Concept Of Physics Introduction to Physics

H C Verma

Concept Of Physics

1.   Find the dimensions of

A. linear momentum,

B. frequency and

C. pressure.

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

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

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

2.   Find the dimensions of

A. angular speed $\omega$

B. angular acceleration $\alpha$

C. torque $\Gamma$ and

D. moment of interia $I$

Some of the equations involving these quantities are

$ \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.

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

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

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

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

3.   Find the dimensions of

A. electric field $ E $,

B. magnetic field $ B $ and

C. magnetic permeability $ \mu_0 $

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

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

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

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

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

4.   Find the dimensions of

A. electric dipole moment $p$ and

B. electric dipole moment $M$

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

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.

$E = hv$ 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

A. the specific heat capacity $c $,

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

C. the gas constant $R$.

Some of the equations involving these quantities are

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

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

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

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

A. density,

B. pressure,

C. momentum and

D. energy.

Taking force, length and time as fundamental quantity

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

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

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

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

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

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

$ 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?

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.

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.

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 ?

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.

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 $.

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

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

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

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

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

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.

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

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

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

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

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

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

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

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.

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

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

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

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

$ \Rightarrow $ $ V $ =$ IR $

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.

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

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

Dimensions of right side,

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

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

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

Equating the dimensions of both sides,

Therefore, $ b + c = 0 $

$ -c + a + b = 0 $

$ -2b = -1 $

Now, solving the equation we get,

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

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 :

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

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

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

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

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

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

LHS = [L]

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

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

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

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

LHS = RHS

So, the relation is correct.

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

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

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

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

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

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

So, the relation is correct.

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

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

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

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

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

So relation is correct.

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

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

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

LHS = RHS

So, the relation is correct

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?

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

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

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

So, the equation is dimensionally incorrect.