# Physics 9702 Doubts | Help Page 73

__Question 392: [Current of Electricity]__The diagram shows a metal filament lamp connected in series with an ammeter, battery and a variable resistor. A high resistance voltmeter is connected across the lamp.

Which graph represents the variation of I, the current through the lamp, with the potential difference V across it?

**Reference:**

*Past Exam Paper – N93 / I / 12*

__Solution 392:__**Answer: A.**

As the p.d. V across the metal filament lamp increases, the temperature of the metal filament increases. This causes it resistance to increase. So, it does not always follow Ohm’s law. [C is incorrect]

Also, since the resistance increases, the current would tend to decrease at higher values of V. [D and E are incorrect] Graph B would be correct if only the lamp was connected in the circuit (given the e.m.f is not too big), without the variable resistor. [B is incorrect]

Let the resistance of the lamp be R

_{L}and the resistance of the variable resistor be R

_{r}. Let the e.m.f. of the supply be E. From the potential divider equation, the p.d. V across the lamp is given by

V = [R

If the p.d. V across the lamp is increasing, it means the R_{L}/ (R_{L}+R_{r})] E_{r}is decreasing since E is constant.

The current I through the lamp is the current in the whole circuit since this is a series connection. The current I is given by I = E / (R

_{L}+R

_{r}).

Since R

_{r}is decreasing, at some point when R

_{r}= 0, the current would tend to I = E / R

_{L}where E would be equal to V. R

_{L}would be given by R

_{L}= V / I. Since R

_{L}is increasing and E (= V) is constant, the current I would be smaller at higher values of V. So, the gradient of the graph should decrease at higher values of V.

__Question 393: [Waves > Stationary waves]__
Diagram shows a standing wave on a
string. The standing wave has three nodes N

_{1}, N_{2}and N_{3}.Which statement is correct?

A All points on the string vibrate in phase.

B All points on the string vibrate with the same amplitude.

C Points equidistant from N

_{2}vibrate with the same frequency and in phase.

D Points equidistant from N

_{2}vibrate with the same frequency and the same amplitude.

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q25 & Paper 12 Q24 & June 2013 Paper 13 Q28*

__Solution 393:__**Answer: D.**

A and B are obviously wrong. The
points on the string (forming a stationary wave) do not vibrate in phase not
with the same amplitude.

Points within a loop (example: N

_{1}-N_{2}loop or N_{2}-N_{3}loop) vibrate in phase, but the points in a loop vibrate out of phase with the points in the adjacent loop. For example, if the points in the loop N_{1}-N_{2}move upwards, then the points in the loop N_{2}-N_{3}will move downwards. Remember that even if a node is shown as an oval (with an upper and lower arc), only one of these arc is present at a particular time. If the upper arc of the loop N_{1}-N_{2}is present, then the lower arc of the loop N_{2}-N_{3}will be present. [A is incorrect]
All the points on the string do not
vibrate with the same amplitude (amplitude is the maximum displacement from the
equilibrium position). Points closer to the nodes have a smaller amplitude and
antinodes are points with maximum amplitude. [B is
incorrect]

The points equidistant from node N

_{2}vibrate with the same frequency and the same amplitude, but not in phase since points in adjacent loops are out of phase. If one such point in loop N_{1}-N_{2}is moving downwards, then the corresponding point equidistant from node N_{2}in loop N_{2}-N_{3}would be moving upwards. [C is incorrect]

__Question 394: [Waves > Stationary waves]__Diagram shows steel wire clamped at one end and tensioned at the other by a weight hung over a pulley.

Vibration generator is attached to wire near the clamped end. Stationary wave with one loop is produced. Frequency of the vibration generator is f.

Which frequency should be used to produce stationary wave with two loops?

A f / 4 B f / 2 C 2f D 4f

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q25 & Paper 12 Q22 & Paper 13 Q24*

**Solution 394:**

**Answer: C.**

Let the length of wire = L

In the case shown (1 loop is formed –
this is called the 1

^{st}harmonic), half a wave is formed (with 2 nodes). So, L = Î»_{1}/ 2 where Î» is the wavelength of the wave formed.
Wavelength Î»

_{1}= 2L
Speed v = f Î»

Therefore, the frequency of the
vibration generator for which the wave shown is produced, f

_{1}= v / Î»_{1}= v / 2L. (this is given as f in the question: f_{1}= f)
The length of wire used is the same
(= L).

For 2 loops to be produced (this is
called the 2

^{nd}harmonic), the length of wire L = Î»_{2}since a complete wave would then be formed with 3 nodes, where is Î»_{2}is the wavelength of the new wave formed.
So, the new frequency f

_{2}= v / Î»_{2}= v / L = 2f_{1}= 2f

__Question 395: [Capacitors]__A 2.0 Î¼F capacitor is charged to a potential difference (p.d.) of 50 V and a 3.0 Î¼F capacitor is charged to a p.d. of 100 V.

Calculate the charge on the plates of each capacitor and the energy stored by each. [4]

The capacitors are then joined together in parallel with their positive plates connected together.

What is the equivalent capacitance of this combination? [1]

Hence calculate the total energy stored by the capacitors after connection. [3]

Suggest why there is a loss of stored energy when the capacitors are connected. [2]

**Reference:**

*Past Exam Paper – Edexcel January 2003 PHY5 Q1*

__Solution 395:__Charge of capacitor, Q = CV

For 1

^{st}capacitor, Q

_{1}= CV = (2.0x10

^{-6}) (50) = 1.0x10

^{-4}C

For 2

^{nd}capacitor, Q

_{2}= CV = (3.0x10

^{-6}) (100) = 3.0x10

^{-4}C

Energy stored in capacitor, W = ½ CV

^{2}

Energy stored in 1

^{st}capacitor, W

_{1}= ½ CV

^{2}= 0.5 (2.0x10

^{-6}) (50)

^{2}= 2.5x10

^{-3}J

Energy stored in 2

^{nd}capacitor, W

_{2}= ½ CV

^{2}= 0.5 (3.0x10

^{-6}) (100)

^{2}= 1.5x10

^{-2}J

The capacitors are now joined together in parallel with their positive plates connected together.

For a parallel combination of capacitors,

Equivalent capacitance, C

_{total}= C

_{1}+ C

_{2}= 2.0 + 3.0 = 5.0 Î¼F

After connection,

Total Charge = (1.0x10

^{-4}) + (3.0x10

^{-4}) = 4.0x10

^{-4 }C

Total energy stored, E

_{total}= ½ (Q

^{2 }/ C) = 0.5 [(4.0x10

^{-4})

^{2}/ (5.0x10

^{-6})] = 0.016 J

There’s loss of stored energy when the capacitors are connected

EITHER due to gain in internal energy during electrical work

OR because of the heating in the wires resulting in the dissipation of thermal heat

OR due to work done in charges

OR because energy is needed to overcome resistance in wires

__Question 396: [Kinematics]__A student, standing on the platform at a railway station, notices that the first two carriages of an arriving train pass her in 2.0s, and the next two in 2.4s. The train is decelerating uniformly. Each carriage is 20m long. When the train stops, the student is opposite the last carriage. How many carriages are there in the train?

**Reference:**

*“Cambridge International As and A level Physics,” 2*

^{nd}edition, by Mike Crundell, Geoff Goodwin and Chris Mee, Chapter 3 (Kinematics) Q2

__Solution 396:__1 carriage is 20m long.

Equation for uniformly accelerated motion: s = ut + ½ at

^{2}

Let the initial speed with which the train starts to pass her = u

The deceleration is uniform.

For 1

^{st}2 carriages: Total distance = 2 (20) = 40m and time taken = 2.0s

40 = u(2) + ½ (a)(2)

^{2}

40 = 2u + 2a

20 = u + a …………….. (1)

For 2

^{nd}2 carriages: We don’t know with what speed the 2

^{nd}2 carriages start to pass her. So, now we consider the motion of the 1

^{st}4 carriages of the train. Total distance = 4 (20) = 80m and total time taken = 2.0 + 2.4 = 4.4s

80 = u(4.4) + ½ (a)(4.4)

^{2}

80 = 4.4u + 9.68a …………….. (2)

From eq(1),

u = 20 – a …………….. (3)

Replace u = 20 – a in eq(2)

80 = 4.4(20 – a) + 9.68a

-8 = 5.28a

Acceleration, a = - 1.52 ms

^{-2}

From eq(3),

Initial speed, u = 20 – (-1.52) = 21.52 ms

^{-1}

Now, let the total length of the carriages = s. Initial speed, u = 21.52ms

^{-1}. Final speed = 0ms

^{-1}(the train stops). Acceleration, a = - 1.52ms

^{-1}.

v

^{2}= u

^{2}+ 2as

0 = (21.52)

^{2}+ 2 (-1.52) s

Total length, s = 152.34m

Number of carriages = 152.34 / 20 = 7.62 = 8

{The number of carriages can only be an integer. Since the student is opposite the last train, the number of carriages = 8.}

oh thankyou whoever you are!!!

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