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1.Units, Dimensions and Measurement
hard
$\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}$ સમીકરણમાં $\mathrm{ab}^{-1}$ નું પારમાણીક સૂત્ર શુ થશે? જ્યાં સંજ્ઞા તેમના પ્રમાણિત અર્થ ધરાવે છે.
A$\left[\mathrm{M}^6 \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$
B$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
C$\left[M^{-1} L^5 T^3\right]$
D$\left[M^6 L^7 T^4\right]$
(JEE MAIN-2024)
Solution
$\because[\mathrm{V}]=[\mathrm{b}]$
$\therefore \text { Dimension of } \mathrm{b}=\left[\mathrm{L}^3\right]$
$\&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right]$
${[\mathrm{a}]=\left[\mathrm{PV}^2\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^6\right]}$
$\text { Dimension of } \mathrm{a}=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]$
$\therefore \mathrm{ab}^{-1}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
$\therefore \text { Dimension of } \mathrm{b}=\left[\mathrm{L}^3\right]$
$\&[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right]$
${[\mathrm{a}]=\left[\mathrm{PV}^2\right]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^6\right]}$
$\text { Dimension of } \mathrm{a}=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]$
$\therefore \mathrm{ab}^{-1}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
Standard 11
Physics
