Units and Measurements

Rearranging the formula for the Gravitational constant $G$:
$G = \frac{F \cdot r^2}{M_1 M_2}$
Substitute the dimensions of each quantity:
Dimensions of Force $[F] = [M^1 L^1 T^{-2}]$
Dimensions of distance $[r] = [L]$
Dimensions of mass $[M_1] = [M_2] = [M]$
$[G] = \frac{[M^1 L^1 T^{-2}] \cdot [L^2]}{[M] \cdot [M]}$
$[G] = \frac{[M^1 L^3 T^{-2}]}{[M^2]}$
$[G] = [M^{-1} L^3 T^{-2}]$

Rearranging the formula for the coefficient of viscosity $\eta$:
$\eta = \frac{F}{6\pi r v}$
Substitute the dimensions of each quantity:
Dimensions of Force $[F] = [M^1 L^1 T^{-2}]$
Dimensions of radius $[r] = [L^1]$
Dimensions of velocity $[v] = [L^1 T^{-1}]$
$[\eta] = \frac{[M^1 L^1 T^{-2}]}{[L^1] \cdot [L^1 T^{-1}]}$
$[\eta] = \frac{[M^1 L^1 T^{-2}]}{[L^2 T^{-1}]}$
$[\eta] = [M^1 L^{-1} T^{-1}]$

Rearranging the formula for specific heat capacity ($S$):
$S = \frac{\Delta Q}{M \Delta T}$
Substitute the dimensions of each physical quantity:
Dimensions of Heat Energy $[\Delta Q] = [M^1 L^2 T^{-2}]$
Dimensions of mass $[M] = [M^1]$
Dimensions of temperature change $[\Delta T] = [K^1]$
$[S] = \frac{[M^1 L^2 T^{-2}]}{[M^1] \cdot [K^1]}$
$[S] = [M^0 L^2 T^{-2} K^{-1}]$

The formula for electric charge is:
$q = I \cdot t$
Substitute the basic dimensions:
Dimension of current $[I] = [A^1]$
Dimension of time $[t] = [T^1]$
$[q] = [A^1] \cdot [T^1]$
$[q] = [M^0 L^0 T^1 A^1]$

The formula for electric potential is:
$V = \frac{W}{q}$
Substitute the dimensions of work done and electric charge:
Dimensions of Work Done $[W] = [M^1 L^2 T^{-2}]$
Dimensions of Charge $[q] = [A^1 T^1]$
$[V] = \frac{[M^1 L^2 T^{-2}]}{[A^1 T^1]}$
$[V] = [M^1 L^2 T^{-3} A^{-1}]$

To find the dimensional formula of resistance ($R$):
1. The dimension of electric current ($I$) is $[\text{A}]$.
2. Charge ($q$) is given by $q = I \cdot t$, so its dimensional formula is $[\text{A}\text{T}]$.
3. Work done ($W$) has the dimension $[\text{M}^1\text{L}^2\text{T}^{-2}]$.
4. Electric potential ($V = \frac{W}{q}$) has the dimensional formula:
$[V] = \frac{[\text{M}^1\text{L}^2\text{T}^{-2}]}{[\text{A}\text{T}]} = [\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-1}]$
5. Using Ohm's law ($R = \frac{V}{I}$):
$[R] = \frac{[\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-1}]}{[\text{A}]} = [\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}]$

To find the dimensional formula of resistance $R$:
1. The dimension of Work $W$ is $[M^1 L^2 T^{-2}]$.
2. The dimension of Electric Current $I$ is $[A^1]$.
3. The dimension of Time $t$ is $[T^1]$.
Substitute these dimensions into the formula $R = \frac{W}{I^2 \cdot t}$:
$[R] = \frac{[M^1 L^2 T^{-2}]}{[A^1]^2 [T^1]}$
$[R] = \frac{[M^1 L^2 T^{-2}]}{[A^2 T^1]}$
$[R] = [M^1 L^2 T^{-3} A^{-2}]$

To find the dimensional formula of resistivity $\rho$:
1. The dimension of Resistance $R$ is $[M^1 L^2 T^{-3} A^{-2}]$.
2. The dimension of Area is $[L^2]$.
3. The dimension of Length $l$ is $[L^1]$.
Substitute these dimensions into $\rho = \frac{R \cdot \text{Area}}{l}$:
$[\rho] = \frac{[M^1 L^2 T^{-3} A^{-2}] \cdot [L^2]}{[L^1]}$
$[\rho] = \frac{[M^1 L^4 T^{-3} A^{-2}]}{[L^1]}$
$[\rho] = [M^1 L^3 T^{-3} A^{-2}]$

To find the dimensional formula of $\alpha$:
1. Kinetic energy $K$ has the dimension of work, which is $[M^1 L^2 T^{-2}]$.
2. Distance $s$ has the dimension of length, which is $[L^1]$.
3. Distance squared $s^2$ has the dimension $[L^2]$.
Using the given relation $K = \alpha s^2$:
$[\alpha] = \frac{[K]}{[s^2]}$
$[\alpha] = \frac{[M^1 L^2 T^{-2}]}{[L^2]}$
$[\alpha] = [M^1 L^0 T^{-2}]$

According to the principle of homogeneity of dimensions, the dimensions of the terms on both sides of a physical equation must be equal.
The dimensional formulas for the given physical quantities are:
Dimensions of Force $[F] = [M^1 L^1 T^{-2}]$
Dimensions of Mass $[M] = [M^1]$
Dimensions of Velocity $[v] = [L^1 T^{-1}]$
Dimensions of Radius $[r] = [L^1]$
Given formula:
$F = \frac{M v^n}{r}$
Writing the dimensions on both sides:
$[M^1 L^1 T^{-2}] = \frac{[M^1] [L^1 T^{-1}]^n}{[L^1]}$
Simplifying the right hand side:
$[M^1 L^1 T^{-2}] = [M^1 L^{n-1} T^{-n}]$
Comparing the exponents of $T$ (or $L$) on both sides:
$-n = -2$
$n = 2$
Thus, the value of the exponent $n$ is 2.

According to the principle of homogeneity of dimensions, both sides of a physically correct equation must have the same dimensions.
Squaring both sides of the given equation:
$v^2 = \frac{G M}{r^n}$
The dimensions of the quantities involved are:
Dimensions of Velocity $[v] = [L T^{-1}]$
Dimensions of $[v^2] = [L^2 T^{-2}]$
Dimensions of Universal Gravitational Constant $[G] = [M^{-1} L^3 T^{-2}]$
Dimensions of Mass $[M] = [M]$
Dimensions of Radius $[r] = [L]$
Substituting the dimensions into the equation:
$[L^2 T^{-2}] = \frac{[M^{-1} L^3 T^{-2}] [M]}{[L]^n}$
Simplifying the right-hand side:
$[L^2 T^{-2}] = [M^0 L^{3-n} T^{-2}]$
Equating the exponents of $L$ on both sides:
$2 = 3 - n$
$n = 3 - 2$
$n = 1$
Therefore, the value of the exponent $n$ is 1.

According to the principle of homogeneity of dimensions, the dimensions of the terms on both sides of a physical equation must be equal.
The dimensional formulas for the given physical quantities are:
Dimensions of Kinetic Energy $[KE] = [M^1 L^2 T^{-2}]$
Dimensions of Force $[F] = [M^1 L^1 T^{-2}]$
Dimensions of Velocity $[v] = [L^1 T^{-1}]$
Dimensions of Time $[t] = [T^1]$
Given relation:
$KE = F v^n t$
Substituting the dimensions into the equation:
$[M^1 L^2 T^{-2}] = [M^1 L^1 T^{-2}] [L^1 T^{-1}]^n [T^1]$
Simplifying the right hand side:
$[M^1 L^2 T^{-2}] = [M^1 L^{1+n} T^{-2-n+1}]$
$[M^1 L^2 T^{-2}] = [M^1 L^{1+n} T^{-n-1}]$
Comparing the exponents of $L$ on both sides:
$1 + n = 2$
$n = 1$
Therefore, the value of the variable exponent $n$ is 1.