Monday, March 16, 2015

Physics 9702 Doubts | Help Page 86

  • Physics 9702 Doubts | Help Page 86



Question 447: [Current of Electricity]
Ammeter reading in the circuit below is I.

Another circuit containing same voltage supply, two switches, an ammeter and two resistors each of resistance R, is shown.

Which row is not correct?

Reference: Past Exam Paper – November 2012 Paper 12 Q37



Solution 447:
Answer: C.
In the first circuit, an ammeter is connected in series with a resistor of resistance R and a supply. The current is I.

The question asks for the combination that is NOT correct.

Choice A: The current flows from the positive terminal to the supply, through resistor R, switch 1, ammeter and finally to negative terminal of supply. This is because the switch has zero resistance, so all current would flow through it instead of the resistance R connected in parallel to it. So, this is similar to the first circuit. The current should be I.

Choice B: Since switch S2 is opened, current will not flow through it and the resistor R connected to it. So again, the current is I.

Choice C: If S1 is opened and S2 is closed, current would flow through both resistors. So, the total resistance in the circuit is twice. The current should be halved (from Ohm’s law, I = V/R). This row is incorrect and hence C is the correct answer.

Choice D: When both switches are opened, the circuit is not complete, so no current would flow.










Question 448: [X-rays]
(a) Typical spectrum of X-ray radiation produced by electron bombardment of metal target is illustrated.

Explain why
(i) Continuous spectrum of wavelengths is produced
(ii) Spectrum has sharp cut-off at short wavelengths

(b) Variation with photon energy E of the linear absorption coefficient μ of X-rays in soft tissue is illustrated.

(i) Explain what is meant by linear absorption coefficient
(ii) For one particular application of X-ray imaging, electrons in X-ray tube are accelerated through potential difference of 50kV.
Use Fig to explain why it is advantageous to filter out low-energy photons from X-ray beam.

Reference: Past Exam Paper – November 2009 Paper 42 Q10



Solution 448:
(a)
(i) E.M. radiation / photons are produced whenever a charged particle is decelerated (at the metal target). The wavelength depends on the magnitude of the deceleration. The electrons have a distribution of decelerations. So, a continuous spectrum of wavelengths is produced.

(ii) EITHER When an electron loses all its energy in one collision OR When the energy of an electron produces a single photon

(b)
(i) For a parallel beam of radiation of initial intensity Io passing through a thickness x of a medium, then the transmitted intensity I is given by
I = Ioexp(-μx)
where μ is the linear absorption coefficient.

(ii)
EITHER Low –energy photons are absorbed (much) more readily OR Low-energy photons are (far) less penetrating.
The low-energy photons do not contribute to the X-ray image.
The low-energy photons could cause tissue damage.










Question 449: [Gravitation]
(a) State Newton’s law of gravitation.

(b) Star and a planet are isolated in space. Planet orbits the star in circular orbit of radius R, as illustrated in Fig.

Angular speed of planet about the star is ω.
By considering circular motion of planet about the star of mass M, show that ω and R are related by the expression
R3ω2 = GM
where G is the gravitational constant. Explain your working.

(c) Earth orbits the Sun in circular orbit of radius 1.5 × 108 km. Mass of the Sun is 2.0 × 1030 kg.
A distant star is found to have planet that has a circular orbit about the star. Radius of the orbit is 6.0 × 108 km and the period of the orbit is 2.0 years.
Use expression in (b) to calculate the mass of the star.

Reference: Past Exam Paper – November 2013 Paper 43 Q1



Solution 449:
(a) Newton’s law of gravitation states that the (gravitational) force (between 2 point masses) is proportional to the product of the two masses and inversely proportional to the square of their separation, provided that their separation is much larger than the ‘size’ of the masses.

(b)
The gravitational force between the masses provides the centripetal force.
{magnitude of Gravitational force = magnitude of Centripetal force}
GMm / R2 = mRω2
where m is the mass of the planet
{m can be eliminated on both sides, and re-arranging gives}
GM = R3ω2

(c)
Angular speed ω = 2π / T
{We will be using the equation GM = R3ω2. G is the gravitational constant and can be used for the gravitational force between any 2 masses. Since G is a constant, for any system it will have the same value. Making G the subject of formula gives G = R3ω2 / M.
Now, ω = 2π / T. 2π is also constant. So, the angular speed is inversely proportional to the period T.}
EITHER
{Here, we are considering the equation G = R3ω2 / M. R3ω2 / M  for the Earth-Sun system is equal to G and G is also equal to R3ω2 / M for the distance star-planet system. Since the angular speed is proportional to the period, G is proportional to R3 / MT3.
For the Sun-Earth system: G is proportional to RSun3 / MSun TSun3. RSun and MSun are given above. The period of Earth around the Sun = TSun is 1 year.
For the distance star-planet system: G is proportional to Rstar3 / Mstar Tstar3. The values are given above.
Since for both system, the equations are equal to G – they are be equated, giving the following}
Mstar / MSun = (Rstar / RSun)3 × (TSun / Tstar)2
{Rstar / RSun = 6.0 × 108 km / 1.5 × 108 km = 4} {TSun / Tstar = 1/2}
Mass of the Star, Mstar = 43 × (½)2 × (2.0 × 1030) = 3.2 × 1031 kg

OR
{G = R3ω2 / M = G = R3(2π)2 / MT2 since ω = 2π / T. 2 years = 2 × 365 × 24 × 3600s}
Mstar = (2π)2 Rstar3 / GT2
Mstar = (2π)2 × (6.0 × 1011)3 / {6.67 × 10–11 × (2×365×24×3600)2} = 3.2 × 1031 kg



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