# 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 = 1.2×10

^{6}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 E

_{K}, 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, E

_{K}= ½ mv^{2}
Gravitational potential energy = mgs

s is the distance travelled.

From the conservation of energy, at
any position, this following equation holds.

E

_{K}= mgs
Since we are examining the graph of
the kinetic energy E

_{K}with distance s, we do not need to expand the formula for kinetic energy (E_{K}= ½ mv^{2})
E

_{K}= mgs
We know that, for the upward motion,
as s increases, E

_{K}decreases. [B is incorrect] The above equation depends only linearly on s (that is the s is not s^{2}or s^{3}or …). So the graph should be a straight line. [C is correct]
Initially, when s = 0, E

_{K}is maximum.
At the maximum height (s = h), E

_{K}= 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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