Question 11: Units , Dimensions and Error Analysis

Question

$$ 	ext { Match the LIST-I with LIST-II } $$
List - I
List - II
A.
Planck's constant
I.
$$
\mathrm{ML}^2 \mathrm{~T}^{-2}
$$
B.
Stopping potential
II.
$$
\mathrm{T}^{-1}
$$
C.
Work function
III.
$$
\mathrm{ML}^2 \mathrm{~T}^{-1}
$$
D.
Threshold frequency
IV.
$$
\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-1}
$$
Choose the correct answer from the options given below:

Accepted Answer

For matching, we find the dimensional formula of each quantity one by one.
A. Planck's constant
Using the relation
$$ E = h
u $$
So,
$$ h = \frac{E}{
u} $$
Dimension of energy is
$$ [E] = \mathrm{ML}^2\mathrm{T}^{-2} $$
Dimension of frequency is
$$ [
u] = \mathrm{T}^{-1} $$
Therefore,
$$ [h] = \frac{\mathrm{ML}^2\mathrm{T}^{-2}}{\mathrm{T}^{-1}} = \mathrm{ML}^2\mathrm{T}^{-1} $$
So, $A 	o 	ext{III}$.
B. Stopping potential
Stopping potential is a potential difference.
We know
$$ V = \frac{	ext{work}}{	ext{charge}} $$
Dimension of work is
$$ \mathrm{ML}^2\mathrm{T}^{-2} $$
Dimension of charge is
$$ \mathrm{AT} $$
Hence,
$$ [V] = \frac{\mathrm{ML}^2\mathrm{T}^{-2}}{\mathrm{AT}} = \mathrm{ML}^2\mathrm{T}^{-3}\mathrm{A}^{-1} $$
So, $B 	o 	ext{IV}$.
C. Work function
Work function is the minimum energy needed to remove an electron. So its dimension is same as energy:
$$ [\phi] = \mathrm{ML}^2\mathrm{T}^{-2} $$
So, $C 	o 	ext{I}$.
D. Threshold frequency
Frequency has dimension
$$ [
u_0] = \mathrm{T}^{-1} $$
So, $D 	o 	ext{II}$.
Final matching
$$ A 	o 	ext{III}, \quad B 	o 	ext{IV}, \quad C 	o 	ext{I}, \quad D 	o 	ext{II} $$
So the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

A-III, B-IV, C-I, D-II

Answer

A-I, B-II, C-III, D-IV

Answer

A-IV, B-III, C-I, D-II

Answer

A-I, B-IV, C-III, D-II

Units , Dimensions and Error Analysis

Choose the correct answer: