# Physics 9702 Doubts | Help Page 45

__Question 271: [Circular motion]__**(a)**Define the radian

The rod is rotated about end C so
that the stone moves in vertical circle of radius 85 cm.

Angular speed ω of rod and stone is
gradually increased from zero until the glue snaps. The glue fixing the stone
snaps when tension in it is 18 N.

For position of the stone at which
the glue snaps,

(i) on dotted circle of Fig, mark
with the letter S the position of the stone

(ii) calculate angular speed ω of
the stone.

**Reference:**

*Past Exam Paper – June 2010 Paper 41 Q1*

__Solution 271:__**(a)**The radian is the angle (subtended) at the centre of a circle by an arc equal in length to the radius.

**(b)**

(i)

{The glue causes the stone
to stay connected to the rod. So, it exerts a force towards the rod (since the
rod is moving in circular motion, this force is always towards the centre.). For
the glue to snap, the force opposing it should be sufficiently large.

When the stone is exactly
below point C, its weight acts vertically downwards while the force exerted by
the glue on the rod (to stay connected) is vertically up.

In any other position, the
weight always acts downwards while the force by the glue is always towards the
centre. So, in such positions, only a component of the weight will oppose the
force due to the glue. Therefore, the force opposing the glue is maximum when
the stone is exactly below C. It is the most likely position that the glue
would snap.}

Point S is shown below C

(ii)

{The resultant force
causes the centripetal acceleration. Resultant force = Tension – Weight (=
Centripetal force). So, tension = Weight + Centripetal force}

(Maximum) Force (tension) = Weight
of stone + Centripetal force

Centripetal force = mrω

^{2}
{18 = 3 - mrω

^{2}}
15 = (3.0 / 9.8) (0.85) ω

^{2})
Angular speed ω = 7.6 rad s

^{-1}

__Question 272: [Current of Electricity > Capacitance]__
Rectified output of sinusoidal
signal generator is connected across a resistor R of resistance 1.5 kΩ, as
shown.

Variation with time t of potential
difference V across R is shown.

**(a)**State how rectification shown in Fig may be achieved.

**(b)**A capacitor is now connected in parallel with resistor R. Resulting variation with time t of potential difference V across R is shown.

(i) Using Fig, determine

1. mean potential difference across
resistor R

2. mean current in resistor

3. time in each cycle during which
capacitor discharges through the resistor.

(ii) Using answers in (i), calculate

1. charge passing through the
resistor during one discharge of the capacitor

2. capacitance of capacitor

**(c)**Second capacitor is now connected in parallel with resistor R and the first capacitor. On Fig, draw a line to show variation with time t of the potential difference V across the resistor.

**Reference:**

*Past Exam Paper – November 2003 Paper 4 Q4*

__Solution 272:__**(a)**By connecting a single diode EITHER in series with resistor R OR in series with an a.c. supply

**(b)**

(i)

1.

{Maximum V = 6.0V. Minimum
V = 4.8V. Mean potential difference = (6.0 + 4.8) / 2 = 5.4V}

(From graph,) 5.4 V

2.

Ohm’s law: V = IR

Current I = 5.4 / (1.5x10

^{3}) = 3.6x10^{-3}A
3.

{As V falls, the capacitor
discharges by supplying charge. So, the cycle during which the capacitor
discharges through the resistor corresponds to the each region where V decreases.}

Time = 0.027s

(ii)

1.

Charge Q = IT = (3.6x10

^{-3}) x 0.027 = 9.72x10^{-5}C
2.

{Since the CHANGE in the
charge has been calculated, we need to substitute the change in potential
difference into the formula for capacitance. Change in V = 6.0 – 4.0 = 1.2V}

Capacitance C = ΔQ / ΔV = (9.72x10

^{-5}) / 1.2 = 8.1x10^{-5}F**(c)**

{If no capacitor was
present (capacitance = 0), then a graph similar to part (a) is obtained. So, by increasing the capacitance, the
decrease in V is less – we have the smoothing effect. So, a greater value of
capacitance causes a greater smoothing effect. Connecting capacitors in
parallel causes the total capacitance in the circuit to increase. Thus, in the
graph, the ripple should be less.}

The line should have a reasonable
shape with less ripple

__Question 273: [Current of Electricity > Resistance]__
What is the equivalent resistance of
this circuit?

__Solution 273:__
First label the different junctions
present in the circuit. Note the junction A and B are the same and E and F are
the same.

The 1Ω
resistor is between B and C but considering the

__upper part__, junction B is the same as D (no component is connected in the upper wire connecting B and D). This means that the 1Ω resistor (currently between B and C) can also be drawn between C and D. So the circuit can also be drawn as:
Similarly, considering the

__bottom__part between C and E (they are the same as no component is connected between them), the circuit becomes
So, the equivalent resistance = [1/1
+ ½ + 1/3]

^{-1}= 6/11

__Question 274: [Current of Electricity > Resistance]__

Circuit contains three similar lamps
A, B and C. Circuit also contains three switches, S

_{1}, S_{2}and S_{3}, as shown.
One of the lamps is faulty. In order
to detect fault, an ohm-meter (meter that measures resistance) is connected
between terminals X and Y. When measuring resistance, ohmmeter causes
negligible current in the circuit.

Fig shows readings of the ohm-meter
for different switch positions.

**(a)**Identify faulty lamp, and nature of the fault

**(b)**Suggest why it is advisable to test the circuit using an ohm-meter that causes negligible current rather than with power supply.

**(c)**Determine resistance of one of the non-faulty lamps, as measured using the ohmmeter.

**(d)**Each lamp is marked 6.0 V, 0.20 A.

Calculate, for one of the lamps
operating at normal brightness,

(i) its resistance

(ii) its power dissipation

**(e)**Comment on answers to (c) and (d)(i).

**Reference:**

*Past Exam Paper – June 2006 Paper 2 Q7*

__Solution 274:__**(a)**

{When all the switches are
open, the meter reads infinity. This is obvious since the circuit is shorted.

In the third test: S

_{1}: open, S_{2}: closed and S_{3}: open, the meter read a resistance of 30Ω. This is because test 2, where only S_{1}is closed, proves that lamp A is NOT faulty. So, the third test proves that lamp B is also NOT faulty.
In the last test, where
only S

_{1}is opened, the resistance read is 15Ω. This proves that lamp C is shorted – that is current passes as if there was no lamp. In this case, it’s as if a wire of negligible resistance is connected in parallel with lamp B.
Current is larger when the
resistance is smaller. [From
Ohm’s law, current I is inversely proportional to resistance R] So, all current would flow in that
wire instead of the lamp B (since lamp B has some resistance). So, the only
resistance encountered by the current is due to lamp A only. So, the resistance read by the meter is 15Ω.

If S

_{1}and S_{2}were closed, and S_{3}was opened, the meter would again read 15Ω since all the current would pass through S_{1}(which has no resistance) instead of passing through lamp B. The current would then flow through lamp A (which is the only lamp that causes a resistance in this case).}
Faulty lamp: Lamp C

Nature of fault: The lamp is shorted

**(b)**The shorted

__lamp A__would cause damage to the supply/lamps / blow fuse in the supply

**(c)**Resistance = 15Ω

**(d)**

(i)

V = IR

Resistance R (= V / R = 6.0 / 0.20)
= 30Ω

(ii)

Power P = VI or I

^{2}R or V^{2}/R
Power P = 1.2W

**(e)**

The filament is cold when measuring
with the ohm-meter in (b). The resistance of the filament rises as the
temperature rises.

Could you please explain the answer to part b of solution 274.

ReplyDeleteFor part b of solution 274, can you please explain the answer.

ReplyDeleteIf lamp A were to be faulted, the overall resistance would be reduced, causing the current flowing to be greater. This would blow the fuse in the supply or case damage to the lamps as these are marked with values of current that should not be exceeded.

DeleteThank you very much :)

Delete