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1.Units, Dimensions and Measurement
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पानी के अन्दर विस्फोट से निर्मित गैस के एक बुलबुले के दोलन का आवर्तकाल $( T ), P ^{ a } d ^{ b } E ^{ c }$ के समानुपाती है जहॉ $P$ स्थैतिक दाब, $d$ पानी का घनत्व और $E$ विस्फोट की ऊर्जा हैं। तो $a, b, c$ है क्रमश:
A$ - \frac{5}{6},\frac{1}{2},\frac{1}{3}$
B$ \frac{1}{2},- \frac{5}{6},\frac{1}{3}$
C$\frac{1}{3},\frac{1}{2},- \frac{5}{6}$
D$1,\, 1,\, 1$
Solution
Time period $\mathrm{T}=\mathrm{P}^{\mathrm{a}} \mathrm{d}^{\mathrm{b}} \mathrm{E}^{\mathrm{c}}$
Taking dimensions of each physical quantity.
$[\mathrm{T}]=\left[\mathrm{ML}^{-1}\mathrm{T}^{-2}\right]^{\mathrm{a}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]^{\mathrm{c}}$
Equating the exponents of $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$ on both the sides,
$\mathrm{M}^{\mathrm{a}+\mathrm{b}+\mathrm{T}} \mathrm{L}^{-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}-2 \mathrm{c}} =\mathrm{T}$
$\mathrm{a}+\mathrm{b}+\mathrm{c} =0$
$-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c} =0$
$-2 \mathrm{a}-2 \mathrm{c} =1$
Solving these equations for $a$, $b$ and $c,$ we get
$a=-\frac{5}{6}, b=\frac{1}{2}$ and $\frac{1}{3}$
Taking dimensions of each physical quantity.
$[\mathrm{T}]=\left[\mathrm{ML}^{-1}\mathrm{T}^{-2}\right]^{\mathrm{a}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{b}}\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]^{\mathrm{c}}$
Equating the exponents of $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$ on both the sides,
$\mathrm{M}^{\mathrm{a}+\mathrm{b}+\mathrm{T}} \mathrm{L}^{-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c}} \mathrm{T}^{-2 \mathrm{a}-2 \mathrm{c}} =\mathrm{T}$
$\mathrm{a}+\mathrm{b}+\mathrm{c} =0$
$-\mathrm{a}-3 \mathrm{b}+2 \mathrm{c} =0$
$-2 \mathrm{a}-2 \mathrm{c} =1$
Solving these equations for $a$, $b$ and $c,$ we get
$a=-\frac{5}{6}, b=\frac{1}{2}$ and $\frac{1}{3}$
Standard 11
Physics
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