CBSE Class 9 Physics All Chapters Worksheet

PHYSICS ASSIGNMENTS

Module : 9P (01 & 02)

Section A

Question 1. Fill in the blanks :-
(i) Velocity of a body is its ________in a particular direction.
(ii) In v - t graph, the area enclosed by the curve and the time - axis, gives the __________ travelled by the body.
(iii) A rubber ball dropped from a certain height is an example of ________ speed.
(iv) 20m/s = ___________ km/hr.
(v) When a body is moving with uniform velocity, its acceleration is_________.
(vi) A ball moving down on inclined plane has uniform _____________.
Answer:
(i) speed
(ii) displacement
(iii) non-uniform
(iv) \(20 \times \frac{18}{5} = 72\)
(v) zero
(vi) acceleration
In simple words: These blanks test fundamental definitions of motion parameters like velocity, acceleration, units conversion, and graphical interpretation.

Exam Tip: Remember to convert m/s to km/h by multiplying by \(\frac{18}{5}\).

Question 2. What type of motion is represented by the following graphs?
Answer:
1. Velocity-Time graph parallel to time axis: Uniform motion (constant velocity, zero acceleration).
2. Slanted straight line downward in Velocity-Time graph: Uniform retardation (uniformly decreasing velocity).
3. Trapezoidal shape in Velocity-Time graph: Motion with varying acceleration, uniform speed, and retardation.
In simple words: Graph slopes and shapes directly show how speed changes over time.

Exam Tip: Always check the axes labels carefully before interpreting any motion graph.

Question 3. State which of the following quantities are scalars and which are vectors:-
a) Velocity b) Speed c) Displacement d) Time e) Area f) Force
Answer:
Scalars: Speed, Time, Area
Vectors: Velocity, Displacement, Force
In simple words: Scalars have magnitude only, while vectors have both magnitude and direction.

Question 4. Interpret the following graphs
Answer:
First graph (Distance-Time line parallel to time axis): The body is at rest.
Second graph (Distance-Time straight line inclined to time axis): The body is moving with uniform speed.
In simple words: Flat line means zero speed, straight slanted line means constant speed.

Exam Tip: The slope of a distance-time graph gives speed.

Section B

Question 1. A body travels a distance of 5m from P to Q and then moves a distance of 10 m at right angle to PQ. Calculate the total distance travelled and the resultant displacement. Draw a diagram using appropriate scale.
Answer:
Total distance = Distance \(PQ + QR = 5\text{ m} + 10\text{ m} = 15\text{ m}\)
Resultant displacement \(PR = \sqrt{PQ^2 + QR^2}\)
\(PR = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}\text{ m} \approx 11.18\text{ m}\)
In simple words: Distance is total path length, while displacement is the straight-line distance from start to end.

Exam Tip: Always use Pythagoras theorem when motion involves right angles.

Question 2. Arrange the following in the increasing order of speed.
(i) A bicycle moving with a speed of 18km/hr.
(ii) A fast runner moving with a speed of 7m/s
(iii) A car moving with a speed of 2000 m/min.
Answer:
Convert all speeds to m/s:
(i) Bicycle speed = \(18 \times \frac{5}{18} = 5\text{ m/s}\)
(ii) Runner speed = 7 m/s
(iii) Car speed = \(\frac{2000\text{ m}}{60\text{ s}} = 33.33\text{ m/s}\)
Increasing order of speed: Bicycle (5 m/s) < Runner (7 m/s) < Car (33.33 m/s), i.e., (i), (ii), (iii).
In simple words: Convert all units to the same system before comparing values.

Exam Tip: To convert km/h to m/s, multiply by \(\frac{5}{18}\).

Question 3. A body is moving in a circle of radius R. The body moves through half the circle. Calculate distance and displacement.
Answer:
Distance = Half of circumference = \(\pi R\)
Displacement = Shortest straight-line distance between initial and final points = Diameter = \(2R\)
In simple words: Distance covers the curved path, displacement is the straight line across the circle center.

Exam Tip: For complete revolution, displacement is zero, but distance is \(2\pi R\).

Question 4. The distance between the house and the market is 7.2 km. If a man takes 10min to reach to market, calculate speed in m/s. Also express it in km/hr.
Answer:
Distance \(d = 7.2\text{ km} = 7200\text{ m}\)
Time \(t = 10\text{ min} = 10 \times 60 = 600\text{ s} = \frac{10}{60}\text{ hr} = \frac{1}{6}\text{ hr}\)
Speed in m/s = \(\frac{7200}{600} = 12\text{ m/s}\)
Speed in km/hr = \(\frac{7.2}{\frac{1}{6}} = 7.2 \times 6 = 43.2\text{ km/hr}\)
In simple words: Divide total distance by total time to find speed.

Exam Tip: Double check your time unit conversions between minutes and seconds.