Units And Dimensions Question 5
Question 5 - 29 January - Shift 1
Match List I with List II :
| List-I (Physical Quantity) |
List-II | ||
|---|---|---|---|
| (Dimensional Formula) | |||
| A | Pressure gradient |
I | $[M^{0} L^{2} T^{-2}]$ |
| B | Energy density | II | $[M^{1} L^{-1} T^{-2}]$ |
| C | Electric Field | III | $[M^{1} L^{-2} T^{-2}]$ |
| D | Latent heat | IV | $[M^{1} L^{1} T^{-3} A^{-1}]$ |
Choose the correct answer from the options given below:
(1) A-III, B-II, C-I, D-IV
(2) A-II, B-III, C-IV, D-I
(3) A-III, B-II, C-IV, D-I
(4) A-II, B-III, C-I, D-IV
Show Answer
Answer: (3)
Solution:
Formula: Dimensional equation
Pressure gradient $=\frac{dp}{dx}=\frac{[ML^{-1} T^{-2}]}{[L]}$
$ =[M^{1} L^{-2} T^{-2}] $
Energy density $=\frac{\text{ energy }}{\text{ volume }}=\frac{[ML^{2} T^{-2}]}{[L^{3}]}$
$ =[M^{1} L^{-1} T^{-2}] $
Electric field $=\frac{\text{ Force }}{\text{ charge }}=\frac{[MLT^{-2}]}{[A . T]}$
$=[M^{1} L^{1} T^{-3} A^{-1}]$
Latent heat $=\frac{\text{ heat }}{\text{ mass }}=\frac{[ML^{2} T^{-2}]}{[M]}$
$=[M^{0} L^{2} T^{-2}]$
