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1.Units, Dimensions and Measurement
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લીસ્ટ $I$ સાથે લીસ્ટ $II$ યોગ્ય રીતે જોડો.
| લીસ્ટ $I$ | લીસ્ટ $II$ |
| $(A)$ યંગનો ગુણાંક $(Y)$ | $(I)$ $\left[ M L ^{-1} T ^{-1}\right]$ |
| $(B)$ શ્યાનતા ગુણાંક $(\eta)$ | $(II)$ $\left[ M L ^2 T ^{-1}\right]$ |
| $(C)$ પ્લાન્ક અચળાંક $(h)$ | $(III)$ $\left[ M L ^{-1} T ^{-2}\right]$ |
| $(D)$ કાર્ય વિધેય $(\phi)$ | $(IV)$ $\left[ M L ^2 T ^{-2}\right]$ |
A$(A)-(II), (B)-(III), (C)-(IV), (D)-(I)$
B$(A)-(III), (B)-(I), (C)-(II), (D)-(IV)$
C$(A)-(I), (B)-(III), (C)-(IV), (D)-(II)$
D$(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$
(JEE MAIN-2023)
Solution
$Y =\frac{\text { Stress }}{\text { Strain }}=\frac{ F / A }{\Delta \ell / \ell}=\frac{\left[ MLT ^{-2}\right]}{\left[ L ^2\right]}=\left[ ML ^{-1} T ^{-2}\right]$
$F =6 \pi \eta rv \Rightarrow \eta=\frac{ F }{6 \pi rv }$
${[\eta]=\frac{\left[ MLT ^{-2}\right]}{[ L ]\left[ LT ^{-1}\right]}=\left[ ML ^{-1} T ^{-1}\right]}$
$E = h v \Rightarrow h =\frac{ E }{v}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ T ^{-1}\right]}=\left[ ML ^2 T ^{-1}\right]$
Work function has same dimension as that of energy, so $[\phi]=\left[ ML ^2 T ^{-2}\right]$
$F =6 \pi \eta rv \Rightarrow \eta=\frac{ F }{6 \pi rv }$
${[\eta]=\frac{\left[ MLT ^{-2}\right]}{[ L ]\left[ LT ^{-1}\right]}=\left[ ML ^{-1} T ^{-1}\right]}$
$E = h v \Rightarrow h =\frac{ E }{v}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ T ^{-1}\right]}=\left[ ML ^2 T ^{-1}\right]$
Work function has same dimension as that of energy, so $[\phi]=\left[ ML ^2 T ^{-2}\right]$
Standard 11
Physics
