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}] $