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Question 74: [Waves > Intensity]

Diagram shows graph of displacement against time for sound wave.

Intensity of sound is halved. Which graph shows displacement of this sound wave?

Reference: Past Exam Paper – November 2012 Paper 11 Q26

Solution 74:

Answer: D.

Intensity, I is proportional to (amplitude, A)².
So, A ∝ √I

If intensity is halved (intensity becomes I / 2),
New amplitude is proportional to √(I/2) = (1 / √2) (√I)
New amplitude becomes (1 / √2) A
{that is, it is reduced by a factor of √2 ≈ 1.41 [1/√2 = 0.71] – so the amplitude is not halved, as suggested by answer C}








Question 75: [Current of Electricity > Charge & Energy]
Battery, with constant internal resistance, is connected to resistor of resistance 250 Ω, as shown.

Current in resistor is 40 mA for a time of 60 s. During this time 6.0 J of energy is lost in internal resistance.
What are the energy supplied to external resistor during the 60 s and e.m.f. of battery?

Reference: Past Exam Paper – June 2014 Paper 11 Q31



Solution 75:
Answer: D.

Energy transferred in external resistor (= Power x time) = I²Rt

Energy transferred = (40x10^-3)² x 250 x 60 = 24 J

Therefore, (since 6J of energy is lost in internal resistance,) the battery must supply a total of (6 + 24 =) 30 J of energy to whole circuit.

Energy supplied by battery, E = QV

where Q is the charge and V is the e.m.f of the battery.

Charge transferred, Q = It = (40x10^-3) x 60 = 2.4C.

So, e.m.f of battery, V = E / Q = 30 / 2.4 = 12.5V.

Question 76: [Conservation of energy > Potential and Kinetic]
Small mass is placed at point P on inside surface of smooth hemisphere. It is then released from rest. When it reaches lowest point T, its speed is 4.0 m s^–1.
Diagram shows speed of mass at other points Q, R and S as it slides down. Air resistance is negligible.

Mass loses potential energy E in falling from P to T.
At which point has the mass lost potential energy E / 4?


A Q                 B R                  C S                  D none of these


Reference: Past Exam Paper – June 2014 Paper 12 Q15



Solution 76:
Answer: B.
This question is more easily seen in terms of kinetic energy.
From conservation of energy,
Potential energy (PE) lost in falling from P to T= Kinetic energy (KE) of mass at T
E = ½ m(4)²     = 8m.

Since the mass is at rest at P, its KE is zero at P (its energy is entirely PE at P) and since T is the lowest point, PE at T is zero. So, the total energy of the mass in the system is E {this is also the total PE at P}.

In general, PE lost at a point = KE at that point (since mass is at rest at P)
(Note that even if the mass has lost some PE at any specific point, it may still have some PE at that point)

From above, E = 8m. So, a PE lost of E /4 = 2m.
At Q, KE = ½ m(1)² = 0.5m = PE lost.
At R,  KE = ½ m(2)² = 2m = PE lost. 

Therefore, when the mass is at R, it has half the final speed and so has gained a kinetic energy of E/4. This is when it has lost a potential energy of E/4. 







Question 77: [Power > Efficiency]
Electric motor has input power P_in, useful output power P_out and efficiency η.

How much power is lost by motor?

Reference: Past Exam Paper – June 2014 Paper 13 Q16



Solution 77:

Answer: D.

Efficiency = output power / input power. (η = P_out / P_in)

So, P_in = P_out / η

Power lost = P_in – P_out = (P_out / η) – P_out = (1/η – 1)P_out

Question 78: [Forces > Moments]
Rigid circular disc of radius r has its centre at X. A number of forces of equal magnitude F act at edge of disc. All the forces are in plane of the disc.

Which arrangement of forces provides moment of magnitude 2Fr about X?

Reference: Past Exam Paper – November 2006 Paper 1 Q14



Solution 78:
Answer: B.
Moment = Force (F) x perpendicular distance (r) of force from pivot X
In this question, moment produced by each force F is Fr
Moments can be either clockwise or anticlockwise.

Ans A: an anticlockwise moment of Fr is produced
Ans B: a total clockwise moment of (Fr + Fr =) 2Fr is produced (this is a couple – the forces (same magnitude) are in opposite direction, but act at different points so that they produce a turning effect - each force produces a moment which is clockwise. The resultant force is however zero.)
Ans C: net moment is zero (since clockwise and anticlockwise moments are both Fr. Resultant force is 2F here, not zero.)
Ans D: a net clockwise moment of (Fr + Fr – Fr = ) Fr is produced