• 9702 November 2015 Paper 21 Worked Solutions | A-Level Physics

Question 4
Fig. 4.1 shows a metal cylinder of height 4.5 cm and base area 24 cm2.



Question 5
A progressive wave transfers energy. A stationary wave does not transfer energy. State two other differences between progressive waves and stationary waves.

An air bubble in a tank of water is rising with constant velocity. The forces acting on the bubble are X, Y and Z as shown.

Question 17

An air bubble in a tank of water is rising with constant velocity. The forces acting on the bubble are X, Y and Z as shown.

What describes the three forces?

A Z is the viscous drag on the bubble, Y is the weight of the bubble, X is the upthrust on the bubble and X = Y + Z.

B Z is the viscous drag on the bubble, Y is the weight of the bubble, X is the upthrust on the bubble and X > Y + Z.

C Z is the weight of the bubble, Y is the viscous drag on the bubble, X is the upthrust on the bubble and X = Y + Z.

D Z is the weight of the bubble, Y is the viscous drag on the bubble, X is the upthrust on the bubble and X > Y + Z.

Reference: Past Exam Paper – June 2015 Paper 13 Q15

Solution:

Answer: A.

The air bubble rises at constant velocity. This means that its acceleration is zero, and thus the resultant force on it is also zero [since F = ma].

Resultant force = 0

Upward force = Downward forces

X = Y + Z                   [Choices B and D are incorrect]

The weight acts downwards. So, either Y or Z is the weight.

BUT the mass of air is very tiny. Its weight will also be very small. So, Y is most likely the weight.

Z is the viscous drag of the water on the bubble. The viscous drag is a resistive force and therefore acts in a direction opposite to the motion.

 

X is the upthrust force which arises due to the difference in pressure at the top and bottom of the air bubble. Its direction is upwards.

An insect jumps with an initial vertical velocity of 1.0 m s-1, reaching a maximum height of 3.5×10-2 m. Assume the deceleration is uniform.

Question 14

An insect jumps with an initial vertical velocity of 1.0 m s^-1, reaching a maximum height of 3.5×10^-2 m. Assume the deceleration is uniform.

What is the magnitude of the deceleration?

A 3.6 m s^-2                        B 9.8 m s^-2                        C 14 m s^-2                         D 29 m s^-2

Reference: Past Exam Paper – June 2015 Paper 11 Q8

Solution: 

Answer: C.

Initial velocity, u = 1.0 m s^-1

Maximum height reached, s = 3.5×10^-2 m

At the maximum height, the insect is momentarily at rest. Its velocity is zero. It cannot have a non-zero velocity as this would allow it to move further up and thus it cannot be called the ‘maximum’ height.

Final velocity, v = 0 m s^-1

v² = u² + 2as

0 = 1² + 2×a×(3.5×10^-2)

Acceleration a = - 1 / 2×3.5×10^-2

Acceleration a = - 14 m s^-2

Deceleration = 14 m s^-2

In the circuit shown, lamp P is rated 250 V, 50 W and lamp Q is rated 250 V, 200 W. The two lamps are connected in series to a 250 V power supply.

Question 3

In the circuit shown, lamp P is rated 250 V, 50 W and lamp Q is rated 250 V, 200 W. The two lamps are connected in series to a 250 V power supply.

Assume that the resistance of each lamp remains constant.

Which statement most accurately describes what happens when the switch is closed?

A Lamp P emits four times as much power as lamp Q.

B Lamp P emits twice as much power as lamp Q.

C Lamp Q emits four times as much power as lamp P.

D Lamp Q emits twice as much power as lamp P.

Reference: Past Exam Paper – November 2015 Paper 13 Q34

Solution:

Answer: A.

Lamp P is rated 250 V, 50 W.

Lamp Q is rated 250 V, 200 W.

Power P = VI = I²R = V² / R

We need to find the resistance of each lamp. Since we do not know the current we have to use the formula

P = V² / R

to calculate the resistances from the ratings.

Resistance R = V² / P

Lamp P: Resistance = 250² / 50 = 1250 Ω

Lamp Q: Resistance = 250² / 200 = 312.5 Ω

Lamp P and Q are connected in series so, the same current flows through them.

Power emitted = VI

Since current I is the same for both lamps, we can conclude that the power emitted is proportional to the p.d. across the lamp.

Power emitted by lamp P: P_P ∝ 1250

Power emitted by lamp Q: P_Q ∝ 321.5

 

Since the resistance of lamp P is (1250 / 312.5 =) 4 times that of lamp Q, the power emitted by lamp P is also 4 times as much as the power emitted by lamp Q.