 # Introduction to Physics

## Concept Of Physics

### H C Verma

1   Find the dimensions of

B. frequency and

A. linear momentum,

C. pressure.

##### Solution :

B. Frequency $\dfrac{1}{t} = [ M^0L^0T^{-1} ]$

A. $mv = [ MLT^{-1} ]$

C. $\dfrac{Force}{Area} = \dfrac{[MLT^{-2}]}{L^2} = [ML^{-1}T^{-2}]$

2   Find the dimensions of

D. moment of interia $I$

A. angular speed $\omega$

$\omega = \dfrac{\theta_2 - \theta_1}{t_2 - t_1} , \alpha = \dfrac{\omega_2 - \omega_1}{t_2 - t_1} , \Gamma = F.r$ and $I = mr^2$ The symbols have standard meanings.

C. torque $\Gamma$ and

Some of the equations involving these quantities are

B. angular acceleration $\alpha$

##### Solution :

D. moment of interia $I = Mr^2 = [M][L^2] = [ML^2T^0]$

A. angular speed $\omega = \dfrac{\theta}{t} = [ M^0 L^0 T^{-1} ]$

C. torque $\Gamma = F r = [MLT^{-2}][L] = [ML^2T^{-2}]$

B. angular acceleration $\alpha = \dfrac{\omega}{t} = \dfrac{M^0 L^0 T^{-2}}{T} = [M^0L^0T^{-3}]$

3   Find the dimensions of

The relevant equations are
$F = qE, F = qvB$, and $B = \dfrac{\mu_0 I}{2\pi \alpha}$

A. electric field $E$,

C. magnetic permeability $\mu_0$

where $F$ is force, $q$ is charge,$v$ is speed, $I$ is current, and $a$ is distance.

B. magnetic field $B$ and

##### Solution :

A. Electric field $E = \dfrac{F}{q} = \dfrac {MLT^{-2}}{[IT]} = [MLT^{-3}{-1}]$

C. Magnetic permeability $\mu_0 = \dfrac {B 2\pi a}{I} = \dfrac {MT^{-2} I^{-1} [L] }{[I]} = [MLT^{-2} I^{-2} ]$

B. Magnetic field $B = \dfrac{F}{qv} = \dfrac{MLT^{-2}}{[IT][LT^{-1}]} = [MT^{-2} I^{-1} ]$

4   Find the dimensions of

A. electric dipole moment $p$ and

The defining equations are $p = q.d$ and $M = IA$; where $d$ is distance, $A$ is area, $q$ is charge and $I$ is current.

B. electric dipole moment $M$

##### Solution :

A. Electric dipole moment $P = qI = [IT] × [L] = [LTI]$

B. Magnetic dipole moment $M = IA = [I] [L^{2}] [L^{2}I]$

5   Find the dimensions of Planck's constant $h$ from the equation $E = hv$ where $E$ is the energy and $v$ is the frequency.

##### Solution :

$E = hv$ where $E =$ energy and $v=$ frequency
$h = \dfrac{E}{v} = \dfrac{[ML^2T^{-2}] }{T^{-1}} = [ML^2 T^{-1}]$

6   Find the dimensions of

Some of the equations involving these quantities are

A.the specific heat capacity $c$,

C.the gas constant $R$.

$Q = mc(T_2 - T_1), l_t=l_0[1+ a(T_2 - T_3)]$ and $PV = nRT.$

B.the coefficient of linear expansion a $\alpha$ and

##### Solution :

A.Specific heat capacity $= C = \dfrac{Q}{m \Delta T} = \dfrac{[ML^2T^{-2}]}{[M][K]} = [L^2T^{-2} K^{-1}]$

C.Gas constant $= R = \dfrac{PV}{nT} = \dfrac{[ML^{-2} T^{-2} ][L^3]}{[(mol)][K]} = [ML^2T^{-2}K^{-1} (mol)^{-1} ]$

B. Coefficient of linear expansion = $\alpha = \dfrac{L_1 - L_2}{L_0 \Delta T} = \dfrac{[L]}{[L][R]} = [K^{-1}]$

7   Taking force, length and time to be the fundamental quantities find the dimensions of

D. energy.

A. density,

C.momentum and

B. pressure,

##### Solution :

C. Momentum $= mv (Force / acceleration) × Velocity = [F / LT^{-2}] × [LT^{-1}] = [FT]$

Taking force, length and time as fundamental quantity

$\dfrac {F}{LT^{-2}} [LT^{-1}]^2 = \dfrac {F}{LT^{-2}} [L^2T^{-2}] = [FL]$

B. Pressure $= \dfrac{F}{A} = \dfrac{ F}{L^2}= [FL^{-2}]$

D. Energy $= \dfrac {1}{2}{mv^2} = \dfrac {Force}{acceleration} {(velocity)^2}$

A. Density $= \dfrac{m}{V} = \dfrac{(force/acceleration)}{Volume} = \dfrac{[F/LT^{-2}]}{[L^2]} = \dfrac{F}{L^4T^{-2}} = [FL^{-4} T^2]$

8   Suppose the acceleration due to gravity at a place is 10 $m/s^2$ . Find its value in $cm/(minute)^2$

##### Solution :

$g=10 \dfrac{meter}{sec^2} = 36 \times 10^5 cm/min^2$

9   If the average speed of a snail is 0.02 $\textit{mile/hr}$ and the average speed of leopard is 70 $\textit{miles/hr}$. Convert these speeds in S.I Units?

##### Solution :

The average speed of snail = 0.02 mile/hr converting to S.I units, $\frac{0.02 X 1.6 X 1000}{3600}$ m/sec $[1 mile = 1.6 km = 1600 m] = 0.0089 ms^{-1}$

the average speed of leopard = 70 miles/hr In S.I Units = 70 miles/hour = $\frac{70 X 1.6 X 1000}{3600}$ = 31 $m/s$

10   The height of mercury column in a barometer in a Calcutta laboratory was recorded to be $75$ cm. Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury =$13.6$, Density of water = $103^{3}$ $kg/m^{3}$, g = $9.8$ $in/s^{2}$ at Calcutta. Pressure =$hpg$ in usual symbols.

##### Solution :

Height h = $75 cm$, Density of mercury = $13600 kg/m^3$, g = $9.8 ms^{-2}$ then, Pressure = $hfg$ = $10 X 10 ^4$ $N/m^2$ (approximately) In C.G.S Units, P = $10 X 10^5$ $dyne/cm$

11   Express the power of a 100 watt bulb in CGS unit.

##### Solution :

12   The normal duration of I.Sc. Physics practical period in Indian colleges is $100$ minutes. Express this period in micro centuries. $1$ micro century = $10^{-6}$ x $100$ years. How many micro centuries did you sleep yesterday ?

##### Solution :

In $1$ micro century = $10^4$ $X$ $100$ years = $10^4$ $X$ $365$ $X$ $24$ $X$ $60$ min . So $100$ min = $10^5/5256$ = $1.9$ micro century.

13   The surface tension of water is $72$ $dyne/cm$. Convert it in SI unit.

##### Solution :

Surface tension of water = $72$ dyne/cm. In S.I Unit, $72$ dyne/cm = $0.072$ N/m

14   The kinetic energy $K$ of a rotating body depends on its moment of inertia $I$ and its angular speed $\omega$. Assuming the relation to be $K= kI^a\omega^b$ where $k$ is a dimensionless constant, find $a$ and $b$. Moment of inertia of a sphere about its diameter is $\frac{2}{5} Mr^2$.

##### Solution :

According to principle to homogeneity of dimensions, $[ML^2T^{-2}]$ = $[ML^2T^{-2}] [T^{-1}]^b$

$K$ = $KI^a \omega^b$ where $k$ = kinetic energy of rotating body and $k$ = dimensionless constant

Dimensions of right side are, $I^a$ = $[ML^2]^a$ , $\omega$ = $[T^{-1}]^b$

Equating the dimensions of both sides, $2$ = $2a$ and $-2$ = $-b$ $\Rightarrow$ $a$ = $1$ and $b$ = $2$

Dimensions of left side are, $K$ = $[ML^2T^{-2}]$

15   Theory of relativity reveals that mass can be converted into energy. The energy $E$ so obtained is proportional to certain powers of mass $m$ and the speed $c$ of light. Guess a relation among the quantities using the method of dimensions.

##### Solution :

So, the relation is $E$ = $KMC^2$

Dimensions of right side, $M^a$ = $[M]^a$ , $[C]^b$ = $[LT^{-1}]^b$

Let energy $E$ $\infty$ $M^aC^b$, where $M$ = Mass, $C$ = Speed of light

$\Rightarrow$ $a$ = $1$ ; $b$ = $2$

Dimensions of the left side, $E$ = $[ML^2T^{-2}]$

Therefore, $[ML^2T^{-2}]$ = $[M]^a[LT^{-1}]^b$

$\Rightarrow$ $E$ = $KM^aC^b$ $K$ = ( Proportionality constant)

16   Let $I$ = current through a conductor, $R$ = its resistance and $V$ = potential difference across its ends. According to Ohm's law, product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for $R$ and $V$ are $ML^{2} I^{-2}T^{-3}$ and $ML^{2} T^{3} I^{-1}$ respectively.

##### Solution :

Therefore, $[ML^2T^3I^{-1}]$ = $[ML^2T^3I^{-2}]$ $[I]$

Dimensional Formulae of $R$ = $[ML^2T^{-3}I^{-2}]$

Dimensional Formulae of $I$ = $[I]$

$\Rightarrow$ $V$ =$IR$

Dimensional Formulae of $V$ = $[ML^2T^{3}I^{-1}]$

17   The frequency of vibration of a string depends on the length $L$ between the nodes, the tension $F$ in the string and its mass per unit length $m$. Guess the expression for its frequency from dimensional analysis.

##### Solution :

Equating the dimensions of both sides,

$-2b = -1$

$L^a$ = $[L^a]$, $F^b$ = $[MLT^{-2}]^b$, $M^c$ = $[ML^{-1}]^c$

Frequency $f$ = $KL^aF^bM^cM$ = Mass/unit length, $L$ = length, $F$ = tension (force)

$a$ = $-1$, $b$ = $\frac{1}{2}$ and $c$ = $\frac{-1}{2}$

$M^0L^0T^{-1}$ = $KM^{b+c} L^{a+b-c} T^{-2b}$

$-c + a + b = 0$

Dimensions of right side,

Now, solving the equation we get,

Therefore, $[T^{-1}]$ = $K[L]^a$ $[MLT^{-2}]^b$ $[ML^{-1}]^c$

Therefore, $b + c = 0$

Dimensions of $f$ = $[T^{-1}]$

Therefore, So frequency $f$ = $KL^{-1}F^\frac{1}{2}M^\frac{-1}{2}$ = $\frac{K}{L}F^\frac{1}{2}M^\frac{-1}{2}$ = $\frac{K}{L}$ = $\sqrt\frac{F}{M}$

18   Test if the following equations are dimensionally correct :

d) $\nu = \frac{1}{2\pi} \sqrt\frac{mgl}{I}$

a) $h = \frac{2 S Cos\theta}{prg}$

c) $V = \frac{\pi P r^4 t}{ 8 \eta \iota }$

where $h$ = height, $S$ = surface tension, $\rho$ = density, $P$ = pressure, $V$ = volume, $\eta$ = coefficient of viscosity, $\nu$ = frequency and $I$ = moment of inertia.

b) $v = \sqrt \frac{p}{\rho}$

##### Solution :

LHS = RHS

c) $V$ = $\frac{(\pi p r^4 t)}{(8\eta I)}$

LHS = dimension of $V$ = $[Lt^{-1}]$

So relation is correct.

Density = $\rho$ = $\frac{M}{V}$ = $[ML^{-3}T^0]$

RHS = $\sqrt{(mgl/I)}$ = $\sqrt\frac{[M][LT^{-2}]}{[ML^{2}]}$ = $[T^{-1}]$

Dimension of $p$ = $[ML^{-1}T^{-2}]$, $r^4$ = $[L^4]$, $t$ = $[T]$

a) $h$ = $\frac{2SCos\theta}{prg}$

Dimension of $p$ = $\frac{F}{A}$ = $[ML^{-1}T^{-2}]$

So, the relation is correct.

RHS = $\frac{2SCos\theta}{prg}$ = $\frac{[MT^{-2}]}{[ML^{-3}T^0][L][LT^{-2}]}$ = $[M^0L^1T^0]$ = $[L]$

b) $v$ = $\sqrt\frac{P}{\rho}$ where $v$ = velocity

So, the relation is correct

RHS = $\frac{\pi p r^4 t}{8\eta I}$ = $\frac{[ML^{-1}T^{-2}][L^4][T]}{[ML^{-1}T^{-2}][L]}$

Surface tension =$S$ = $F/I$ = $\frac{MLT^{-2}}{L}$ = $[MT^{-2}]$

RHS = $\sqrt\frac{P}{\rho}$ = $\sqrt\frac{[ML^{-1}T^{-2}]}{[ML^{-3}]}$ = $[L^2T^{-2}]^{\frac{1}{2}}$ = $[LT^{-1}]$

LHS = Dimesion of $v$ = $[T^{-1}]$

LHS = Dimension of $V$ = $[L^3]$

LHS = dimension of $V$ = $[LT^{-1}]$

Radius = $r$ = $[L]$, $g$ = $[LT^{-2}]$

So, the relation is correct.

LHS = RHS

Coefficient of viscosity = $[ML^{-1}T^{-1}]$

LHS = [L]

d) $v$ = $\frac{1}{2\pi}\sqrt{(mgl/l)}$

Dimension of $\rho$ = $\frac{m}{V}$ = $[ML^{-3}]$

19   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 :

So, the equation is dimensionally incorrect.

Dimensions of the left side = $\int\frac{dx}{\sqrt{(a^2 - x^2)}}$ = $\int\frac{L}{\sqrt{(L^2 - L^2})}$ = $[L^0]$

So, dimensions of $\int\frac{dx}{\sqrt{(a^2 - x^2)}}$ $\ne$ $\frac{1}{a} Sin{-1}\big(\frac{a}{x})$

Dimensions of the right side = $\frac{1}{a} Sin^{-1}\big(\frac{a}{x})$ = $[L^{-1}]$