Class 9 Physics Chapter 1 numerical problems are according to the new syllabus of the Punjab Board. Chapter 1 is related to Physical Quantities and Measurements in which you will study Physics, applications of Physics, and daily used measuring instruments.
Class 9 Physics Chapter 1 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
micro = µ = 10-6
nano = n = 10-9
pico = p = 10-12
kilo = k = 103
mega = M = 106
mili = m = 10-3
one day = 24 x 60 x 60 seconds
Numerical Problems Chapter 1: Physical Quantities and Measurements
1.1 Calculate the number of seconds in a (a) day (b) week (c) month and state your answers using SI prefixes.
Solution:
a) Seconds in a day:
- Hours in a day = 24
- Minutes in an hour = 60
- Seconds in a minute = 60 24 × 60 × 60 = 86,400 seconds = 8.64 × 10⁴ s
b) Seconds in a week:
- Days in a week = 7
- Seconds in a week = 86,400 × 7 = 604,800 seconds = 6.048 × 10⁵ s
c) Seconds in a month (assuming 30-day month):
- Days in a month = 30
- Seconds in a month = 86,400 × 30 = 2,592,000 seconds = 2.592 × 10⁶ s
1.2 This problem is already answered in the scientific notation above:
Solution:
a) 8.64 × 10⁴ s
b) 6.048 × 10⁵ s
c) 2.592 × 10⁶ s
1.3 Solve the addition/subtraction problems in scientific notation:
Solution:
a) 4 × 10–⁴ kg + 3 × 10–⁵ kg
First, convert to same exponent: 3 × 10–⁵ = 0.3 × 10-4
Now add: 4 × 10-4 + 0.3 × 10-4 = 4.3 × 10-4 kg
b) 5.4 × 10-6 m – 3.2 × 10-5 m
Convert to same exponent: 5.4 × 10-6 = 0.54 × 10-5
Now subtract: 0.54 × 10-5 – 3.2 × 10-5 = -2.66 × 10-5 m
1.4 Solve the following multiplication or division. State your answers in scientific notation.
(a) (5 × 10⁴ m) × (3 × 10–² m)
(b) 6 × 10⁸ kg / 3 × 10⁴ m³
Solution:
a) (5 × 10⁴ m) × (3 × 10⁻² m)
Multiply numbers: 5 × 3 = 15
Add exponents: = 104×10-2 = 104-2 = 102
Result: 15 × 10²
Scientific Notation: 1.5 × 10³ m²
b) (6 × 10⁸ kg)/(3 × 10⁴ m³)
Divide numbers: 6 ÷ 3 = 2
Subtract exponents: 10⁸ ÷ 10⁴= 108-4 = 104
Result: 2.0 × 10⁴ kg/m³
1.5 Calculate the following and state your answer in scientific notation.
(3 × 10² kg) × (4.0 km) ÷ 5 × 10² s²
Solution:
Calculate (3 × 10² kg) × (4.0 km) ÷ 5 × 10² s²
First, convert km to m: 4.0 km = 4.0 × 10³ m
Now multiply:
- Numbers: 3 × 4.0 ÷ 5 = 2.4
- Exponents: 102 × 103 ÷ 102= 102+3-2 = 102
Result: 2.4 × 102 kg⋅m
1.6 State the number of significant digits in each measurement.
(a) 0.0045 m (b) 2.047 m (c) 3.40 m (d) 3.420 × 104m
Solution:
a) 0.0045 m → 2 significant digits (45)
b) 2.047 m → 4 significant digits (2047)
c) 3.40 m → 3 significant digits (340)
d) 3.420 × 10³ m → 4 significant digits (3420)
1.7 Write in scientific notation:
(a) 0.0035 m (b) 206.4 × 10² m
Solution:
a) 0.0035 m
0.0035 m = 3.5 × 10-3 m
b) 206.4 × 102
206.4 × 10² = 2.064 × 102 × 102
Therefore: 2.064 × 102+2 m
= 2.064 × 104m
1.8 Write using correct prefixes:
(a) 5.0 × 10⁴ cm (b) 580 × 10² g (c) 45 × 10–4 s
Solution:
a) 5.0 × 104 cm
First, convert to base unit (meters):
= 5.0 × 10⁴ cm
= 5.0 × 104 ÷ 102 m (Divided by 102)
= 5.0 × 104-2 m
= 5.0 × 10² m
Convert to km:
= 5.0 × 102 ÷ 103 km (Divided by 103)
= 5.0 × 102-3 km
= 5.0 × 10-1 km
= 0.5 km
b) 580 × 10² g
= 580 × 102 g = 580 × 100 g
= 58,000 g = 58 × 1000 g
= 58 × 103 g
= 58 kg (103 = kilo)
c) 45 × 10–4 s
= 45 × 10–4 s = 4.5 × 101 × 10–4 s
= 4.5 × 101–4 s = 4.5 × 10–3 s
= 4.5 ms (10–3 = mili)
1.9 Light year is a unit of distance used in Astronomy. It is the distance covered by light in one year. Taking the speed of light as 3.0 × 10⁸ m s-1, calculate the distance.
Solution:
Speed of light = 3.0 × 10⁸ m/s
Time = 1 year = 365.25 × 24 × 60 × 60 seconds
Time= 31,557,600 seconds
Time= 3.16 × 10⁷ s
Using formula,
Distance = Speed × Time
Distance = (3.0 × 10⁸ m/s) × (3.16 × 10⁷ s)
Distance = 9.46 × 10¹⁵ m
1.10 Express the density of mercury given as 13.6 g cm-3
in kg m-3
Solution:
= 13.6 g cm⁻³
To convert to kg m⁻³:
- 1 g = 10⁻³ kg
- 1 cm³ = 10⁻⁶ m³
Therefore: 13.6 g/cm³ = 13.6 × (10⁻³/10⁻⁶)
= 13.6 × 10³ kg/m³
= 1.36 × 10⁴ kg/m³
Class 9 Physics Full New Book Numerical Problems with Solution
