Units And Dimensions Question 1
Question 1 - 24 January - Shift 1
Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Planck’s constant (h) | I. | $[M^{1} L^{2} T^{-2}]$ |
| B. | Stopping potential (Vs) | II. | $[M^{1} L^{1} T^{-1}]$ |
| C. | Work function $(\varnothing)$ | III. | $[M^{1} L^{2} T^{-1}]$ |
| D. | Momentum (p) | IV. | $[M^{1} L^{2} T^{-3} A^{-1}]$ |
(1) A-III, B-I, C-II, D-IV
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-IV, C-III, D-I
(4) A-I, B-III, C-IV, D-II
Show Answer
Answer: (2)
Solution:
Formula: Dimensional equation
(A) Planck’s constant
$h v=E$
$h=\frac{E}{v}=\frac{M^{1} L^{2} T^{-2}}{T^{-1} \text{ hong }}=M^{1} L^{2} T^{-1}$
(B) $E=qV$
$ V=\frac{E}{q}=\frac{M^{1} L^{2} T^{-2}}{A^{1} T^{1}}=M^{1} L^{2} T^{-3} A^{-1}(IV) $
(C) $\phi($ work function $)=$ energy
$ =M^{1} L^{2} T^{-2} $
(D) Momentum (p) = F.t
$ \begin{aligned} & =M^{1} L^{1} T^{-2} T^{1} \\ & =M^{1} L^{1} T^{-1} \end{aligned} $
