Python Fundamentals
Class 11 Python Chapter 3 MCQs
Class 11 Python Chapter 3 MCQs
The Adventures of Toto (RTC-3)
Read the passage carefully:
Even Grandfather realised that. We were not well-to-do, and could not afford the frequent loss of dishes, clothes, curtains and wall paper. So Grandfather found the Tonga-driver and sold Toto back to him – for only three rupees.
Even Grandfather realised that. We were not well-to-do, and could not afford the frequent loss of dishes, clothes, curtains and wall paper. So Grandfather found the Tonga-driver and sold Toto back to him – for only three rupees.
The Adventures of Toto (RTC-2)
Read the passage carefully:
The poor man was taken aback; but, with great presence of mind and much to Grandfather’s annoyance, he said, “Sir, you have a dog with you. You’ll have to pay for it accordingly.”
The poor man was taken aback; but, with great presence of mind and much to Grandfather’s annoyance, he said, “Sir, you have a dog with you. You’ll have to pay for it accordingly.”
The Adventures of Toto (RTC-1)
Read the passage carefully:
His presence in the house still a secret, Toto was now transferred to a big cage in the servants’ quarters. There a number of Grandfather’s pets lived very sociably together – a tortoise, a pair of rabbits, a tame squirrel and, for a while, my pet goat.
His presence in the house still a secret, Toto was now transferred to a big cage in the servants’ quarters. There a number of Grandfather’s pets lived very sociably together – a tortoise, a pair of rabbits, a tame squirrel and, for a while, my pet goat.
The Lost Child (RTC-3)
Read the passage carefully:
The child turned his nose away from the basket and reiterated his sob, “I want my mother. I want my father!”
The child turned his nose away from the basket and reiterated his sob, “I want my mother. I want my father!”
The Lost Child RTC-2
Read the passage carefully:
“I want that burfi,” he slowly murmured. But he half knew as he begged that his plea would not be heeded because his parents would say he was greedy. So, without waiting for an answer, he moved on.
“I want that burfi,” he slowly murmured. But he half knew as he begged that his plea would not be heeded because his parents would say he was greedy. So, without waiting for an answer, he moved on.
The Lost Child (RTC-1)
Read the passage carefully:
“Come, child,” called his parents, as he lagged behind, fascinated by the toys in the shops that lined the way. He hurried towards his parents, his feet obedient to their call, his eyes still lingering on the receding toys.
“Come, child,” called his parents, as he lagged behind, fascinated by the toys in the shops that lined the way. He hurried towards his parents, his feet obedient to their call, his eyes still lingering on the receding toys.
Class 9 Work and Energy Numericals
Class 9 Work and Energy Numericals
Work and energy solved numericals
Numerical 1: Work Done by a Constant Force
Q1. A force of 12 N acts on an object and moves it through a distance of 4 m in the direction of the force. Find the work done.
Solution:
Given:
Force, F = 12 N
Displacement, s = 4 m
Formula:
Work done, W = F × s
Calculation:
W = 12 × 4
W = 48 J
Answer: Work done = 48 joules
Numerical 2: Zero Work Done
Q2. A person pushes a wall with a force of 200 N for 5 minutes, but the wall does not move. Find the work done.
Solution:
Displacement, s = 0 (wall does not move)
Formula:
W = F × s
W = 200 × 0 = 0 J
Answer: Work done = 0 joule
Because displacement is zero
Numerical 3: Work Done Against Gravity
Q3. A boy lifts a bag of mass 10 kg to a height of 2 m. Calculate the work done.
(Take g = 10 m/s²)
Solution:
Given:
m = 10 kg
h = 2 m
g = 10 m/s²
Formula:
W = mgh
Calculation:
W = 10 × 10 × 2
W = 200 J
Answer: Work done = 200 joules
Numerical 4: Kinetic Energy
Q4. Find the kinetic energy of a car of mass 1000 kg moving with a speed of 20 m/s.
Solution:
Given:
m = 1000 kg
v = 20 m/s
Formula:
KE = ½ mv²
Calculation:
KE = ½ × 1000 × (20)²
KE = 500 × 400
KE = 2,00,000 J
Answer: Kinetic energy = 2 × 10⁵ J
Numerical 5: Change in Kinetic Energy
Q5. A ball of mass 5 kg is initially at rest. It starts moving with a speed of 6 m/s. Find the work done on the ball.
Solution:
Initial velocity, u = 0
Final velocity, v = 6 m/s
Mass, m = 5 kg
Formula:
Work done = Change in kinetic energy
KE = ½ mv²
KE = ½ × 5 × 36
KE = 90 J
Answer: Work done = 90 joules
Numerical 6: Potential Energy
Q6. Calculate the potential energy of an object of mass 8 kg placed at a height of 5 m.
(g = 10 m/s²)
Solution:
Given:
m = 8 kg
h = 5 m
g = 10 m/s²
Formula:
PE = mgh
Calculation:
PE = 8 × 10 × 5
PE = 400 J
Answer: Potential energy = 400 joules
Numerical 7: Law of Conservation of Energy
Q7. A stone of mass 2 kg is dropped from a height of 10 m.
Find its kinetic energy just before hitting the ground.
(g = 10 m/s²)
Solution:
Initial potential energy = mgh
PE = 2 × 10 × 10 = 200 J
According to law of conservation of energy,
Final KE = Initial PE
Answer: Kinetic energy = 200 joules
Numerical 8: Power
Q8. A man lifts a load of 500 N to a height of 4 m in 10 seconds. Find his power.
Solution:
Work done = Force × Height
W = 500 × 4 = 2000 J
Time, t = 10 s
Formula:
Power = Work / Time
P = 2000 / 10
P = 200 W
Answer: Power = 200 watts
Numerical 9: Electrical Energy Consumption
Q9. An electric bulb of 100 W glows for 5 hours. Find the energy consumed in joules.
Solution:
Power = 100 W
Time = 5 hours = 5 × 3600 = 18000 s
Formula:
Energy = Power × Time
Energy = 100 × 18000
Energy = 18,00,000 J
Answer: Energy consumed = 1.8 × 10⁶ joules
Numerical 10: Velocity from Kinetic Energy
Q10. The kinetic energy of an object is 450 J and its mass is 10 kg. Find its velocity.
Solution:
KE = ½ mv²
450 = ½ × 10 × v²
450 = 5v²
v² = 90
v = √90 ≈ 9.5 m/s
Answer: Velocity ≈ 9.5 m/s
Exam Central
Numerical 11: Combined Concept (PE → KE)
Q11. A stone of mass 2 kg is thrown vertically upward with a speed of 10 m/s.
Find the maximum height reached by the stone.
(Take g = 10 m/s²)
Solution:
Given:
m = 2 kg
u = 10 m/s
g = 10 m/s²
v = 0 (at maximum height)
Using energy conservation:
Initial KE = Final PE
½mu² = mgh
½ × 2 × (10)² = 2 × 10 × h
100 = 20h
h = 5 m
Answer: Maximum height = 5 metres
Numerical 12: Work Done by Retarding Force
Q12. A car of mass 1200 kg moving at 20 m/s is brought to rest by applying brakes.
Find the work done by the braking force.
Solution:
Given:
m = 1200 kg
u = 20 m/s
v = 0
Work done = Change in kinetic energy
W = ½m(v² − u²)
W = ½ × 1200 × (0 − 400)
W = −240000 J
Answer: Work done = –2.4 × 10⁵ J
Negative sign shows work done against motion
Numerical 13: Power While Climbing Stairs (Very Important)
Q13. A student of mass 50 kg climbs a staircase of 30 steps, each step being 20 cm high, in 12 seconds.
Find the power developed by the student.
(Take g = 10 m/s²)
Solution:
Mass, m = 50 kg
Height of one step = 20 cm = 0.2 m
Total height = 30 × 0.2 = 6 m
Weight = mg = 50 × 10 = 500 N
Work done = mgh
W = 50 × 10 × 6 = 3000 J
Time, t = 12 s
Power = Work / Time
P = 3000 / 12
P = 250 W
Answer: Power developed = 250 watts
Numerical 14: Energy at Halfway Point
Q14. An object of mass 4 kg is dropped from a height of 20 m.
Find its kinetic energy when it is at a height of 10 m above the ground.
(Take g = 10 m/s²)
Solution:
Initial potential energy = mgh
= 4 × 10 × 20
= 800 J
Potential energy at 10 m height:
= 4 × 10 × 10
= 400 J
Using conservation of energy:
KE = Total energy − Remaining PE
KE = 800 − 400
KE = 400 J
Answer: Kinetic energy = 400 joules
Numerical 15: Multiple Appliances Energy Consumption
Q15. Three electrical appliances of power 500 W, 1000 W, and 1500 W work for 2 hours each.
Find the total energy consumed in kilowatt-hour (kWh) and in joules.
Solution:
Total power = 500 + 1000 + 1500
Total power = 3000 W = 3 kW
Time = 2 hours
Energy in kWh = Power × Time
Energy = 3 × 2
Energy = 6 kWh
Conversion:
1 kWh = 3.6 × 10⁶ J
Energy in joules = 6 × 3.6 × 10⁶
= 21.6 × 10⁶ J
Answer:
Energy consumed = 6 kWh
= 2.16 × 10⁷ joules