Physical World Units and Measurements - Result Question 23
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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
======= ####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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/physical-world-units-and-measurements/physical-world-units-and-measurements—result-question-23.md (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.
