- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
medium
જો બળનો એકમ $100\,N$, લંબાઈનો એકમ $10\,m$ અને સમયનો એકમ $100\,s$ હોય, તો નવી એકમ પદ્ધતિમાં દ્રવ્યમાનનો એકમ શું હશે ?
Option A
Option B
Option C
Option D
Solution
$[\mathrm{F}]=100 \mathrm{~N}$
$\left[\mathrm{~L}^{1}\right]=10 \mathrm{~m}$
$\left[\mathrm{~T}^{\mathrm{l}}\right]=100 \mathrm{~s}$
$[\mathrm{~F}]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]=100 \mathrm{~N}$
સમીકરણ $(2)$ અને $(3)$ની કિંમત મુક્તા,
$\mathrm{M}^{1} \times 10 \times(100)^{-2}=100$
$\mathrm{M}^{1}=\frac{(1)}{10 \times(100)^{-2}}=\frac{10^{6}}{10}$
$\therefore \mathrm{M}^{1}=10^{5} \mathrm{~kg}$
$\therefore 2=10^{5} \mathrm{~kg}$
$\left[\mathrm{~L}^{1}\right]=10 \mathrm{~m}$
$\left[\mathrm{~T}^{\mathrm{l}}\right]=100 \mathrm{~s}$
$[\mathrm{~F}]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]=100 \mathrm{~N}$
સમીકરણ $(2)$ અને $(3)$ની કિંમત મુક્તા,
$\mathrm{M}^{1} \times 10 \times(100)^{-2}=100$
$\mathrm{M}^{1}=\frac{(1)}{10 \times(100)^{-2}}=\frac{10^{6}}{10}$
$\therefore \mathrm{M}^{1}=10^{5} \mathrm{~kg}$
$\therefore 2=10^{5} \mathrm{~kg}$
Standard 11
Physics
