The Question is from neet-physics , and it was created by the author VENKATESWARARAO

Question

Given the equation $P = \frac{a}{V} - \frac{b}{V^2} + cT$, where $P$ is pressure, $V$ is volume, and $T$ is temperature.  What are the dimensions of $abc$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Dimensional Analysis
	**Related Topics**: Units and Measurements, Physical Quantities
	**Prerequisites**: Understanding of basic physical quantities (pressure, volume, temperature) and their dimensions.
## SOLUTION PATHWAY
**Given Information**:
	* Equation: $P = \frac{a}{V} - \frac{b}{V^2} + cT$
	* $P$ is pressure, $V$ is volume, $T$ is temperature
**Method & Steps**:
	1.	Determine the dimensions of pressure, volume and temperature.
		Reasoning: We need to express these quantities in terms of fundamental dimensions: mass (M), length (L), time (T), and temperature (Θ).
		Formula/Rule: Pressure (P) = Force / Area = (MLT⁻²) / L² = ML⁻¹T⁻², Volume (V) = L³, Temperature (T) = Θ
	2.	Determine the dimensions of a, b, and c.
		Working: From the equation, $\frac{a}{V}$ must have the same dimensions as $P$. Thus, $[a] = [P][V] = (ML^{-1}T^{-2})(L^3) = ML^2T^{-2}$. Similarly, $\frac{b}{V^2}$ has the same dimensions as $P$, so $[b] = [P][V^2] = (ML^{-1}T^{-2})(L^6) = ML^5T^{-2}$.  Finally, $cT$ has the same dimensions as $P$, so $[c] = \frac{[P]}{[T]} = \frac{ML^{-1}T^{-2}}{Θ} = ML^{-1}T^{-2}Θ^{-1}$.
		Key Point: Each term in an addition or subtraction must have the same dimensions.
	3.	Determine the dimensions of abc.
		Result: $[abc] = [a][b][c] = (ML^2T^{-2})(ML^5T^{-2})(ML^{-1}T^{-2}Θ^{-1}) = M^3L^6T^{-6}Θ^{-1}$
		Verification: The resulting dimension matches option A.
 ## OPTION ANALYSIS
[A] ✓: The dimensions of $abc$ were calculated to be $M^3L^6T^{-6}Θ^{-1}$, which matches this option.
[B] :x: This option does not match the calculated dimension for $abc$.
[C] :x: This option does not match the calculated dimension for $abc$.
[D] :x: This option does not match the calculated dimension for $abc$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly determining the dimensions of pressure. Make sure to use the definition of pressure as force per unit area.
		2.[Error 2]: Not ensuring all terms in the equation have the same dimensions. Each term must have the same dimensions as pressure.
	**Quick Tips**:
		1.[Helpful shortcut]: Always start by writing down the fundamental dimensions of basic physical quantities like pressure, volume, and temperature.
		2.[Memory aid]: Remember that addition and subtraction can only be done with quantities of the same dimension.

VENKATESWARARAO · Answer

The correct Option is "M^3L^6T^{-6}Θ^{-1}"

VENKATESWARARAO · Answer

M^2L^5T^{-4}Θ^{-1}

VENKATESWARARAO · Answer

M^2L^4T^{-5}Θ

VENKATESWARARAO · Answer

M^3L^7T^{-5}Θ^{-1}

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