- Home
- Standard 11
- Physics
1.Units, Dimensions and Measurement
medium
ઓહમના નિયમના પ્રયોગમાં જુદાં જુદાં અવલોકનો દરમિયાન એક અજ્ઞાત અવરોધનું મૂલ્ય $4.12 \Omega , 4.08 \Omega , 4.22 \Omega $ અને $ 4.14 \Omega$ મળે છે. અવલોકનમાં નિરપેક્ષ ત્રુટિ અને સાપેક્ષ ત્રુટિ અનુક્રમે ....... મળે.
A$0.04, 0.0096$
B$0.4, 0.096$
C$0.004, 0.96$
D$0.004, 0.00096$
Solution
${\rm{(i)}}\,\,{\rm{\bar R}}\, = \frac{{{{\rm{R}}_{\rm{1}}}\, + {{\rm{R}}_{\rm{2}}}\, + {{\rm{R}}_{{\rm{3}}\,\,}} + {{\rm{R}}_{\rm{4}}}}}{{\rm{4}}}$
$\, = \,\,\frac{{{\rm{4}}{\rm{.12}} + {\rm{4}}{\rm{.08}} + {\rm{4}}{\rm{.22}} + {\rm{4}}{\rm{.14}}}}{{\rm{4}}}\,\, = \,\,\frac{{{\rm{16}}{\rm{.56}}}}{{\rm{4}}}\,\, = \,\,{\rm{4}}{\rm{.14}}$
${\rm{\Delta }}{{\rm{R}}_{\rm{1}}}\, = \,{\rm{4}}{\rm{.14}} – \,{\rm{4}}{\rm{.12}}\, = \,{\rm{0}}{\rm{.02,}}$
$\,{\rm{\Delta }}{{\rm{R}}_{\rm{2}}}\, = {\rm{4}}{\rm{.14}}\, – {\rm{4}}{\rm{.08}}\, = \,{\rm{0}}{\rm{.06,}}\,$
$\,{\rm{\Delta }}{{\rm{R}}_{\rm{3}}}\, = {\rm{4}}{\rm{.14}}\, – {\rm{4}}{\rm{.22}}\, = \,{\rm{0}}{\rm{.08,}}\,$
${\rm{\Delta }}{{\rm{R}}_{\rm{4}}} = \,{\rm{4}}{\rm{.14}}\, – \,{\rm{4}}{\rm{.14}}\, = \,{\rm{0}}{\rm{.00}}$
${\rm{\Delta \bar R}}\,\, = \,\frac{{{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{1}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{2}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{3}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{4}}}\,{\rm{|}}}}{{\rm{4}}}\,\,\,$
$\, = \,\,\frac{{{\rm{|}}\,\,{\rm{0}}{\rm{.02|}} + {\rm{|0}}{\rm{.06|}} + {\rm{|}}\, – \,{\rm{0}}{\rm{.08}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{0}}{\rm{.00}}\,{\rm{|}}}}{{\rm{4}}}$
$\, = \,\frac{{{\rm{0}}{\rm{.16}}}}{{\rm{4}}}\,\,\, = \,\,{\rm{0}}{\rm{.04}}$
નિરપેક્ષ ત્રુટિ $\Delta \bar R\,\, = \,\,0.04$
સાપેક્ષ ત્રુટિ $\frac{{{\rm{\Delta \bar R}}}}{{{\rm{\bar R}}}}\, = \,\frac{{{\rm{0}}{\rm{.04}}}}{{{\rm{4}}{\rm{.14}}}}\,\,\, = \,{\rm{0}}{\rm{.00966}}\,\, = \,\,\,{\rm{0}}{\rm{.0096}}$
$\, = \,\,\frac{{{\rm{4}}{\rm{.12}} + {\rm{4}}{\rm{.08}} + {\rm{4}}{\rm{.22}} + {\rm{4}}{\rm{.14}}}}{{\rm{4}}}\,\, = \,\,\frac{{{\rm{16}}{\rm{.56}}}}{{\rm{4}}}\,\, = \,\,{\rm{4}}{\rm{.14}}$
${\rm{\Delta }}{{\rm{R}}_{\rm{1}}}\, = \,{\rm{4}}{\rm{.14}} – \,{\rm{4}}{\rm{.12}}\, = \,{\rm{0}}{\rm{.02,}}$
$\,{\rm{\Delta }}{{\rm{R}}_{\rm{2}}}\, = {\rm{4}}{\rm{.14}}\, – {\rm{4}}{\rm{.08}}\, = \,{\rm{0}}{\rm{.06,}}\,$
$\,{\rm{\Delta }}{{\rm{R}}_{\rm{3}}}\, = {\rm{4}}{\rm{.14}}\, – {\rm{4}}{\rm{.22}}\, = \,{\rm{0}}{\rm{.08,}}\,$
${\rm{\Delta }}{{\rm{R}}_{\rm{4}}} = \,{\rm{4}}{\rm{.14}}\, – \,{\rm{4}}{\rm{.14}}\, = \,{\rm{0}}{\rm{.00}}$
${\rm{\Delta \bar R}}\,\, = \,\frac{{{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{1}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{2}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{3}}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{\Delta }}{{\rm{R}}_{\rm{4}}}\,{\rm{|}}}}{{\rm{4}}}\,\,\,$
$\, = \,\,\frac{{{\rm{|}}\,\,{\rm{0}}{\rm{.02|}} + {\rm{|0}}{\rm{.06|}} + {\rm{|}}\, – \,{\rm{0}}{\rm{.08}}\,{\rm{|}}\, + \,{\rm{|}}\,{\rm{0}}{\rm{.00}}\,{\rm{|}}}}{{\rm{4}}}$
$\, = \,\frac{{{\rm{0}}{\rm{.16}}}}{{\rm{4}}}\,\,\, = \,\,{\rm{0}}{\rm{.04}}$
નિરપેક્ષ ત્રુટિ $\Delta \bar R\,\, = \,\,0.04$
સાપેક્ષ ત્રુટિ $\frac{{{\rm{\Delta \bar R}}}}{{{\rm{\bar R}}}}\, = \,\frac{{{\rm{0}}{\rm{.04}}}}{{{\rm{4}}{\rm{.14}}}}\,\,\, = \,{\rm{0}}{\rm{.00966}}\,\, = \,\,\,{\rm{0}}{\rm{.0096}}$
Standard 11
Physics