• Physics 9702 Doubts | Help Page 141

Question 695: [Sensing devices]

(a) State name of an electrical sensing device that will respond to changes in

(i) length

(ii) pressure

(b) Relay is sometimes used as the output of a sensing circuit.

Output of particular sensing circuit is either +2V or -2V.

On Fig.1, draw symbols for relay and any other necessary component so that the external circuit is switched on only when the output from the sensing circuit is +2V.

Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q10

Solution 695:

(a)

(i) Strain gauge

(ii) Piezo-electric / quartz crystal / transducer

(b) For the circuit, a coil of relay is connected between the sensing circuit output and earth and the switch is across the terminals of the external circuit.

A diode is connected in series with the coil with a correct polarity for the diode (downwards). A second diode with correct polarity is connected (upwards and parallel to coil of relay).

{Note that the following explanations have been taken directly from the application booklet.

The diode D₁ conducts only when the output is positive with respect to earth and thus the relay coil is energized only when the output is positive. When the current in the relay coil is switched off, a back e.m.f. is generated in the coil that could damage the sensing circuit. A diode D₂ is connected across the coil to protect the sensing circuit from this back e.m.f.}

Question 696: [Current of Electricity]

(a) A wire has length 100 cm and diameter 0.38 mm. Metal of the wire has resistivity 4.5 × 10^–7 Ω m.

Show that resistance of the wire is 4.0 Ω.

(b) The ends B and D of the wire in (a) are connected to a cell X, as shown in Fig.1.

Cell X has electromotive force (e.m.f.) 2.0 V and internal resistance 1.0 Ω.

A cell Y of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to the wire at points B and C, as shown in Fig.1.

Point C is distance l from point B. The current in cell Y is zero. Calculate

(i) current in cell X,

(ii) potential difference (p.d.) across the wire BD,

(iii) distance l.

(c) Connection at C is moved so that l is increased. Explain why the e.m.f. of cell Y is less than its terminal p.d.

Reference: Past Exam Paper – November 2014 Paper 23 Q6

Solution 696:

Go to
A wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10–7 Ω m.

Question 697: [Measurement > Prefixes]

A signal has frequency of 2.0 MHz.

What is period of the signal?

A 2 μs                         B 5 μs                         C 200 ns                                  D 500 ns

Reference: Past Exam Paper – November 2010 Paper 11 Q1 & Paper 13 Q4

Solution 697:

Answer: D.

Frequency f = 2.0 MHz = 2.0×10⁶ Hz

Period = 1 / f = 1 / (2.0×10⁶) = 0.5×10^-6 = 500×10^-9 = 500ns

Question 698: [Kinematics + Dynamics]

A helicopter has a cable hanging from it towards the sea below, as shown in Fig.1.

A man of mass 80 kg rescues child of mass 50.5 kg. The two are attached to the cable and are lifted from sea to the helicopter. Lifting process consists of an initial uniform acceleration followed by a period of constant velocity and then completed by a final uniform deceleration.

(a) Calculate combined weight of the man and child.

(b) Calculate tension in the cable during

(i) initial acceleration of 0.570 m s^–2,

(ii) period of constant velocity of 2.00 m s^–1.

                                                      

(c) During final deceleration the tension in the cable is 1240 N. Calculate this deceleration.

(d) (i) Calculate time over which the man and child are

1. moving with uniform acceleration,

2. moving with uniform deceleration.

(ii) Time over which the man and child are moving with constant velocity is 20 s. On Fig.2, sketch a graph to show variation with time of the velocity of the man and child for the complete lifting process.

Reference: Past Exam Paper – June 2011 Paper 23 Q3

Solution 698:

(a) {Combined mass = 80 + 50.5 = 130.5kg}

Combined weight = m × g = 130.5 × 9.81 = 1280 N

(b)

(i)

Resultant force F = ma

{Tension is upwards while the weight acts downwards. The resultant force / acceleration is upwards.}

T – 1280 = 130.5 × 0.57

Tension T = 1280 + 74.4 = 1350 N

(ii) Tension T = 1280 N

{For constant velocity, the resultant force is zero. Thus, tension = weight.}

                                                      

(c)

{Resultant force F = ma

For a deceleration, the direction of the resultant force should be opposite to the direction of motion. Motion is upwards, so the resultant force is downwards.

Tension – Weight = Resultant force = ma      (a is negative here)}

1240 – 1280 = 130.5 × a

a = (–) 0.31 m s^–2

(d) (i)

1.

{Initial acceleration a = 0.570 m s^–2. Initial velocity, u = 0. Final velocity, v = 2.00 m s^–1.

a = (v – u) / t}

Time t {= (v – u) / a = 2.00 / 0.570} = 3.5s

2.

{Deceleration a = - 0.31 m s^–2. Initial velocity, u = 2.00ms^-1. Final velocity, v= 0.

a = (v – u) / t}

Time t {= (v – u) / a = -2.00 / -0.31} = 6.5s

(ii) The graph has a basic shape with the correct points.

Question 699: [Alternating Current]

An ideal transformer has 5000 turns on its primary coil. It is to be used to convert mains supply of 230 V r.m.s. to an alternating voltage having a peak value of 9.0 V.

(a) Calculate number of turns on the secondary coil.

(b) Output from the transformer is to be full-wave rectified. Fig.1 shows part of the rectifier circuit.

On Fig.1, draw

(i) diode symbols to complete diagram of the rectifier such that terminal A of the resistor R is positive with respect to terminal B,

(ii) symbol for a capacitor connected to provide smoothing of potential difference across the resistor R.

(c) Fig.2 shows variation with time t of the smoothed potential difference V across the resistor R.

(i) State interval of time during which the capacitor is being charged from the transformer.

(ii) Resistance of the resistor R is doubled. On Fig.2, sketch variation with time t of the potential difference V across the resistor.

Reference: Past Exam Paper – June 2007 Paper 4 Q4

Solution 699:

(a)

r.m.s. output = 9/√2 or peak input = 230√2

N_S / N_P = V_S / V_P

{N_S / 5000 = (9/√2) / 230        or N_S / 5000 = 9 / 230√2}

N_S = 138 → 140 turns

(b)

(i) The four diodes should be correctly positioned (regardless of output polarity) giving correct output polarity (all ‘point to left’)

{The diodes should be drawn such that current always passes point A first, before passing point B.}

(ii) The capacitor should be shown in parallel with R

 

(c)

(i) time t₁ to time t₂

(ii) The sketch has the same peak values with the ripple reduced and a reasonable shape