Question 171: Units , Dimensions and Error Analysis

Question

If surface tension (S), Moment of inertia (I) and
Planck's constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be :-

Accepted Answer

We know,
surface tension (S) = $${F \over L}$$ = $${{\left[ {ML{T^{ - 2}}} ight]} \over {\left[ L ight]}}$$
$$ 	herefore $$ [S] = $${\left[ {M{T^{ - 2}}} ight]}$$
Moment of inertia (I) = mr
2
$$ 	herefore $$ [I] = $${\left[ {M{L^2}} ight]}$$
Planck's constant (h) = $${E \over f}$$ = Et
$$ 	herefore $$ [h] = $${\left[ {M{L^2}{T^{ - 1}}} ight]}$$
Also linear momentum (p) = mv = $${\left[ {ML{T^{ - 1}}} ight]}$$
Now we have to express p in terms of s, I and h.
$$ 	herefore $$ Let, [P] = [S
a
I
b
h
c
]
$$ \Rightarrow $$ $${\left[ {ML{T^{ - 1}}} ight]}$$ = $${\left[ {M{T^{ - 2}}} ight]}$$
a
$${\left[ {M{L^2}} ight]}$$
b
$${\left[ {M{L^2}{T^{ - 1}}} ight]}$$
c
$$ \Rightarrow $$ $${\left[ {ML{T^{ - 1}}} ight]}$$ = [ M
a + b + c
L
2b +2c
T
- 2a - c
]
By comparing the dimensions of both sides, we get
a + b + c = 1  .........(1)
2b +2c = 1  ..............(2)
- 2a - c = -1   ...................(3)
By solving those three equations we get,
a = $${1 \over 2}$$
b = $${1 \over 2}$$
c = 0
$$ 	herefore $$  linear momentum [p] = [S
1/2
I
1/2
h
0
]

Answer

S
1/2
I
1/2
h
0

Answer

S
3/2
I
1/2
h
0

Answer

S
1/2
I
1/2
h
-1

Answer

S
1/2
I
3/2
h
-1

Units , Dimensions and Error Analysis

If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be :-

Choose the correct answer: