SATHEE: Physical World Units and Measurements

Physical World Units and Measurements - Result Question 23

25. An equation is given as : $(P + \frac{a}{V^{2}}) = b \frac{θ}{V}$ where $P =$ Pressure, $V =$ Volume $Misplaced &$ Absolute temperature. If $a$ and $b$ are constants, then dimensions of $a$ will be

(a) $[M L^{5} T^{- 2}]$

(b) $[M^{- 1} L^{5} T^{2}]$

(d) $[M L^{5} T^{1}]$

[1996]

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Answer:

Correct Answer: 25. (a)

Solution:

(a) $(P + \frac{a}{V^{2}}) = b \frac{θ}{V}$

According to the principle of homogeneity quantity with same dimension can be added or subtracted.

Hence, Dimension of $P =$ Dimension of $\frac{a}{V^{2}}$

$\Rightarrow$ Dimension of $\frac{Force}{Area} =$ Dimension of $\frac{a}{V^{2}}$

$\Rightarrow [\frac{M L T^{- 2}}{L^{2}}] = \frac{a}{[L^{3}]^{2}} \Rightarrow a = [M L^{5} T^{- 2}]$

To get the dimensions of physical constant, we write any formula or equation incorporating the given constant and then by substituting the dimensional formula of all other quantities, we can find the dimensions of the required constant or coefficients.

Physical World Units and Measurements - Result Question 23

25. An equation is given as : $(P + \frac{a}{V^{2}}) = b \frac{θ}{V}$ where $P =$ Pressure, $V =$ Volume $Misplaced &$ Absolute temperature. If $a$ and $b$ are constants, then dimensions of $a$ will be

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Physical World Units and Measurements - Result Question 23

25. An equation is given as : (P+aV2)=bθV where P= Pressure, V= Volume Misplaced &Misplaced & Absolute temperature. If a and b are constants, then dimensions of a will be

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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25. An equation is given as : $(P + \frac{a}{V^{2}}) = b \frac{θ}{V}$ where $P =$ Pressure, $V =$ Volume $Misplaced &$ Absolute temperature. If $a$ and $b$ are constants, then dimensions of $a$ will be