Class 9 - Unit # 01 : Physical Quantities & Measurements - Solved Numericals/Structured Questions


1. Matching Actions with Branches of Physics

Below is the correct classification of everyday physical actions and their corresponding branches of physics.

Column A: Action Column B: Branch of Physics
Cooking / Bar B.Q. Thermodynamics
Turning the bulb on Electricity
Riding a bicycle Mechanics
Looking for Giant Galaxies Astrophysics
Producing a loud sound Sound / Acoustics
Describing an atom Atomic Physics
Obtaining physical energy from the Earth Geophysics

2. Physical Quantities and S.I. Units

Physical Quantity S.I. Unit Type
Electric Current Ampere (A) Base
Volume m3 Derived
Time Second (s) Base
Temperature Kelvin (K) Base
Force Newton (N) Derived
Density kg/m3 Derived
Acceleration m/s2 Derived

3. Conversion of Units

Convert the following physical values into their corresponding formats:

a) 230 cm = 2.3 m
b) 250 g = 0.25 kg
c) 0.5 s = 500 ms
d) 0.8 m = 800 mm
e) 350 ms = 0.35 s
f) 1.2 kg = 1200 g

4. Measurement and Instruments

a) What is the measurement of the width of the aluminum sheet shown on the Vernier caliper?

1. Main Scale Reading (MSR): 3.9 cm

2. Vernier Scale Reading (VSR): 1 × L.C = 1 × 0.01 cm = 0.01 cm

Total Width Measurement = MSR + VSR = 3.9 cm + 0.01 cm = 3.91 cm

b) Which gives more precise measurement: Vernier caliper, Screw Gauge or meter rule?

The Screw Gauge provides the most precise measurements because it possesses the smallest Least Count (typically 0.01 mm or 0.001 cm), which is much smaller than that of a Vernier caliper (0.1 mm) or a meter rule (1 mm).

5. Measuring Pendulum Oscillations

i) What would be the most accurate way of measuring the time for one oscillation of a swinging pendulum?

a) Record time for 10 oscillations and multiply by 10
• b) Record time for 10 oscillations and divide by 10 (Correct Answer)
c) Record time for one oscillation
d) Record time from X to Y and double it

Reason: Recording time for multiple oscillations and dividing reduces the human reaction error associated with pressing the start/stop buttons.

ii) Suggest an instrument for measuring the time period more accurately.

A Digital Stopwatch (or smart photogate system). A digital stopwatch is highly accurate and can measure time intervals up to 0.01 seconds, minimizing mechanical and parallax errors.

6. Prefix and Metric Notation

Rewrite the following quantities using appropriate metric prefixes:

a) 75000 m = 75 km (Kilometers)
b) 2/1000 s = 2 ms (Milliseconds)
c) 1/1,000,000 g = 1 μg (Micrograms)
d) 1,000,000,000 m = 1 Gm (Gigameters)

7. Standard and Scientific Notation

Convert the values into scientific or standard notation forms:

a) Radius of 1st orbit of Hydrogen atom (r = 0.53 Å):
Scientific notation = 5.3 × 10-1 Å (or 5.3 × 10-11 m)

b) 1 Light Year distance parameter (2,628,000,000,000 m):
Scientific notation = 2.628 × 1012 m

c) Vacuum pressure of 2.7 × 10-4 torr:
Standard notation = 0.00027 torr

8. Calculating Volumes of Geometric Shapes

Given parameters: Length (l) = Radius (r) = Height (h) = 2 m.

a) Sphere:
V =
4
3
π r3 =
4
3
× 3.1416 × (2)3
V = 33.51 m3
b) Cube:
V = l3 = (2)3
V = 8.00 m3
c) Cylinder:
V = π r2 h = 3.1416 × (2)2 × 2
V = 25.13 m3
d) Pyramid (Square Base):
V =
1
3
× Area of Base × h =
1
3
× (2 × 2) × 2
V = 2.67 m3

9. Mass, Volume and Density Relation

Find the density of wood in spherical form (V = 33.51 m3) and cube form (V = 8.00 m3) if the mass is 1 kg. Is there any change in density due to shape?

1. Density as a Sphere (ρsphere):
ρ =
Mass (m)
Volume (V)
=
1 kg
33.51 m3
ρsphere ≈ 0.0298 kg/m3
2. Density as a Cube (ρcube):
ρ =
1 kg
8.00 m3
ρcube = 0.125 kg/m3
Physical Conclusion:

No, density does not change due to shape alone. Density is an intrinsic physical property of a material (Wood). However, because these wood samples have different volumes for the same 1 kg mass, their densities differ because they represent different structures or states of compression, not purely because of their shapes.

10. Immersing Stone in a Measuring Cylinder

A measuring cylinder is filled with 500 cc of water. A stone of mass 20 g is immersed such that the water level rises up to 800 cc. Which statement is correct?

a) The difference between the readings gives the density of the stone.
• b) The difference between the readings gives the volume of the stone. (Correct Answer)
c) The final reading gives the density of the stone.
d) The final reading gives the volume of the stone.

Reason: The displacement method dictates that the volume of water displaced equals the physical volume of the submerged object:
Volume = 800 cc - 500 cc = 300 cc (or cm3).

11. Determining Significant Figures

Determine the number of significant figures in the following measurements:

a) 980 has 2 significant figures.
b) 91.60 has 4 significant figures.
c) 10010.100 has 8 significant figures.
d) 0.0086 has 2 significant figures.

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