• Physics 9702 Doubts | Help Page 135

Question 671: [Radioactivity]

(a) Describe structure of an atom of the nuclide ²³⁵₉₂U.

(b) Deflection of α-particles by a thin metal foil is investigated with the arrangement shown in Fig.1. All the apparatus is enclosed in a vacuum.

The detector of α-particles, D, is moved around path labelled WXY.

(i) Explain why apparatus is enclosed in a vacuum.

(ii) State and explain readings detected by D when it is moved along WXY.

(c) A beam of α-particles produces a current of 1.5 pA. Calculate number of α-particles per second passing a point in the beam.

Reference: Past Exam Paper – November 2012 Paper 21 Q6

Solution 671:

(a) An atom of the nuclide ²³⁵₉₂U has 92 protons in the nucleus and 92 electrons around nucleus. It also has 143 neutrons (in the nucleus).

(b)

(i) The α-particle travels short distance in air.

(ii) There is a very small proportion in the backwards direction / at large angles. The majority pass through the metal foil with no /small deflections.

EITHER most of the mass is in a very small volume (nucleus) and is charged

OR most of atom is empty space

(c)

Current I = Q / t

{The current is due to a number n of α-particles. An α-particle is a helium nucleus and thus has a charge of +2e.

Charge q of 1 α-particle = 2 × (1.6 × 10^–19) C

Total charge Q = nq

Current I = (nq) / t = (n/t) q

(n/t): number of α-particles per second passing a point in the beam

(n/t) = I / q}

n / t = (1.5 × 10^–12) /( 2 × 1.6 × 10^–19) = 4.7 × 10⁶ s^–1

Question 672: [Work, Energy and Power]

A man has mass of 80 kg. He ties himself to one end of a rope which passes over a single fixed pulley. He pulls on other end of the rope to lift himself up at an average speed of 50 cm s^–1.

What is average useful power at which he is working?

A 40 W                       B 0.39 kW                   C 4.0 kW                     D 39 kW

Reference: Past Exam Paper – June 2011 Paper 13 Q17

Solution 672:

Answer: B.

Average speed = 50 cm s^–1 = 0.5 m s^–1

Since we are considering an average speed, the acceleration on the system is zero. So, the resultant force on the rope is zero.

The weight of the man (= mg) acts downwards. So, the man should pull on the rope with an equal and opposite force, for the resultant to be zero.

Power = Fv = (80×9.81) × 0.5 = 392 W = 0.39 kW

Question 673: [Work, Energy and Power]

(a) State what is meant by work done.

(b) Trolley of mass 400 g is moving at a constant velocity of 2.5 m s^–1 to the right as shown in Fig.1.

Show that kinetic energy of the trolley is 1.3 J.

(c) Trolley in (b) moves to point P as shown in Fig.2.

At point P speed of the trolley is 2.5 m s^–1.

A variable force F acts to the left on trolley as it moves between points P and Q.

Variation of F with displacement x from P is shown in Fig.3.

Trolley comes to rest at point Q.

(i) Calculate distance PQ.

(ii) On Fig.4, sketch variation with x of velocity v for the trolley moving between P and Q.

Reference: Past Exam Paper – November 2013 Paper 22 Q3

Solution 673:

(a) Work done on a body is the product of the force applied on the body multiplied by the distance moved (displacement) by the body in the direction of the force.

(b)

Mass = 400g = 0.4 kg

Kinetic energy = ½ mv² = ½ × 0.4 × (2.5)² = 1.25J ≈ 1.3J

(c)

(i)

Acceleration is not constant. Equation for uniformly accelerated motion cannot be used.

Let x be the distance between P and Q.

(from previous part) Work done = 1.3 J (or 1.25J)

Also, Work done = area under graph = ½ Fx

1.3 = (14x) / 2

So, x = 0.19m

Or, if Work done is taken to be 1.25J, x = 0.179m = 0.18m

(ii)

{From Fig.3, the force increases with distance, x. So, a straight line (which represents a constant acceleration) is incorrect.

A connection should be made between the increase in the opposing force, F with the distance. This is shown by an increase in gradient in the in variation of velocity with distance. That is, the speed decreases by a greater value over the same amount of distance as x increases.

Note that in a graph, a vertical straight line corresponds to a gradient = infinity while a horizontal straight line corresponds to a gradient = 0. Recall that gradient = ∆y/∆x.}

Take the origin to be at P.

At P, v = 2.5ms^-1, x = 0m. And at Q, v = 0ms^-1, x = 0.19m.

Since force increases with distance, the curve will have an increasing gradient.

Question 674: [Direct Sensing]

An electronic sensor may be represented by the block diagram of Fig.1.

(a) State function of the processing unit.

(b) Student designs a sensing unit for temperature change. A 4 V supply, a fixed resistor of resistance 2.5 kΩ and thermistor are available. Thermistor has resistance 3.0 kΩ at 6 °C and resistance 1.8 kΩ at 20 °C.

Complete circuit diagram of Fig.2 to show how the resistor and the thermistor are connected to provide an output that is greater than 2 V at 6 °C and less than 2 V at 20 °C. Mark clearly the output V_OUT.

(c) Suggest two uses of relay as part of an output device.

Reference: Past Exam Paper – November 2013 Paper 41 & 42 Q9

Solution 674:

(a) The processing unit operates on/takes signal from a sensing device so that it gives an output voltage.

(b)

The thermistor and resistor should be in series between +4V line and earth.

V_out shown clearly across thermistor

{The lower line (Earth) is at 0V and the upper line is at +4V. As we go from the upper line towards the Earth symbol, the potential should decrease.

The resistance of the thermistor 1.8 kΩ at 20 °C and 3.0 kΩ at 6 °C. The resistor has a fixed resistance of 2.5 kΩ. By connecting the 2 resistors in series, a potential divider circuit is formed.

The circuit is to be drawn such that at 20 °C, the output voltage is less than 2V. At 20 °C, the temperature of the thermistor is less than at 6 °C, so the p.d. across it will be less. Actually, it would be [1.8 / (1.8+2.5)] × 4 = 1.67. This is only a difference in potential (p.d.).

At this temperature, if the thermistor is connected to the upper line of +4V, then the upper terminal of the thermistor would be at +4V and its lower terminal (which is connected to the fixed resistor) would be at a potential of 4 – 1.67 = 2.33V. This is greater than 2V, so the output cannot be drawn from the upper line (of +4V) to the lower terminal of the thermistor. Also, the upper terminal of the fixed resistor is at 2.33V and its lower terminal (connected to Earth symbol) is at 0V. So, the output cannot be drawn between these 2 points too.

The correct way is to connect the thermistor to Earth. Since the p.d. across it at the temperature of 20 °C is 1.67V, the upper terminal of the thermistor would be at 1.67V and its lower terminal at 0V. So, the output can be drawn across these 2 points.}

(c) Choose any 2:

Remote sensing

Switching large current by means of a small current

Isolating circuit from high voltage

Switching high voltage by means of a small voltage/current