Physics 1 Marks Question

$[\text{h}]=[\text{ML}^2\text{T}^{-1}],[\text{c}]=[\text{LT}^{-1}],\text[{G}]=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$

1. Let $\text{m}\propto\text{c}^\text{x}\text{h}^\text{v}\text{G}^\text{z}$

$\Rightarrow\text{m}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{i})$

Where, k is a dimensionless constant of proportionality.

Substituting dimensions of each term in Eq. (i), we get,

$[\text{ML}^0\text{T}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$

Comparing powers of same terms on both sides, we get,

$\text{b}-\text{c}=1\ \ \ ...(\text{ii})$

$\text{a}+2\text{b}+3\text{c}=0\ \ \ \ ...(\text{iii})$

$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{iv})$

Adding Eqs. (ii), (iii) and (iv), we get,

$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$

Substituting value of b in Eq. (ii), we get,

$\text{c}=-\frac{1}{2}$

From Eq. (iv)

$\text{a}=-\text{b}-2\text{c}$

Substituting values of b and c, we get,

$\text{a}=-\frac{1}{2}-2\Big(-\frac{1}{2}\Big)=\frac{1}{2}$

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

$\text{m}=\text{kc}^\frac{1}{2}\text{h}^\frac{1}{2}\text{G}^{-\frac{1}{2}}=\text{k}\sqrt{\frac{\text{ch}}{\text{G}}}$

2. Let $\text{L}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$

$\Rightarrow\text{L}=\text{kc}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{v})$

Where k is a dimensionless constant.

Substituting dimensions of each term in Eq. (v), we get

$[\text{M}^0\text{LT}^0]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$

$=[\text{M}^{\text{b}-\text{c}\ }\text{L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$

On comparing powers of same terms, we get,

$\text{b}-\text{c}=0\ \ \ ...(\text{vi})$

$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{vii})$

$-\text{a}-\text{b}-2\text{c}=0\ \ \ \ ...(\text{viii})$

Adding Eqs. (vi), (vii) and (viii), we get,

$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$

Substituting value of b in Eq. (vi), we get,

$\text{c}=\frac{1}{2}$

From Eq. (viii), $\text{a}=-\text{b}-2\text{c}$

Substituting values of b and c, we get,

$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)=-\frac{3}{2}$

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

$\text{L}=\text{kc}^{-\frac{3}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^3}}$

3. Let $\text{T}\propto\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}$

$\Rightarrow\text{T}=\text{c}^\text{a}\text{h}^\text{b}\text{G}^\text{c}\ \ \ \ ...(\text{ix})$

Where, k is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$[\text{M}^0\text{L}^0\text{T}^1]=[\text{LT}^{-1}]^\text{a}\times[\text{ML}^2\text{T}^{-1}]^\text{b}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$

$=[\text{M}^{\text{b}-\text{c}}\text{ L}^{\text{a}+2\text{b}+3\text{c}}\ \text{T}^{-\text{a}-\text{b}-2\text{c}}]$

On comparing powers of same terms, we get,

$\text{b}-\text{c}=0 \ \ \ ...(\text{x})$

$\text{a}+2\text{b}+3\text{c}=1\ \ \ ...(\text{xi})$

$-\text{a}-\text{b}-2\text{c}=1\ \ \ \ ...(\text{xii})$

Adding Eqs. (x), (xi) and (xii), we get,

$2\text{b}=1\Rightarrow\text{b}=\frac{1}{2}$

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

$\text{c}=\text{b}=\frac{1}{2}$

From Eq. (xii),

$\text{a}=-\text{b}-2\text{c}-1$

Substituting values of b and c, we get,

$\text{a}=-\frac{1}{2}-2\Big(\frac{1}{2}\Big)-1=-\frac{5}{2}$

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

$\text{T}=\text{kc}^{\frac{-5}{2}}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}=\text{k}\sqrt{\frac{\text{hG}}{\text{c}^5}}$