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1.Units, Dimensions and Measurement
medium
Match List $I$ with List $II$
| List $I$ | List $II$ |
| $(A)$ Young's Modulus $(Y)$ | $(I)$ $\left[ M L ^{-1} T ^{-1}\right]$ |
| $(B)$ Co-efficient of Viscosity $(\eta)$ | $(II)$ $\left[ M L ^2 T ^{-1}\right]$ |
| $(C)$ Planck's Constant $(h)$ | $(III)$ $\left[ M L ^{-1} T ^{-2}\right]$ |
| $(D)$ Work Function $(\phi)$ | $(IV)$ $\left[ M L ^2 T ^{-2}\right]$ |
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
Similar Questions
Match List$-I$ with List$-II$
| List$-I$ | List$-II$ |
| $(a)$ $h$ (Planck's constant) | $(i)$ $\left[ M L T ^{-1}\right]$ |
| $(b)$ $E$ (kinetic energy) | $(ii)$ $\left[ M L ^{2} T ^{-1}\right]$ |
| $(c)$ $V$ (electric potential) | $(iii)$ $\left[ M L ^{2} T ^{-2}\right]$ |
| $(d)$ $P$ (linear momentum) | $( iv )\left[ M L ^{2} I ^{-1} T ^{-3}\right]$ |
