# Physics 9702 Doubts | Help Page 14

__Question 74: [Waves > Intensity]__
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.**

^{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) =

**I**^{2}Rt
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**(since mass is at rest at P)

__lost__at a point = KE at that point*(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**= (P_{in}– P_{out}_{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

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