# Physics 9702 Doubts | Help Page 166

__Question 830: [Current of Electricity > Resistance]__
Copper wire is stretched so that its
diameter is reduced from 1.0 mm to a uniform 0.5 mm.

Resistance of the unstretched copper
wire is 0.2 Ω.

What will be the resistance of the
stretched wire?

A 0.4 Ω B 0.8 Ω C
1.6 Ω D 3.2 Ω

**Reference:**

*Past Exam Paper – November 2012 Paper 13 Q33*

__Solution 830:__**Answer: D.**

Resistance R of a wire = ρL / A

Cross-sectional area A = πd

^{2}/ 4
The resistance R is inversely
proportional to the area, and thus is also inversely proportional to the
diameter squared.

When the wire is stretched, the
diameter is reduced by half. This causes the area to be reduced by 4. So, the
resistance R is increased by a factor of 4 due to the cross-sectional area A.

But, the volume of the wire is
always constant.

Volume V = AL where L is the length of the wire

The stretched wire has ¼ the area of
the unstretched wire. For its volume to remain constant, its length must have
been increased by a factor of 4.

Volume V = (A/4) (4L)

So, the length of the stretched wire
is 4L.

Resistance R of unstretched wire = ρL / A = 0.2 Ω

Resistance R of unstretched wire = ρ(4L) / (A/4) = 16 (ρL / A) = 16 (0.2) = 3.2 Ω

In such a question, sketching a
diagram would be useful.

__Question 831: [Work, Energy and Power > Efficiency]__
A crane is being used to lift
containers off a ship. One container has mass of 14 000 kg and is being lifted
vertically with a speed of 3.2 m s

^{–1}.
Electric motor being used to supply
the power to lift the container is using a current of 240 A at a potential
difference of 2200 V.

What is the efficiency of the
system?

A 8.1 % B 8.5 % C
48 % D 83 %

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q21*

__Solution 831:__**Answer: D.**

Power P = VI

(Input) Power from electric motor =
VI = 2200 × 240
= 5.28×10

^{5 }W
Power = Energy / time = Force × distance / time = Force × speed

Force = Weight (= mg)

(Output) Power from crane = (14000 × 9.81)
× 3.2
= 4.39488×10

^{5 }W
Efficiency = Output power / Input
power = (4.39488×10

^{5}) / (5.28×10^{5}) = 0.83 = 83%

__Question 832: [Dynamics + Energy]__
Two trolleys are placed together on
horizontal runway with a compressed spring between them.

When they are released, the 2 kg
trolley moves to the left at 2 m s

^{–1}.
How much energy was stored in the
spring?

A 4 J B 6 J C
8 J D 12 J

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q15 & Paper 12 Q14*

__Solution 832:__**Answer: D.**

From the conservation of momentum,

Sum of momentum before collision =
Sum of momentum after collision

Before collision, the sum of momentum
is zero since the trolleys are stationary.

After collision,

2kg (2ms

^{-1}) + 1kg (v) = 0
Velocity v = -4ms

^{-1}= 4ms^{-1}(considering magnitude only).
As the trolleys move, they have
kinetic energy. This kinetic energy was initially stored in the spring.

Kinetic energy = ½ mv

^{2}
Energy stored = Sum of kinetic
energies of trolleys = ½ (2)(2)

^{2}+ ½ (1)(4)^{2}= 12J

__Question 833: [Work, Energy and Power]__
An escalator is 60 m long and lifts
passengers through vertical height of 30 m, as shown.

To drive the escalator against
forces of friction when there are no passengers requires a power of 2.0 kW.

The escalator is used by passengers
of average mass 60 kg and the power to overcome friction remains constant.

How much power is required to drive
the escalator when it is carrying 20 passengers and is travelling at 0.75 m s

^{–1}?
A 4.4 kW B 6.4 kW C
8.8 kW D 10.8 kW

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q16*

__Solution 833:__**Answer: B.**

**EITHER**

Total weight of passengers = W =
20(mg) = 20 × 60 × 9.81
N

Each passenger are moved a distance
of 60m as the escalator goes up. The weight of the passengers acts vertically
downwards. There will be a component of weight acting down the slope of the
escalator.

Component of weight along slope of
the escalator = W sinθ

where θ is the angle the escalator
makes with the horizontal

sinθ = O / H = 30 / 60 = 0.5

Work needs to be done against
forces. Additionally, work should be done against the component of weight acting
down the slope.

Power = Energy / time

Power = 2000 + (W.D/t) = 2000+ Fv

Power = 2000 + (20×60×9.81×0.5)(0.75)
= 6414.5W = 6.4kW

**OR**

The question can also be worked out
by considering the gain in potential energy of the 20 passengers.

Total weight of passengers = W =
20(mg) = (20 × 60 × 9.81)
N

Gain in potential energy = mgh

Gain in potential energy per second
= 20(mgv)

The gain in potential energy per
second depends on vertical distance moved per second.

sinθ = O / H = 30 / 60 = 0.5

where θ is the angle the escalator
makes with the horizontal

Vertical component of velocity, v =
0.75 × sinθ = 0.75 × 0.5 = 0.375 ms

^{-1}
Gain in potential energy per second
= (20 × 60 × 9.81)
× 0.375 = 4414.5W = 4.4 kW

Total power required = Power to move
against friction + Gain in potential energy per second

Total power required = 2.0 + 4.4 =
6.4 kW

Here, 2 common mistakes are to use
60 m rather than 30 m vertical distance and forgetting to add on the 2.0 kW.

__Question 834: [Measurement > Graphs]__
Diagram shows a calibration curve
for a thermistor, drawn with an unusual scale on the vertical axis.

What is the thermistor resistance
corresponding to a temperature of 40 °C?

A 130 Ω B 150 Ω C
400 Ω D 940 Ω

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q5*

__Solution 834:__**Answer: C.**

We need to divide the range of
values between the successive values displayed (on the y-axis) by number of
lines in the range, including those corresponding to the 2 values.

Consider the mark ‘100’ and ‘1000’.
Including the lines at 100 and 1000, we have 10 lines. Between these 10 lines,
there are 9 intervals – one between each 2 lines.

9 intervals represent a range of
1000 – 100 = 900

1 interval correspond to 900 / 9 =
100 Ω

So, the horizontal temperature axis (the
bottom-most horizontal line) corresponds to 100 Ω. The line just above it
corresponds to 200 Ω. The next line is 300 Ω. Then, 400, 500, 600, ….

At 40°, the curve is at the 4

^{th}line from the bottom.
EITHER This corresponds to 400 Ω.

OR There are 3 intervals from that
line to the bottom-most horizontal line.

So, reading = 100 + 3(100) = 400 Ω

physics paper may june 2013 paper 12 question 6

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/07/9702-june-2013-paper-12-worked.html

For Q830, why cant I directly use this equation 0.2(22/7)(1/4)= R(22/7)(0.25/4) R=0.8 ? Thank you.

ReplyDeleteCould you explain the formula. How you reached there?

DeleteAlso, the correct solution is 3.2, not 0.8.