Units and Measurements class 11

Best online coaching for JEE

Notes • Solved Problems

What is Physics?

Physics is the study of nature and the laws that govern natural phenomena. It involves observing events, identifying patterns, formulating laws, and using them to predict outcomes.

SI System of Units

Units must be:

invariable,
easily reproducible,
internationally accepted.

All international unit decisions are made by CGPM (General Conference on Weights and Measures).

SI (International System of Units) has 7 base quantities + 2 supplementary quantities (plain angle and solid angle).

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	$$\theta$$
Amount of substance	mole	mol
Luminous intensity	candela	cd
Plane angle	radian	rad
Solid angle	steradian	sr

SI Prefixes

(Important for JEE)

Power of 10	Prefix	Symbol
$$10^{18}$$	exa	E
$$10^{15}$$	peta	P
$$10^{12}$$	tera	T
$$10^{9}$$	giga	G
$$10^{6}$$	mega	M
$$10^{3}$$	kilo	k
$$10^{2}$$	hecto	h
$$10^{1}$$	deka	da
$$10^{-1}$$	deci	d
$$10^{-2}$$	centi	c
$$10^{-3}$$	milli	m
$$10^{-6}$$	micro	$\mu$
$$10^{-9}$$	nano	n
$$10^{-12}$$	pico	p
$$10^{-15}$$	femto	f
$$10^{-18}$$	atto	a

Note: These prefixes appear in numerical problems and unit conversions.

Units and Dimensions

Dimensions express a physical quantity in terms of powers of base dimensions.

Base dimensions:

$$ M \text{ (mass)},\quad L \text{ (length)} $$
$$ T \text{ (time)}, I \text{ (current)} $$
$$ K \text{ (temperature)},\quad \text{mol},\quad \text{cd} $$

Examples:

$$ [F] = MLT^{-2}$$
$$ [E] = ML^2 T^{-2}$$

Magnitudes and constants (like $2$, $\tfrac12$, $\pi$) do not appear in dimensional formulas.

Uses of Dimensional Analysis

1. Checking Dimensional Homogeneity

All terms in an equation must have the same dimensions.

Example:

\[ x = ut + \frac12 at^2 \]

\[ [x] = L,\qquad [ut] = L\]

\[ [at^2] = L \]

Important:

If an equation is dimensionally incorrect, it is definitely wrong.
If dimensionally correct, it may or may not be physically correct.

2. Conversion of Units

Dimensional formula helps convert units of derived quantities.

Example:

\[1\,\text{m} = 10^2\,\text{cm}\]

3. Deriving Relations among Quantities

Example: Time period of pendulum:

\[ t = k l^{a} g^{b} \]

Dimensional comparison gives:

\[ t = k\sqrt{\frac{l}{g}} \]

($k$ is dimensionless constant.)

Note: Works only in multiplicative dependence.

Limitations of Dimensional Analysis

Can not give dimensionless constants (2, $\tfrac12$, $2\pi$).
Works only for multiplicative relationships (multiplication).
Cannot handle equations involving addition/subtraction.
Cannot derive relations containing trigonometric/exponential/logarithmic functions.

Order of Magnitude

A quantity is written as:

\[ a \times 10^x,\qquad 1 \le a < 10 \]

has an order of magnitude $x$.

Example:

Diameter of Sun $\approx 1.39 \times 10^9$ m → order $\approx 10^9$ m.

Used for quick estimation and approximation.

Units and Dimensions JEE PYQ

Suppose the acceleration due to gravity at a place is $9\ \text{m/s}^2$. Find its value in $\text{cm}/(\text{minute})^2$. \[ \begin{aligned} g &= 9\ \text{m s}^{-2} \\ &= 9 \times 100\ \text{cm s}^{-2} \\ &= 900\ \text{cm s}^{-2} \\[6pt] &= 900\ \text{cm} \left(\frac{60\ \text{s}}{1\ \text{min}}\right)^2\\ &= 900 \times 3600\ \text{cm min}^{-2} \\[4pt] &= 3{,}240{,}000\ \text{cm min}^{-2} \\[4pt] &= 3.24 \times 10^6\ \text{cm min}^{-2}. \end{aligned} \] So, \[ g = 3.24 \times 10^6\ \text{cm}/(\text{minute})^2. \]

$$ \bigl(1\ \text{m} = 100\ \text{cm}\bigr)$$ $$\bigl(1\ \text{s}^{-2} = 60^2\ \text{min}^{-2}\bigr)$$

Express the power of a $500$ watt bulb in CGS unit.
Solution:
\[ 1\ \text{watt} = 1\ \text{joule/second} \]
In CGS, the unit of energy is the erg and
\[ 1\ \text{J} = 10^{7}\ \text{erg}. \]
Therefore,
\[ \begin{aligned} 500\ \text{W} &= 500\ \frac{\text{J}}{\text{s}} \\ &= 500 \times 10^{7}\ \frac{\text{erg}}{\text{s}} \\[2pt] &= 5 \times 10^{9}\ \frac{\text{erg}}{\text{s}}. \end{aligned} \]
So, the power of a $500$ W bulb in CGS units is
\[ P = 5 \times 10^{9}\ \text{erg s}^{-1}. \]
The surface tension of water is $10\ \text{dyne/cm}$. Convert it in SI unit.
Solution:

In CGS system:
\[ 1\ \text{dyne} = 10^{-5}\ \text{N}, \]
\[1\ \text{cm} = 10^{-2}\ \text{m}. \]
Therefore,
\[ 1\ \frac{\text{dyne}}{\text{cm}} = \frac{10^{-5}\ \text{N}}{10^{-2}\ \text{m}} = 10^{-3}\ \frac{\text{N}}{\text{m}}. \]
Hence,
\[ \begin{aligned} 10\ \frac{\text{dyne}}{\text{cm}} &= 10 \times 10^{-3}\ \frac{\text{N}}{\text{m}} \\ &= 10^{-2}\ \frac{\text{N}}{\text{m}} \\ &= 0.01\ \frac{\text{N}}{\text{m}}. \end{aligned} \]
Thus, the surface tension in SI units is
\[ S = 0.01\ \text{N/m}. \]
Test if the following equations are dimensionally correct:
1. $h = \dfrac{2S \cos\theta}{\rho r g}$,
2. $v = \sqrt{\dfrac{P}{\rho}}$,
3. $V = \dfrac{\pi P r^{4} t}{8 \eta l}$,
4. $v = \dfrac{1}{2\pi} \sqrt{\dfrac{mgl}{I}}$,
where $h =$ height, $S =$ surface tension, $\rho =$ density, $P =$ pressure, $V =$ volume, $\eta =$ viscosity, $v =$ frequency, $I =$ moment of inertia.

Solution:

Useful dimensions:
\[ [h] = L, \] \[ [S] = \frac{\text{force}}{\text{length}} = \frac{MLT^{-2}}{L} \] \[ [S] = MT^{-2}, \] \[ [\rho] = ML^{-3},\quad [r] = L \] \[ [g] = LT^{-2}, \] \[ [P] = \frac{\text{force}}{\text{area}} = \frac{MLT^{-2}}{L^{2}} \] \[ [P] = ML^{-1}T^{-2}, \] \[ [V] = L^{3} \] \[ [\eta] = \text{(viscosity)} = ML^{-1}T^{-1}, \] \[ [v] = T^{-1},\quad [I] = ML^{2} \] \[ [m] = M,\quad [l] = L. \]
(a)
\[ h = \frac{2S\cos\theta}{\rho r g}. \]
$\cos\theta$ is dimensionless. So
\[ \left[\frac{S}{\rho r g}\right] = \frac{MT^{-2}}{(ML^{-3})\cdot L\cdot (LT^{-2})}\] \[ = \frac{MT^{-2}}{ML^{-1}T^{-2}} = L. \]
Hence RHS has dimension $L$, same as $[h]$.
(a) is dimensionally correct.

(b)
\[ v = \sqrt{\frac{P}{\rho}}. \] \[ \left[\frac{P}{\rho}\right] = \frac{ML^{-1}T^{-2}}{ML^{-3}} = L^{2}T^{-2}. \]
Thus
\[ \left[\sqrt{\frac{P}{\rho}}\right] = LT^{-1}, \]
which is dimension of velocity. So
(b) is dimensionally correct.

(c)
\[ V = \frac{\pi P r^{4} t}{8 \eta l}. \]
Ignoring numerical constants ($\pi$, $8$),
\[ [Pr^{4}t] = (ML^{-1}T^{-2})\cdot L^{4}\cdot T \] \[ [Pr^{4}t] = ML^{3}T^{-1}. \] \[ [\eta l] = (ML^{-1}T^{-1})\cdot L = MT^{-1}. \]
So
\[ \left[\frac{Pr^{4}t}{\eta l}\right] = \frac{ML^{3}T^{-1}}{MT^{-1}} = L^{3} = [V]. \]
(c) is dimensionally correct.

(d)
\[ v = \frac{1}{2\pi}\sqrt{\frac{mgl}{I}}. \]
Inside the root:
\[ [mgl] = M \cdot (LT^{-2}) \cdot L = ML^{2}T^{-2} \] \[ [I] = ML^{2}. \]
So
\[ \left[\frac{mgl}{I}\right] = \frac{ML^{2}T^{-2}}{ML^{2}} = T^{-2}. \]
Hence
\[ \left[\sqrt{\frac{mgl}{I}}\right] = T^{-1}. \]
Thus RHS $\sim T^{-1}$, same as frequency $v$.
(d) is dimensionally correct.
Let $x$ and $a$ stand for distance. Is \[ \int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \frac{1}{a} \sin^{-1} \frac{a}{x} \] dimensionally correct?
Solution:
\[ [x] = [a] = L. \]
Left-hand side integrand:
\[ \frac{dx}{\sqrt{a^{2} - x^{2}}} \]
has
\[ [dx] = L,\quad [\sqrt{a^{2} - x^{2}}] = \sqrt{L^{2}} \] \[ [dx] = L \]
So the integrand is dimensionless, and the integral itself is also dimensionless.

Right-hand side:
\[ \frac{1}{a}\sin^{-1}\left(\frac{a}{x}\right). \]
Inside $\sin^{-1}(\,\cdot\,)$:
\[ \left[\frac{a}{x}\right] = \frac{L}{L} = 1, \]
so the argument is dimensionless; hence $\sin^{-1}(\cdot)$ is dimensionless. Thus:
\[ \left[\frac{1}{a}\sin^{-1}\left(\frac{a}{x}\right)\right] = \frac{1}{L} \times 1 = L^{-1}. \]
So:
\[ \text{LHS is dimensionless}\] \[ \text{RHS has dimension } L^{-1}. \]
Therefore, the equation is dimensionally incorrect.