• 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 = fλ

Electromagnetic waves travel at the speed of 3.0x10⁸ms^-1.

Frequency, f = 1.00GHz = 1x10⁹Hz

Wavelength, λ = v / f = (3x10⁸) / (1x10⁹) = 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.]