# 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
print("(a) One degree =",one_degree,"rad")
print("(b) One minute =",one_minute,"rad")
print("(c) One second =",one_second,"rad")
# 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
AC=AB/math.tan(math.radians(θ))
# Result
print("The distance of the tower C from his original position A =",round(AC,0),"m")
# 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")
# 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")
# 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")
# 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")
# 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 %")
# 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")
# 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,"%")
# 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")
# 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)")
# 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")
# 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")
# 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")
# 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")
# 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")
# 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)")