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1.Units, Dimensions and Measurement
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$1$ $joule$ ઉર્જાને નવી પધ્ધતિમાં ફેરવવામાં આવે છે જ્યાં લંબાઈ $10\, m$, દળ $10\, kg$ અને સમય $1$ $minute$ માં માપવામાં આવે છે. તો નવી પધ્ધતિમાં $1\, J$ નું મૂલ્ય કેટલું થાય?
A$36 \times 10^{-4}$
B$36 \times 10^{-3}$
C$36 \times 10^{-2}$
D$36 \times 10^{-1}$
Solution
$1 \mathrm{J}=2 \quad$ New system
$\mathrm{n}_{2}=\mathrm{n}_{1}\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{2}\left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{-2}$
$\quad=1\left[\frac{\mathrm{kg}}{10 \mathrm{kg}}\right]\left[\frac{\mathrm{m}}{10 \mathrm{m}}\right]^{2}\left[\frac{10}{60}\right]^{-2}$
$\quad=\left(\frac{1}{10}\right)^{1} \times\left(\frac{1}{10}\right)^{2} \times(60)^{2}=36 \times 10^{-1}$
$\mathrm{n}_{2}=\mathrm{n}_{1}\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{2}\left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{-2}$
$\quad=1\left[\frac{\mathrm{kg}}{10 \mathrm{kg}}\right]\left[\frac{\mathrm{m}}{10 \mathrm{m}}\right]^{2}\left[\frac{10}{60}\right]^{-2}$
$\quad=\left(\frac{1}{10}\right)^{1} \times\left(\frac{1}{10}\right)^{2} \times(60)^{2}=36 \times 10^{-1}$
Standard 11
Physics
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