Solutions to Exercises on Introduction to Physics HC Verma's Concepts of Physics Part 1 (10-14)

 Q#10

The height of the 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 kg/m³, g = 9.8 m/s² at Calcutta. Pressure = hρg in usual symbols.      

Answer:
Here h = 75 cm =0.75 m

Density of mercury, ρ = specific gravity x Density of water = 13.6 x 10³ kg/m³
g = 9.8 m/s²

Hence the pressure at the base of mercury column  in SI units = hρg

= 0.75 x 13.6 x 10³ x 9.8 N/m²
= 99960 N/m²                
= 10 x 10⁴ N/m²

In CGS units, units of mass = g, length = cm and time = s. We know that 1 kg = 1000 g, 1 m = 100 cm.

Hence, 1 N = 1 kgm/s² = (1000 g)(100 cm)/s²

= 10⁵ gcm/s² = 10⁵ dyne

So, 1 N/m² = (10⁵ dyne)/(100 cm)²

= 10 dyne/cm²

So, in CGS unit pressure = 10x10⁴ N/m²

= (10 x 10⁴)(10 dyne/cm²)

= 10 x 10⁵ dyne/cm²

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

Answer:
(a) 1 watt = 1 joule/second = 1 Nm/s

But 1 N = 1 kgm/s² = 1000 x 100 gcm/s² = 10⁵ dyne

Hence 1 watt = 10⁵ (100 dynecm/s) = 10⁷ erg/s

So, 100 watt = 100 x 10⁷ erg/s = 10⁹ erg/s

Q#12
The normal duration of I.Sc. physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury = 10⁻⁶x100 years. How many microcenturies did you sleep yesterday?            

Answer:
(a) 10⁻⁶ x 100 years = 1 microcentury

{1 year = 365 x 24 x 60 mimutes}

So, 10⁻⁶ x 100 x 365 x 24 x 60 minutes = 1 microcentury

100 minutes = 10⁶/(365 x 24 x 60) microcenturies = 1.90 microcenturies.    

I slept for 6 hours yesterday, i.e. = 6 x 60 min = 360 minutes = 360 x (1.9/100) micocenturies
= 6.84 microcenturies

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

Answer:
(a) The surface tension of water = 72 dyne/cm

{But 1 dyne = 1 gcm/s² = 1/(1000 x 100) kgm/s² = 10⁻⁵ N}

= 72 x 10⁻⁵ x 100 N/m = 0.072 N/m    

Q#14
The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ⍵. Assuming the relation to be K = kIaωb where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is (2/5)Mr².  

Answer:
(a) The dimensions of the moment of inertia, I = [ML²], and of angular speed, ⍵ = [T⁻¹]
The dimensions of KE, K = [ML²T⁻²]

For the given relation to be dimensionally correct,

 [ML²T⁻²] = [ML²]^a[T⁻¹]^b = [M^aL²^aT⁻^b]

Equating the dimensions on both sides,

a = 1

And, -b = -2 or b = 2

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