Physical World Units and Measurements - Result Question 23
25. An equation is given as : $(P+\frac{a}{V^{2}})=b \frac{\theta}{V}$ where $P=$ Pressure, $V=$ Volume $& \theta=$ Absolute temperature. If $a$ and $b$ are constants, then dimensions of $a$ will be
(a) $[ML^{5} T^{-2}]$
(c) $[ML^{-5} T^{-1}]$
(b) $[M^{-1} L^{5} T^{2}]$
(d) $[ML^{5} T^{1}]$
[1996]
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Answer:
Correct Answer: 25. (a)
Solution:
- (a) $(P+\frac{a}{V^{2}})=b \frac{\theta}{V}$
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 $\frac{\text{ Force }}{\text{ Area }}=$ Dimension of $\frac{a}{V^{2}}$
$\Rightarrow[\frac{MLT^{-2}}{L^{2}}]=\frac{a}{[L^{3}]^{2}} \Rightarrow a=[M L^{5} T^{-2}]$
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.
