# Chapter 2 : Units and Measurement¶

## Example 2.1 , page : 19¶

In [1]:
# Importing module

import math

# Variable delaration

three_sixty_degree=2*math.pi

# Calculation

# Since 360° = 2π rad
one_degree=three_sixty_degree/360
# Since 1° = 60′
one_minute=one_degree/60
# Since 1′ = 60″
one_second=one_minute/60

# Result


(a) One degree = 0.017453292519943295 rad
(b) One minute = 0.0002908882086657216 rad
(c) One second = 4.84813681109536e-06 rad


## Example 2.2 , page : 19¶

In [2]:
# Importing module

import math

# Variable delaration

θ=40                     # Parallax angle in degree
AB=100                   # Distance between A and Bin m

# Calculation

# AB = AC tan θ

# Result

print("The distance of the tower C from his original position A =",round(AC,0),"m")

The distance of the tower C from his original position A = 119.0 m


## Example 2.3 , page : 19¶

In [3]:
# Importing module

import math

# Variable delaration

# Since, 1°54′ = 114′
θ=114                    # The angle θ  subtended at the moon by the two directions of observation
b=1.276*10**7            # Diameter of the Earth

# Calculation

θ=114*60*4.85*10**6
D=b/θ                    # The earth-moon distance

# Result

print("The distance of the moon from the Earth =",D,"m")

The distance of the moon from the Earth = 0.0003846385723759571 m


## Example 2.4 , page : 19¶

In [4]:
# Importing module

import math

# Variable declaration

α=1920                   # Sun's angular diameter in minutes
D=1.496*10**11           # The distance of the Sun from the Earth

# Calculation

α=1920*4.85*10**-6       # Sun's angular diameter in radians
d=α*D                    # Sun's diameter

#Result

print("Sun's diameter =",d,"m")

Sun's diameter = 1393075199.9999998 m


## Example 2.5 , page : 20¶

In [5]:
# Calculation

# The size of a nucleus is in the range of 10**-15 m and 10**-14 m.
# The tip of a sharp pin is taken to be in the range of 10**5 m and 10**4 m.
# Thus we are scaling up by a factor of 1010. An atom roughly of size 1010 m will be scaled up to a size of 1 m.

# Result

print("Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long")

Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long


## Example 2.6 , page : 25¶

In [6]:
# Calculations

# The range of variation over the seven days of observations is 162 s for clock 1, and 31 s for clock 2.
# The average reading of clock 1 is much closer to the standard time than the average reading of clock 2.
# The important point is that a clocks zero error is not as significant for precision work as its variation, because a zero-error can always be easily corrected.

# Result

print("Clock 2 is to be preferred to clock 1")

Clock 2 is to be preferred to clock 1


## Example 2.7 , page : 25¶

In [7]:
# Importing module

import math

# Variable delaration

p=[2.63,2.56,2.42,2.71,2.80]  # The readings of period of oscillation of a simple pendulum

# Calculation

T=sum(p)/len(p)
p[:] = [x - T for x in p]
q=[abs(x) for x in p]
DT=sum(q)/len(p)
δa=(round(DT,1)/round(T,1))*100

# Result

print("The absolute errors =",q)
print("A more correct way will be to write, T =",round(T,1),"±", round(DT,1)," s ")

print("The relative error or the percentage error =",δa,"≈ 4 %")

The absolute errors = [0.005999999999999783, 0.06400000000000006, 0.20400000000000018, 0.08599999999999985, 0.1759999999999997]
A more correct way will be to write, T = 2.6 ± 0.1  s
The relative error or the percentage error = 3.8461538461538463 ≈ 4 %


## Example 2.8 , page : 26¶

In [8]:
# Importing module

import math

# Variable delaration

t1=20                    # Temperature of first body in degree Celsius
Δt1=.5                   # Error in temperature of first body degree Celsius
t2=50                    # Temperature of second body degree Celsius
Δt2=.5                   # Error in temperature of first body degree Celsius

# Calculation

t=t2 - t1
Δt=max((Δt1 + Δt2),(Δt1 - Δt2))

# Result

print("The temperature difference wth the error =",t,"\u00b0 C ±",Δt,"\u00b0 C")

The temperature difference wth the error = 30 ° C ± 1.0 ° C


## Example 2.9 , page : 27¶

In [9]:
# Importing module

import math

# Variable delaration

V=5                      # The percentage error in voltage
I=2                      # The percentage error in current

# Calculation

R=V+I

# Result

print("The total error in Resistance =",R,"%")

The total error in Resistance = 7 %


## Example 2.10 , page : 27¶

In [10]:
# Importing module

import math

# Variable delaration

R1=100                   # Resistance of first resistor in ohm
ΔR1=3                    # Error in Resistance of first resistor in ohm
R2=200                   # Resistance of second resistor wth error term in ohm
ΔR2=4                    # Error in Resistance of second resistor in ohm

# Calculation

R=R1+R2
ΔR=ΔR1+ΔR2
R_prim=R1*R2/(R1+R2)
ΔR_prim=(R_prim/R1)**2*ΔR1 + (R_prim/R2)**2*ΔR2

# Result

print("The equivalent resistance of the series combination =",R,"±",ΔR,"ohm")
print("The equivalent resistance of the parallel combination =",round(R_prim,1),"±",round(ΔR_prim,1),"ohm")

The equivalent resistance of the series combination = 300 ± 7 ohm
The equivalent resistance of the parallel combination = 66.7 ± 1.8 ohm


## Example 2.11 , page : 27¶

In [11]:
# Since the relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.

# Result

print("The relative error in Z is  ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)")

The relative error in Z is  ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)


## Example 2.12 , page : 27¶

In [12]:
# Importing module

import math

# Variable delaration

L=20                     # Length in cm
ΔL=1                     # Eror in lengthin mm
t=90                     # Total time in s
Δt=1                     # Error in time in s
n=100                    # Number of oscillations

# Calculation

# The period of oscillation of a simple pendulum is, T = 2Π √(L/g)
# Hence, g = 4π²L/T²

ΔL=ΔL*10**-1
T=n/t
ΔT_div_T=Δt/t                               # Error in T
Δg_div_g= (ΔL/L) + 2*(ΔT_div_T)             # Error in g
per_g= 100*(ΔL/L) + 2*100*(ΔT_div_T )

# Result

print("The percentage error in g =",per_g,"≈ 3")

The percentage error in g = 2.7222222222222223 ≈ 3


## Example 2.13 , page : 30¶

In [13]:
# Importing module

import math

# Variable delaration

a=7.203                  # Side of the cube in m

# Calculation
# The number of significant figures in the measured length is 4.
# Hence the calculated area and the volume should therefore be rounded off to 4 significant figures.

SA=6*a**2                # Surface area of the cube
V=a**3                   # Volume of the cube

# Result

print("Surface area of the cube =",round(SA,1),"m²")
print("Volume of the cube =",round(V,1),"m^3")

Surface area of the cube = 311.3 m²
Volume of the cube = 373.7 m^3


## Example 2.14 , page : 30¶

In [14]:
# Importing module

import math

# Variable delaration

m=5.74                   # Mass of the substancein g
v=1.2                    # Volume of the substance in cm^3

# Calculation

# There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured  volume.
# Hence the density should be expressed to only 2 significant figures.
D=m/v

# Result

print("Density =",round(D,1),"g/cm^3")

Density = 4.8 g/cm^3


## Example 2.15 , page : 33¶

In [15]:
# Consider the equation (1/2)mv²=mgh
# where
# m : Mass of the body
# v : Velocity of the body
# g : Acceleration due to gravity
# h : Height

#Calculation

print("The dimensions of LHS are :[M][L T-1 ]²  = [M][L² T-²] = [M L² T-²]")
print("The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]")

# Result

print("The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct")

The dimensions of LHS are :[M][L T-1 ]²  = [M][L² T-²] = [M L² T-²]
The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct


## Example 2.16 , page : 33¶

In [1]:
# Calculation

# Every correct formula or equation must have the same dimensions on both sides of the equation.
# Also, only quantities with the same physical dimensions can be added or subtracted.
# The dimensions of the quantity on the right side are
# [M² L^3 T^-3] for (a)
# [M L² T-²] for (b)and (d)
# [M L T-²] for (c)
# The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added.
# Since the kinetic energy K has the dimensions of [M L2 T2], formulas (a), (c) and (e) are ruled out.
# Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula.
# For this, one must turn to the actual definition of kinetic energy.
# The correct formula for kinetic energy is given by (b).

# Result

print("The correct formula for kinetic energy is, K = (1/2)mv²")

The correct formula for kinetic energy is, K = (1/2)mv²


## Example 2.17 , page : 33¶

In [2]:
# Calculation

# The dependence of time period T on the  quantities l, g and m as a product may be written as :
# T = k l^x g^y m^z
# where k is dimensionless constant and x, y and z are the exponents.
# By considering dimensions on both sides, we have [L^0 M^0 T^1] = [L^1]^x [L^1 T^-2]^y [M^1]^z  = L^(x+y) T^(-2y) M^z
# On equating the dimensions on both sides, we have: x + y = 0; 2y = 1; and z = 0
# So that x = (1/2); y = -(1/2)  and  z = 0
# T = k * l^(1/2) * g^-(1/2)
# In other word,   T = k √(l/g)
# The value of constant k can not be obtained by the method of dimensions.
# Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
# Actually, k = 2π so that T = 2π √(l/g)

# Result

print("The expression for the time period of a simple pendulum is, T = 2π √(l/g)")

The expression for the time period of a simple pendulum is, T = 2π √(l/g)