Physical Quantities

For the 1st Student:

For the 2nd Student:

The 1st student has a smaller range, so his data are more precise. The 2nd student’s average ($6.09 \text{ mm}$) is closer to the accurate value of $6.10 \text{ mm}$, so the 2nd student’s data are more accurate.

In addition and subtraction, the final result should be retained to as many decimal places as there are in the number with the least decimal places.

For example: $3.24 + 0.2 = 3.44$ ($\times$ Incorrect) But $3.4$ is the correct answer as there is only one digit after the decimal in $0.2$.

The final result should retain as many significant figures as there are in the original number with the least significant figures.

For example: $$1.234 \text{ (4 SF)} \times 2.0 \text{ (2 SF)} = 2.468 \text{ (normal multiplication)}$$ $$\rightarrow 2.5 \text{ (multiplication with correct SF)}$$

Solution: $y = a + bx + cx^2$    $y – a – bx – cx^2 = 0$

Now, Dimension of $a$ = Dimension of $y$  $\Rightarrow$  $a = [M^0L^1T^0]$

And, Dimension of $bx$ = Dimension of $y$  $\Rightarrow$  Dimension of $b = \frac{[L]}{[L]} = [M^0L^0T^0]$

And, Dimension of $cx^2$ = Dimension of $y$  $\Rightarrow$  $c [L^2] = [L]$  $\Rightarrow$  $c = \frac{[L]}{[L^2]} = [L^{-1}]$

Therefore, the dimensions of $a, b,$ and $c$ are $[L]$, $[M^0L^0T^0]$, and $[L^{-1}]$ respectively.

Solution: $P = ae^{bt}$    $P – ae^{bt} = 0$

Now, Dimension of $bt = 0$ (exponent/power is dimensionless)  $\Rightarrow$  Dimension of $b = \frac{1}{[T]} = [T^{-1}]$

And, Dimension of $ae^{bt}$ = Dimension of $P$  $\Rightarrow$  Dimension of $a = [ML^{-1}T^{-2}]$ (Since $e^{bt}$ is dimensionless, because $e$ is a constant and dimensionless).

Solution: $F – A \cos(Bx) – C \sin(Dt) = 0$

Dimension of $Bx = 0$  $\Rightarrow$  $B = [L^{-1}]$

Dimension of $Dt = 0$  $\Rightarrow$  $D = [T^{-1}]$

(It is because they are dimensionless; $Bx$ and $Dt$ have some numerical value that is dimensionless).

Also, Dimension of $A \cos(Bx)$ = Dimension of $F$  $\Rightarrow$  Dimension of $A = [MLT^{-2}]$ (Since $\cos(Bx)$ gives a dimensionless numeric value).

And, Dimension of $C \sin(Dt)$ = Dimension of $F$  $\Rightarrow$  Dimension of $C = [MLT^{-2}]$ (Same as above).

Now, Dimension of $D/B = \frac{[T^{-1}]}{[L^{-1}]} = [M^0L^1T^{-1}]$

Dimension of $A/C = \frac{[MLT^{-2}]}{[MLT^{-2}]} = [M^0L^0T^0]$

We will use the principle of homogeneity to check the relation.

Solution: Given: $v^2 = u^2 + 2as$

Dimension of LHS = $[v^2] = [LT^{-1}]^2 = [L^2T^{-2}]$

Dimension of RHS = $[u^2] + [2as] = [LT^{-1}]^2 + [LT^{-2}][L] = [L^2T^{-2}] + [L^2T^{-2}]$

Since Dimension of LHS = Dimension of RHS, the given equation is dimensionally correct.

(a) Escape velocity $v = \sqrt{\frac{2RG}{M}}$ where $R$ and $M$ are Radius and mass of earth respectively and $G$ is the universal gravitational constant.

Solution: Dimension of LHS = $[M^0L^1T^{-1}]$

Dimension of $G = \frac{F \times d^2}{m_1 \cdot m_2} = \frac{[MLT^{-2}] \times [L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

Dimension of RHS = $\sqrt{\frac{[L][M^{-1}L^3T^{-2}]}{[M]}} = \sqrt{[M^{-2}L^4T^{-2}]} = [M^{-1}L^2T^{-1}]$

Since $[LHS] \neq [RHS]$, the given equation is dimensionally incorrect.

(b) Time period of Pendulum, $T = 2\pi \sqrt{\frac{l}{g}}$ where $l$ is length and $g$ is acceleration due to gravity.

Solution: Dimension of LHS = $[T]$

Dimension of RHS = $\sqrt{\frac{[L]}{[LT^{-2}]}} = \sqrt{[T^2]} = [T]$

Since $[LHS] = [RHS]$, the given equation is dimensionally correct.

Qn) Derive the dimensional relation of the time period of a pendulum with mass, length, and acceleration due to gravity.

Solution: Let us assume $t \propto m^a l^b g^c$ — (i)

Where: $t = $ time period of pendulum, $l = $ length of pendulum, $m = $ mass of bob, $g = $ acceleration due to gravity

Equation (i) can be written as: $t = k m^a l^b g^c$ (where $k$ is a dimensionless constant)

Writing dimensions on both sides, we get: $[T] = [M]^a [L]^b [LT^{-2}]^c$   $[T] = [M^aL^{b+c}T^{-2c}]$

On equating, we get:

Putting values we get: $t = k m^0 l^{1/2} g^{-1/2}$   $t = k \sqrt{\frac{l}{g}}$

Experimentally: $k = 2\pi$. The required relation is $t = 2\pi\sqrt{\frac{l}{g}}$. It is noted that the time period of a simple pendulum does not depend on the mass of the bob.

Question: Using the method of dimensions, derive an expression for the centripetal force ($F$) acting on a particle of mass ($m$) moving with velocity ($v$) in a circle of radius ($r$).

Let $F \propto m^a v^b r^c$   $F = k m^a v^b r^c$, where $k$ is a dimensionless constant.

Writing the dimensions of LHS and RHS in the equation, we get: $[M L T^{-2}] = k [M]^a [L T^{-1}]^b [L]^c$   $[M L T^{-2}] = k [M^a L^{b+c} T^{-b}]$

Using the principle of homogeneity: $a = 1$,   $-b = -2 \implies b = 2$,   $b + c = 1 \implies 2 + c = 1 \implies c = -1$

So, the required expression is: $F = k m^1 v^2 r^{-1}$  $\Rightarrow$  $F = \frac{mv^2}{r}$ (where $k=1$)

Solution: The dimension of energy (Joule) is $[M^1 L^2 T^{-2}]$.

We have $N_1 U_1 = N_2 U_2$   $\Rightarrow$   $N_2 = N_1 \left[\frac{U_1}{U_2}\right]$

$N_2 = 10 \times \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$

$N_2 = 10 \times \left[\frac{1\text{ kg}}{100\text{ g}}\right]^1 \left[\frac{1\text{ m}}{10\text{ cm}}\right]^2 \left[\frac{1\text{ s}}{30\text{ s}}\right]^{-2}$

$N_2 = 10 \times \left[\frac{1000\text{ g}}{100\text{ g}}\right] \left[\frac{100\text{ cm}}{10\text{ cm}}\right]^2 [30\text{ s}]^2$

$N_2 = 10 \times 10 \times (10)^2 \times (30)^2$   $\Rightarrow$   $N_2 = 9 \times 10^6$

Therefore, $10 \text{ Joules} = 9 \times 10^6$ new units.

Solution: The dimension of density is $[M L^{-3} T^0]$.

We have $n_1 U_1 = n_2 U_2$   $\Rightarrow$   $n_2 = n_1 \left[\frac{U_1}{U_2}\right]$

$n_2 = 11.3 \times \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^{-3} \left[\frac{T_1}{T_2}\right]^0$

$n_2 = 11.3 \times \left[\frac{1\text{ g}}{1\text{ kg}}\right]^1 \left[\frac{1\text{ cm}}{1\text{ m}}\right]^{-3} \times 1$

$n_2 = 11.3 \times \left[\frac{1\text{ g}}{1000\text{ g}}\right] \times \left[\frac{1\text{ cm}}{100\text{ cm}}\right]^{-3}$

$n_2 = 11.3 \times 10^{-3} \times (10^{-2})^{-3} = 11.3 \times 10^{-3} \times 10^{+6} = 11.3 \times 10^3$

Value in the SI system = $11.3 \times 10^3\text{ kg/m}^3$.

Solution: Here, $\text{kg/m}^3$ is the unit of density with the dimension $[M^1 L^{-3} T^0]$.

Using $n_1 U_1 = n_2 U_2$   $\Rightarrow$   $n_2 = n_1 \times \left[\frac{U_1}{U_2}\right]$

$n_2 = 5 \times \left[\frac{1\text{ kg}}{1\text{ g}}\right]^1 \left[\frac{1\text{ m}}{1\text{ cm}}\right]^{-3} \left[\frac{1\text{ s}}{1\text{ s}}\right]^0$

$n_2 = 5 \times \left[\frac{1000\text{ g}}{1\text{ g}}\right] \left[\frac{100\text{ cm}}{1\text{ cm}}\right]^{-3} \times 1$

$n_2 = 5 \times 1000 \times 1 \times 10^{-6} = 5 \times 10^{-3}$

Therefore, $5\text{ kg/m}^3 = 5 \times 10^{-3}\text{ g/cm}^3$.

Solution: Here, Watt is the SI unit of power. Power ($P$) = $\frac{\text{Work}}{\text{Time}} = [M L^2 T^{-3}]$.

We have $n_1 U_1 = n_2 U_2$   $\Rightarrow$   $n_2 = n_1 \times \left[\frac{U_1}{U_2}\right]$

$n_2 = 10 \times \left[\frac{1\text{ kg}}{1\text{ g}}\right]^1 \left[\frac{1\text{ m}}{1\text{ cm}}\right]^2 \left[\frac{1\text{ s}}{1\text{ s}}\right]^{-3}$

$n_2 = 10 \times \left[\frac{1000\text{ g}}{1\text{ g}}\right] \left[\frac{100\text{ cm}}{1\text{ cm}}\right]^2 \times 1$

$n_2 = 10 \times 1000 \times 10000 = 1 \times 10^8$

Therefore, $10\text{ Watts} = 1 \times 10^8\text{ ergs/sec}$.

Solution: We have $F = \eta A \frac{dv}{dx}$   $\Rightarrow$   $\eta = \frac{F}{A \left(\frac{dv}{dx}\right)}$

Dimension of $F = [M L T^{-2}]$; Dimension of $A = [L^2]$; Dimension of $\frac{dv}{dx} = [T^{-1}]$

Now, by the principle of homogeneity, the dimension of LHS must equal the dimension of RHS. So, the dimension of $\eta$ is:

$$[\eta] = \frac{[M L T^{-2}]}{[L^2][T^{-1}]}$$ $$[\eta] = [M L^{-1} T^{-1}]$$

Answer: No, a dimensionally correct equation may not be physically correct. This is because dimensional analysis does not account for constant values. For example, the equations $v = u + at$ and $v = 2u + 3at$ both have the same dimensions $[L T^{-1}]$, but the latter is not physically correct. However, a dimensionally wrong equation is always physically wrong.