^{-2}} = LT^{-1}$, which matches the dimensions of speed.
[D] :x:  The dimensions of $\frac{ho}{E}$ are $\frac{ML^{-3}}{ML^{-1}T^{-2}} = L^{-2}T^{2}$, which does not match the dimensions of speed.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly determining the dimensions of physical quantities. [How to avoid]: Refer to standard tables or derive dimensions from fundamental definitions.
		2.[Error 2]: Making mistakes in solving the system of equations. [How to avoid]: Double-check your work and ensure all equations are used correctly.
	**Quick Tips**:
		1.[Helpful shortcut]: When using dimensional analysis, remember that constants do not affect dimensions.
		2.[Memory aid]: Use the mnemonic 'MLT' to remember the fundamental dimensions (Mass, Length, Time).

ABDUL JILANI BASHA · Answer

$v \propto \sqrt{E\rho}$

ABDUL JILANI BASHA · Answer

$v \propto \frac{E}{\rho}$

ABDUL JILANI BASHA · Answer

The correct Option is "$v \propto \sqrt{\frac{E}{\rho}}$"

ABDUL JILANI BASHA · Answer

$v \propto \frac{\rho}{E}$

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