Wednesday, June 3, 2015

Physics 9702 Doubts | Help Page 161

  • Physics 9702 Doubts | Help Page 161


Question 800: [Kinematics]
Graph shows the variation with time of the speed of a raindrop falling vertically through air.

Which statement is correct?
A The acceleration decreases to produce a steady speed.
B The acceleration increases as the speed increases.
C The air resistance decreases as the speed increases.
D The resultant force increases as the speed increases.

Reference: Past Exam Paper – June 2014 Paper 12 Q11



Solution 800:
Answer: A.
The gradient of a speed-time graph gives the acceleration.

Acceleration of the falling raindrop decreases (gradient decreases) with time to produce a steady speed (gradient zero).

The air resistance acting on a body increases as its speed increases.

The resultant force on a body depends on its acceleration. The speed may increase while the acceleration is constant. In this case, the resultant force on the body is constant, even if its speed increases.










Question 801: [Current of Electricity]
The e.m.f. of battery is 9.0 V. Reading on the high-resistance voltmeter is 7.5 V.

What is the current I?
A 0.10 A                     B 0.50 A                     C 0.60 A                     D 2.0 A

Reference: Past Exam Paper – November 2008 Paper 1 Q36



Solution 801:
Answer: B.
The high-resistance voltmeter, which is connected across the 9.0V battery, reads a p.d. of 7.5V. This means that there is some lost volts in the battery and this is due to its internal resistance.
Lost volts in battery = 9.0 – 7.5 = 1.5V

From Kirchhoff’s law, the sum of p.d. in the circuit is equal to the e.m.f. in the circuit.
p.d. across the 15Ω resistor = 7.5V

Ohm’s law: V = IR
Current I through the 15Ω resistor = V / R = 7.5 / 15 = 0.5A









Question 802: [Current of Electricity]
A cylindrical piece of soft, electrically-conducting material has resistance R. It is rolled out so that its length is doubled but its volume stays constant.
What is its new resistance?
A R / 2                        B R                              C 2R                            D 4R

Reference: Past Exam Paper – June 2011 Paper 12 Q33



Solution 802:
Answer: D.
Resistance R = ρL / A

Volume V = AL

If the length L is doubled, then the area A should be halved so that the volume remains constant.

New resistance, R = ρ (2L) / 0.5A = 4ρL / A = 4R










Question 803: [Work, Energy and Power > Hydroelectricity]
Water from a reservoir is fed to turbine of a hydroelectric system at a rate of 500 kg s–1. The reservoir is 300 m above level of the turbine.
The electrical output from the generator driven by turbine is 200 A at a potential difference of 6000 V.
What is the efficiency of the system?
A 8.0 %                       B 8.2 %                       C 80 %                        D 82 %

Reference: Past Exam Paper – November 2012 Paper 11 Q21



Solution 803:
Answer: D.
The water, being at a height of 300 m has gravitational potential energy. This energy is converted into kinetic energy of the turbine which then produces electrical energy.

Gravitational potential energy = mgh
Input power = Energy / time = mgh / t = (m/t)gh

Water from a reservoir is fed to turbine of a hydroelectric system at a rate of 500 kg s–1.
So, m/t = 500 kg s–1

Input power = 500 × 9.81 × 300 = 1 471 500 W
(Use g = 9.81 m s–2 and not 10 m s–2)


The electrical output from the generator driven by turbine is (I =) 200 A at a potential difference of (V =) 6000 V.
Output power = VI = 6000 × 200 = 1 200 000 W = 1.2×106 W

Efficiency = Output power / Input energy = (1471500 / 1200000) ×100% = 81.5% = 82%










Question 804: [Current of Electricity]
A high-resistance voltmeter connected across battery reads 6.0 V.
When battery is connected in series with a lamp of resistance of 10 Ω, the voltmeter reading falls to 5.6 V.
Which statement explains this observation?
A The electromotive force (e.m.f.) of the battery decreases because more work is done across its internal resistance.
B The e.m.f. of the battery decreases because work is done across the lamp.
C The potential difference (p.d.) across the battery decreases because more work is done across its internal resistance.
D The p.d. across the battery decreases because work is done across the lamp.

Reference: Past Exam Paper – November 2010 Paper 12 Q32



Solution 804:
Answer: C.
‘A high-resistance voltmeter connected across battery reads 6.0 V.’
This gives the e.m.f. of the battery. e.m.f. = 6.0V. The e.m.f. is a property of the battery – it does not increase or decrease. [A and B are incorrect]

The potential difference across a battery (difference in potential between the 2 terminals of the battery) is equal to its e.m.f. if its internal resistance is zero.

‘When battery is connected in series with a lamp of resistance of 10 Ω, the voltmeter reading falls to 5.6 V.’
This means that there is some lost volts in the battery, which is due to its internal resistance. Since there is some internal resistance, the p.d. across the battery is not equal to its e.m.f. The potential difference (p.d.) across the battery decreases because more work is done across its internal resistance.










Question 805: [Kinematics > Linear motion]
A stone, thrown vertically upwards from the ground level, rises to a height h and then falls to its starting point.
Assuming that air resistance is negligible, which of the following graphs best shows how EK, the kinetic energy of the stone, varies with s, the distance travelled?


Reference: Past Exam Paper – N89 / I / 2



Solution 805:
Answer: C.
As the stone is thrown up, it is given kinetic energy by the thrower. This kinetic energy is being converted to gravitational potential energy as the ball moves up.

The acceleration due to gravity is downwards. So, the speed of the stone decreases as it moves upward until it becomes zero at the maximum height h.

Kinetic energy, EK = ½ mv2
Gravitational potential energy = mgs
s is the distance travelled.

From the conservation of energy, at any position, this following equation holds.
EK = mgs
Since we are examining the graph of the kinetic energy EK with distance s, we do not need to expand the formula for kinetic energy (EK = ½ mv2)

EK = mgs
We know that, for the upward motion, as s increases, EK decreases. [B is incorrect] The above equation depends only linearly on s (that is the s is not s2 or s3 or …). So the graph should be a straight line. [C is correct]

Initially, when s = 0, EK is maximum.
At the maximum height (s = h), EK = 0.
Then, as the stone falls down again, distance travelled s keeps on increasing but since the stone is now losing height (gravitational potential energy), it gains kinetic energy again.

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