StudentAidUnit 1: Measurements — Numericals
11th Class Physics · Unit 1: Measurements
1.1.Astronomers usually measure astronomical distances in light years. One light year is the distance that light travels in one year. If speed of light is 3 \times 10^{8} m s^{-1}, what is one light year in metres?
Given
Speed of light
c = 3 × 108 m/s
Time in 1 year
t = 365 × 24 × 60 × 60 = 31,536,000 s = 3.156 × 107 s
To find
Distance covered in one light year (in metres)
S = ?
Formula
Distance formula
Distance = Speed × time
Working
Substituting values
S = vt = 3 × 108 × 3.1536 × 107 s
Result
One light year
= 9.5 × 1015 m
1.2.Write the estimated answer of the following in standard form. (a) How many seconds are there in 1 year? (b) How many years are in 1 second?
Given
Part (a): Time conversion factors
1 year = 365 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds
To find
Number of seconds in 1 year
? s
Calculation
365 × 24 × 60 × 60 = 31,536,000 s = 3.1536 × 107 s
Result
Part (a) result
3.15 × 107 s
Given
Part (b): From part (a)
1 year = 3.15 × 107 seconds
To find
Number of years in 1 second
? years
Calculation
1 second = frac{1}{3.15 × 107 Years = 3.1 × 10-8 Years
Result
Part (b) result
3.1 × 10-8 years
1.3.The length and width of a rectangular plate are measured to be 18.3 cm and 14.60 cm, respectively. Find the area of the plate and state the answer to correct number of significant figures.
Given
Length
L = 18.3 cm
Width
W = 14.60 cm
To find
Area of the plate (considering significant figures)
A = ?
Formula
Area formula
A = L × W
Working
Substituting values
A = 18.3 × 14.60 = 267.18 cm2
Determining significant figures
Smallest significant figures = 3 (from 18.3)
Result
Area of the plate
A = 267 cm2
1.4.Find the sum of the masses given in kg up to appropriate precision: (i) 3.197, (ii) 0.068, (iii) 13.9, (iv) 3.28
Given
Masses
m1 = 3.197 kg, m2 = 0.068 kg, m3 = 13.9 kg, m4 = 3.28 kg
To find
Sum of masses with correct precision
m = ?
Addition
m1 + m2 + m3 + m4 = 3.197 + 0.068 + 13.9 + 3.28 = 20.445 kg
Determining decimal places
Smallest decimal place is 1 (from 13.9)
Result
Sum of masses
= 20.4 kg
1.5.The diameter and length of a metal cylinder measured with the help of a Vernier Callipers of least count 0.01 cm are 1.22 cm and 5.35 cm respectively. Calculate its volume and uncertainty in it.
Given
Diameter
D = 1.22 cm
Length
L = 5.35 cm
Least count
L.C. = 0.01 cm
To find
Volume of cylinder and uncertainty
V = ?
Radius calculation
r = D2 = 1.222 = 0.61 cm
Formula
Volume formula
V = pi r2 L
Working
Substituting values
V = 3.14 × (0.61)2 × 5.35 = 3.14 × 0.3721 × 5.35 = 6.25090 cm3
Significant figures
= 6.2 cm3 (correct significant figs)
Uncertainty calculation
V = pi × left(d2right)2 × L
Total uncertainty in volume
Total uncertainty = 2(%uncertainty in diameter) + (%uncertainty in length)
Percentage uncertainty in diameter
0.011.22 × 100 = 0.82%
Percentage uncertainty in length
0.015.35 × 100 = 0.19%
Total percentage uncertainty
2(0.82%) + 0.19% = 1.64% + 0.19% = 1.83% ≈ 1.8%
Absolute uncertainty
1.8 × 6.2100 = 11.16100 = 0.1 cm3
Result
Volume with uncertainty
V = 6.2 ± 0.1 cm3
1.6.Show that the expression v_{f}^{2} - v_{i}^{2} = 2aS is dimensionally correct, where v_i is the initial velocity, a is the acceleration and v_f is the velocity after covering a distance S.
Given
Equation
vf2 - vi2 = 2aS
To find
Check dimensional correctness
Formula
Dimensions of velocity squared
Dimensions of v2 = [L2T-2]
Dimensions of acceleration
Dimensions of a = [LT-2]
Dimensions of distance
Dimensions of S = [L]
Dimensions of right side
aS = [LT-2] × [L] = [L2T-2]
Result
Conclusion
Both sides have same dimensions. So, equation is dimensionally correct.
1.7.Show that the famous "Einstein equation" E = mc^{2} is dimensionally consistent.
Given
Equation
E = mc2
To find
Check dimensional consistency
Recall
W = E = Fd
Dimensions of energy
Dimensions of E = [ML2T-2]
Dimensions of mass
Dimensions of m = [M]
Dimensions of speed of light
Dimensions of c = [LT-1]
Dimensions of right side
mc2 = [M][L2T-2] = [ML2T-2]
Result
Conclusion
Both sides match. Equation is dimensionally consistent.
1.8.Derive a formula for the time period of a simple pendulum using dimensional analysis. The various possible factors on which the time period T may depend are: (i) length of the pendulum \ell, (ii) mass of the bob m, (iii) angle \theta which the thread makes with the vertical, (iv) accelerates due to gravity g.
Given
Possible dependence factors
T depends on ell, m, θ, and g
To find
Formula for time period T
Formula
Assume general form
T ∝ ma × ellb × θc × gd
Or with constant
T = constant × ma × ellb × θc × gd quad (1)
Task
Find the values of powers a, b, c and d
Write dimensions of both sides
[T] = constant × [M]a [L]b [1]c [LT-2]d
Compare dimensions for T
[T] = [T]-2d ⇒ 1 = -2d ⇒ d = -12
Compare dimensions for M
[M]0 = [M]a ⇒ a = 0
Compare dimensions for L
[L]0 = [L]b+d ⇒ 0 = b + d ⇒ b = -d = 12
For angle θ
θ is dimensionless, so c can be any value but θc = 1
Working
Substitute values of a, b, θ and d into Eq. (1)
T = constant × m0 × ell1/2 × 1 × g-1/2
Result
Final formula
T = constant sqrt{ellg