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Question 444: (Gravitation)

Isolated spherical planet has diameter of 6.8 × 10⁶ m. Its mass of 6.4 × 10²³ kg may be assumed to be a point mass at centre of the planet.

(a) Show that gravitational field strength at surface of the planet is 3.7 N kg⁻¹.

(b) Stone of mass 2.4 kg is raised from surface of the planet through a vertical height of 1800 m.

Use value of field strength given in (a) to determine change in gravitational potential energy of the stone.

Explain your working.

(c) A rock, initially at rest at infinity, moves towards planet. At point P, its height above surface of the planet is 3.5 D, where D is the diameter of the planet, as shown.

Calculate speed of the rock at point P, assuming that change in gravitational potential energy is all transferred to kinetic energy.

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q1



Solution 444:
Go to
An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet.









Question 445: [Photoelectric effect]
A 60 W light bulb converts electrical energy to visible light with an efficiency of 8%. Calculate the visible light intensity 2 m away from the light bulb.              [3]
The average energy of the photons emitted by the light bulb in the visible region is 2 eV. Calculate the number of these photons received per square metre per second at this distance from the light bulb.                                                                         [2]



Solution 445:
For a (light) wave,
Intensity, I = P / 4πx²
where P is the power of the source and x is the distance from the source

Total power of light bulb = 60W
However, it is stated that the efficiency is only 8%. So, only 8% of the 60W will be useful.
Useful power from light bulb = 0.8 (60)

Distance x from light bulb = 2m

Intensity, I = 0.8 (60) / 4π(2)² = 0.1 Wm^-2  


Let the number of photons be N.
So, each of the N photons emitted by the light bulb has an average energy of 2eV. Thus, the N photons, with energy 2eV, causes an intensity of 0.1Wm^-2 (as calculated before).

Note that 2eV = 2 x (1.6x10^-19) = 3.2x10^-19 J
The units of intensity may also be written as Wm^-2 = Js^-1m^-2.

An intensity of 0.1 Jm^-2s^-1 is produced by N photons, each of energy 3.2x10^-19J. {Look how the units are obtained – the J cancel out.}
N (3.2x10^-19) = 0.1
Number of photons, N = 0.1 / (3.2x10^-19) = 3.125x10¹⁷ = 3x10¹⁷ m^-2s^-1










Question 446: [Waves > Double Slit]
Double slit experiment, using light of wavelength 600 nm, results in fringes being produced on screen. Fringe separation is found to be 1.0 mm.
When distance between the double slits and viewing screen is increased by 2.0 m, the fringe separation increases to 3.0 mm.
What is separation of the double slits producing the fringes?
A 0.4 mm                    B 0.6 mm                    C 0.9 mm                    D 1.2 mm

Reference: Past Exam Paper – November 2010 Paper 12 Q27



Solution 446:

Answer: B.

Wavelength λ = xd / nL

where

x: distance from the central fringe,

d: slit separation / distance between slits,

n: order of fringe

and L: distance between the slits and the screen.

Wavelength λ = 600nm. Distance from central fringe, x = 1.0mm.

So,

(600x10^-9) = (1x10^-3)d / nL.

For comparison, n, the order of the fringe being considered, should be equal in both cases, so it can be omitted.

Thus,

(600x10^-9)L = (1x10^-3)d          ----- equation (1)

When the distance from the slits to the screen, L becomes (L+2) {it is increased by 2}, the separation from the central fringe, x = 3.0mm.

Therefore,

(600x10^-9) = (3x10^-3)d / (L+2)

(600x10^-9)L + 2(600x10^-9) = (3x10^-3)d           -----equation (2)

Taking equation (2) – equation (1),

(2x10^-3)d = 2(600x10^-9)

Slit separation, d = 0.0006m = 0.6mm