# Physics 9702 Doubts | Help Page 18

__Question 95: [Force > Moment]__Diagram shows plan view of door which requires a moment of 12Nm to open it.

What is the minimum force that must be applied at door’s midpoint to ensure it opens?

A 4.8 N B 9.6 N C 15 N D 30 N

**Reference:**

*Past Exam Paper – November 2007 Paper 1 Q14*

__Solution 95:__**Answer: D.**

Moment = Force x
perpendicular distance of force from pivot

For minimum force, the force should
be applied perpendicularly to the door.

Midpoint = 0.80 /2 = 0.40m

Moment = F x d = 12Nm

F x 0.80 = 12

Minimum force, F = 12 / 0.40 = 30N

__Question 96: [Current of Electricity > Potential difference and Resistance]__Power supply and solar cell are compared using potentiometer circuit shown.

e.m.f. produced by solar cell is measured on potentiometer.

Potentiometer wire PQ is 100.0 cm long and has resistance of 5.00 Ω. Power supply has e.m.f. of 2.000 V and solar cell has e.m.f. of 5.00 mV.

Which resistance R must be used so that galvanometer reads zero when PS = 40.0 cm?

A 395 Ω B 795 Ω C 995 Ω D 1055 Ω

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q35*

__Solution 96:__**Answer: B.**

When PS = 40.0cm, the potential difference across the length of wire PS is given by

**2.000 V × (0.400 × 5.00 Ω / [R + 5.00 Ω]**). (Potential divider equation)

{Since the total resistance of the 1m wire = 5.00Ω, the resistance of PS = 0.4 x 5.00Ω. [R + 5.00Ω] is the total resistance in the circuit.}

(A galvanometer is a type of sensitive ammeter: an instrument for detecting electric current.)

For the galvanometer to read zero, the

__e.m.f of the solar cell must be equal to that potential difference (given above)__such that the

*potential difference across the galvanometer is zero*.

So, by equating this p.d. to the e.m.f. of solar cell (

**= 5.00mV**), resistance R can be obtained by rearranging the equation.

2.000 V × (0.400 × 5.00 Ω / [R + 5.00 Ω] = 5 × 10

^{-3}V

2.000 × (0.400 × 5.00) = (5 × 10

^{-3}) [R + 5.00]

4 = (5 × 10

^{-3}) [R + 5.00]

R + 5.00 = 4 / (5 × 10

^{-3}) = 800

R = 800 – 5 = 795 Ω

Resistance R is found to be 795 Ω.

__Question 97: [Waves > Stationary waves]__Transmitter of electromagnetic waves is placed 45 cm from reflective surface.

Emitted waves have frequency of 1.00 GHz. Stationary wave is produced with node at the transmitter and node at the surface.

How many antinodes are in the space between transmitter and surface?

A 1 B 2 C 3 D 4

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q27*

__Solution 97:__**Answer: C.**

**Speed of wave, v = f**

**λ**

Electromagnetic waves
travel at the speed of 3.0x10

^{8}ms^{-1}.
Frequency, f = 1.00GHz = 1x10

^{9}Hz
Wavelength, λ = v / f = (3x10

^{8}) / (1x10^{9}) = 0.3m
Distance between transmitter and
reflective surface = 45cm = 0.45m

No. of periods present between
transmitter and reflective surface = 0.45/0.3 = 1.5

An antinode is a point
with maximum displacement.

1 period contains 2 antinodes + 1
antinode is present in the 0.5 (half) of a period.

Total number of antinodes = 2 + 1 =
3

__Question 98: [Dynamics]__Supermarket trolley, total mass 30 kg, is moving at 3.0 m s

^{–1}. Retarding force of 60 N is applied to trolley for 0.50 s in opposite direction to trolley’s initial velocity.

What is trolley’s new velocity after application of the force?

A 1.0 m s

^{–1}B 1.5 m s

^{–1}C 2.0 m s

^{–1}D 2.8 m s

^{–1}

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q10 & Paper 12 Q9*

__Solution 98:__**Answer: C.**

Force F on trolley= Δp / Δt

where Δp is the change in momentum
of the trolley in during a time Δt

Assuming v and u are in
same direction,

Δp = m(v – u) = 30(v – 3)

Time of application of force, Δt =
0.5s

Force is in opposite direction and
so, is taken as negative.

**Δp / Δt = F**

30(v – 3) / 0.5 = - 60.

New velocity, v = (0.5 x -60 / 30) +
3 = 2ms

^{-1}

__Question 99: [Current of Electricity > Potentiometer]__In potentiometer circuit below, moveable contact is placed at N on the bare wire XY, such that galvanometer shows zero deflection.

Resistance of variable resistor is now increased.

What is the effect of this increase on potential difference across wire XY and on position of moveable contact for zero deflection?

**Reference:**

*Past Exam Paper – November 2002 Paper 1 Q35*

__Solution 99:__**Answer: D.**

Let the resistance of the wire from
point X to Y = W and let the resistance of the variable resistor = R.

e.m.f. of driver
cell = p.d. across variable resistor + p.d. across wire XY

As the resistance of the variable
resistor, R is increased, the p.d. across it also increases (V = IR) and hence,
the p.d. across the wire decreases. [A and B are incorrect]

For the galvanometer to
show zero deflection, the potential

__difference__across it should be zero (no current flows through it). That is, the__potential__at one side of the galvanometer should be equal to the__potential__at the other side (thus the potential difference is zero).
Initially (before
the resistance of the variable resistor is increased), the galvanometer showed
zero deflection. So, the potential difference across it was zero. In that case,
the p.d. across XN was equal to the e.m.f. of the cell.

But an increase in the resistance of
the variable resistor causes the p.d. across XY to decrease. This reduced p.d.
across is that across the whole of the wire. Hence, the same length of wire XN
(as before) now corresponds to a p.d. less than before. A deflection would be
noted on the galvanometer as this new (reduced) p.d. across XN is lower than
the e.m.f. of the cell.

To cause the galvanometer to show
zero deflection again, the p.d. across wire XN should be increase so that it
becomes equal to the e.m.f. of the cell again. This is done by moving the
contact closer to end Y.

[V = IR and R = ρL/A. So, V ∝ L. The longer the wire XN, the greater the p.d. across it.]
in question 99, how is the galvanometer measuring the potential difference between XN and variable resistor?isn't the p.d, that it measures, equivalent to the p.d. JUST ACROSS XN?

ReplyDeleteFirst, do not forget that potential and potential difference are different things.

ReplyDeleteXN is not a point, but a part of the wire. The galvanometer measures p.d. across 2 points.

One end of the galvanometer is connected to the negative terminal of the battery and so, is at a potential of zero. The other end is connected at point N. So, the galvanometer shows the p.d. between these 2 points.

The length XN of the wire is the part if the wire that is contributing some resistance to the circuit. Part XY of the wire is not contributing any resistance. So, point N and Y may be considered to be at the same potential.

Now, the potential at N depends on both the resistance of the variable resistor and the length of wire being considered. Remember that N is moveable and any new position is still called N.

what do you mean when you say "The length XN of the wire is the part if the wire that is contributing some resistance to the circuit. Part XY of the wire is not contributing any resistance. So, point N and Y may be considered to be at the same potential.Now, the potential at N depends on both the resistance of the variable resistor and the length of wire being considered. Remember that N is moveable and any new position is still called N."

DeleteThe galvanometer measures the p.d(in this case)between the negative terminal of the cell and N. So for the p.d. to be zero, the potential at N should also be 0, the same as that at the -ve terminal of cell, but this is not possible. p.d. at N can never be zero.

Along with this, please do explain q37 and q29 in s13/paper 11.

First of all, there was an error in the explanation I provided at solution 99. For zero deflection, the p.d. across XN is equal to the E.M.F. of the cell to which the galvanometer is connected, NOT equal to the p.d. across the variable resistor. I already corrected it. The end connected to the terminal of the cell is actually connected to the cell itself, so the potential there would be equal to the e.m.f of the cell.

DeleteBy the way, there's another question on potentiometer above, you may read it if you want.

As for the sentence you quoted, well this is the basis of potentiometer - that's how a potentiometer works.

both paper 11 questions are available at

Deletehttp://physics-ref.blogspot.com/2015/02/9702-june-2013-paper-11-worked.html

for question 96, I dont know why Im having trouble equating, can you show me how

ReplyDeleteThe workings have been included

Deletein question 95 is this equation of any value F x 0.80 = 12?

ReplyDeleteNo, this is for the minimum value of F. If the force is not perpendicular or the distance has been changed, but it would have a different value.

DeleteSir, how to do 9702/12/M/J/12 no 14 ? Thanks.

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/06/9702-june-2012-paper-12-worked.html

Please solve Q1,2(c),3 of 9702/41/M/J/17

ReplyDeleteQ1 is explained at

Deletehttp://physics-ref.blogspot.com/2017/11/explain-how-satellite-may-be-in.html