# 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 S

_{2}is opened, current will not flow through it and the resistor R connected to it. So again, the current is I.

__Choice C:__If S

_{1}is opened and S

_{2}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 Î¼ 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 I

_{o}passing through a thickness x of a medium, then the transmitted intensity I is given by
I = I

_{o}exp(-Î¼x)
where Î¼ 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 Ï‰.

By
considering circular motion of planet about the star of mass M, show that Ï‰ and
R are related by the expression

R

^{3}Ï‰^{2}= GM
where
G is the gravitational constant. Explain your working.

**(c)**Earth orbits the Sun in circular orbit of radius 1.5 × 10

^{8}km. Mass of the Sun is 2.0 × 10

^{30}kg.

A
distant star is found to have planet that has a circular orbit about the star.
Radius of the orbit is 6.0 × 10

^{8}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
/ R

^{2}= mRÏ‰^{2}
where
m is the mass of the planet

{m can be eliminated on both sides, and re-arranging gives}

GM
= R

^{3}Ï‰^{2}**(c)**

Angular
speed Ï‰ = 2Ï€ / T

{We will be using the equation GM = R

^{3}Ï‰^{2}. 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 = R^{3}Ï‰^{2}/ M.
Now, Ï‰ = 2Ï€ / T. 2Ï€ is also constant. So, the angular speed is
inversely proportional to the period T.}

EITHER

{Here, we are considering the equation G = R

^{3}Ï‰^{2}/ M. R^{3}Ï‰^{2}/ M for the Earth-Sun system is equal to G and G is also equal to R^{3}Ï‰^{2}/ M for the distance star-planet system. Since the angular speed is proportional to the period, G is proportional to R^{3}/ MT^{3}.
For the Sun-Earth system: G is proportional to R

_{Sun}^{3}/ M_{Sun}T_{Sun}^{3}. R_{Sun}and M_{Sun}are given above. The period of Earth around the Sun = T_{Sun}is 1 year.
For the distance star-planet system: G is proportional to R

_{star}^{3}/ M_{star}T_{star}^{3}. The values are given above.
Since for both system, the equations are equal to G – they are be
equated, giving the following}

M

_{star}/ M_{Sun}= (R_{star}/ R_{Sun})^{3}× (T_{Sun}/ T_{star})^{2}
{R

_{star}/ R_{Sun}= 6.0 × 10^{8}km / 1.5 × 10^{8}km = 4} {T_{Sun}/ T_{star}= 1/2}
Mass
of the Star, M

_{star}= 4^{3}× (½)^{2}× (2.0 × 10^{30}) = 3.2 × 10^{31}kg
OR

{G = R

^{3}Ï‰^{2}/ M = G = R^{3}(2Ï€)^{2}/ MT^{2}since Ï‰ = 2Ï€ / T. 2 years = 2 × 365 × 24 × 3600s}
M

_{star}= (2Ï€)^{2}R_{star}^{3}/ GT^{2}
M

_{star}= (2Ï€)^{2}× (6.0 × 10^{11})^{3}/ {6.67 × 10^{–11}× (2×365×24×3600)^{2}} = 3.2 × 10^{31}kg
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