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Karnataka Board PUCPUC Science Class 11

If velocity of light c, Planck’s constant h and gravitational contant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

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Question

If velocity of light c, Planck’s constant h and gravitational contant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Long Answer
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Solution

We have to apply the principle of homogeneity to solve this problem. The principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be the same, i.e., dimensions of LHS and RHS should be equal,

We know that, the dimensions of `[h] = [ML^2T^-1], [c] = [LT^-1], [G] = [M^-1L^3T^-2]`

(i) Let `m ∝ c^x h^c G^z`

⇒ `m = kc^ah^bG^c`  ......(i)

Where k is a dimensionless constant of proportionality.

Substituting the dimensions of each term in equation (i), we get

`[ML^0T^0] = [LT^-1]^a xx [ML^2T^-1]^b [M^-1L^3T^2]^c`

Comparing powers of same terms on both sides, we get

`b - c` = 1  ......(ii)

`a + 2b + 3c` = 0 ......(iii)

`-a - b - 2c` = 0 ......(iv)

Adding equations (ii), (iii) and (iv), we get

`2b` = 1 ⇒ `b = 1/2`

Substituting the value of b in equation (ii), we get

`c = - 1/2`

From equation (iv)

`a = - b - 2c`

Substituting values of b and c, we get

`a = - 1/2 - 2(-1/2) = 1/2`

Putting values of a, b and c in equation (i), we get

`m = kc^(1/2) h^(1/2) G^(-1/2) = k sqrt((ch)/G)`

(ii) Let `L ∝ c^a h^b G^c`

⇒ `L = kc^ah^bG^c`  ......(v)

Where k is a dimensionless constant.

Substituting the dimensions of each term in equation (v), we get

`[M^0LT^0] = [LT^-1]^a xx [ML^2T^-1]^b [M^-1L^3T^-2]^c`

= `[M^(b-c) L^(a+2b+3c) T^(-a-b-2c)]`

On comparing powers of the same terms, we get

`b - c` = 0  ......(vi)

`a + 2b + 3c` = 1 ......(vii)

`-a - b - 2c` = 0 ......(viii)

Adding equations (vi), (vii) and (viii), we get

`2b` = 1 ⇒ `b = 1/2`

Substituting the value of b in equation (vi), we get

`c = 1/2`

From equation (viii)

`a = - b - 2c`

Substituting values of b and c, we get

`a = - 1/2 - 2(1/2) = -3/2`

Putting values of a, b and c in equation (v), we get

`L = kc^(-3/2) h^(1/2) G^(1/2) = k sqrt((hG)/c^3)`

(iii) Let `T ∝ c^a h^b G^c`

⇒ `T = c^ah^bG^c`  ......(ix)

Where k is a dimensionless constant.

Substituting the dimensions of each term in equation (ix), we get

`[M^0L^0T^1] = [LT^-1]^a xx [ML^2T^-1]^b xx [M^-1L^3T^-2]^c`

= `[M^(b-c) L^(a+2b+3c) T^(-a-b-2c)]`

On comparing powers of the same terms, we get

`b - c` = 0  ......(x)

`a + 2b + 3c` = 1 ......(xi)

`-a - b - 2c` = 0 ......(xii)

Adding equations (x), (xi) and (xii), we get

`2b` = 1 ⇒ `b = 1/2`

Substituting the value of b in equation (x), we get

`c = b = 1/2`

From equation (xii)

`a = - b - 2c - 1`

Substituting values of b and c, we get

`a = - 1/2 - 2(1/2) - 1 = -5/2`

Putting values of a, b and c in equation (ix), we get

`T = kc^(-5/2) h^(1/2) G^(1/2) = k sqrt((hG)/c^5)`

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Chapter 2: Units and Measurements - Exercises [Page 11]

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NCERT Exemplar Physics Exemplar [English] Class 11
Chapter 2 Units and Measurements
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