Units And Dimensions Question 4
Question 4 - 25 January - Shift 2
| List | List II | ||
|---|---|---|---|
| A. | Young’s Modulus $(Y)$ | I. | $[M L^{-1} T^{-1}]$ |
| B. | Co-efficient of Viscosity $(\eta)$ | II. | $[M L^{2} T^{-1}]$ |
| C. | Planck’s Constant $(h)$ | III. | $[M L^{-1} T^{-2}]$ |
| D. | Work Function $(\phi)$ hong | IV. | $[M L^{2} T^{-2}]$ |
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-III, B-I, C-II, D-IV
(3) A-I, B-III, C-IV, D-II
(4) A-I, B-II, C-III, D-IV
Show Answer
Answer: (2)
Solution:
Formula: Dimensional equation
$Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{F / A}{\Delta \ell / \ell}=\frac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1} T^{-2}]$
$F=6 \pi \eta rv \Rightarrow \eta=\frac{F}{6 \pi rv}$
$[\eta]=\frac{[MLT^{-2}]}{[L][LT^{-1}]}=[ML^{-1} T^{-1}]$
$E=h \nu \Rightarrow h=\frac{E}{v}=\frac{[ML^{2} T^{-2}]}{[T^{-1}]}=[ML^{2} T^{-1}]$
Work function has same dimension as that of energy, so $[\phi]=[ML^{2} T^{-2}]$
