Units And Dimensions Question 3
Question 3 - 25 January - Shift 1
Match List I with List II
| List - I | List - II | ||
|---|---|---|---|
| A | Surface tension | I. | $Kg m^{-1} s^{-1}$ |
| B | Pressure | II. | $Kg ms^{-1}$ |
| C | Viscosity | III. | $Kg m^{-1} s^{-2}$ |
| D | Impulse | IV. | $Kg s^{-2}$ |
Choose the correct answer from the options given below:
(1) A-IV, B-III, C- II, D- I
(2) A-IV, B-III, C-I, D-II
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-I, C-III, D-IV
Show Answer
Answer: (2)
Solution:
Formula: Dimensional equation
(A) Surface Tension $=\frac{F}{\ell}=\frac{MLT^{-2}}{L}=ML^{-1} T^{-2}$
$ =Kgs^{-2}(IV) $
(B) Pressure $=\frac{F}{A}=\frac{MLT^{-2}}{L^{2}}$
$ =kg m^{-1} s^{-2} \text{ (III) } $
(C) Viscosity $==\frac{F}{A(\frac{dV}{dz})}=\frac{\text{ MLT }^{-2}}{L^{2}(\frac{LT^{-1}}{L})}$
$ =ML^{-1} T^{-1}=kg m^{-1} s^{-1}(I) $
(D) Impulse $=\int Fdt=MLT^{-2} \times T$
$ =MLT^{-1}=Kgms^{-1} \text{ (II) } $
