# Physics 9702 Doubts | Help Page 85

__Question 444: (Gravitation)__
Isolated spherical planet has
diameter of 6.8 × 10

^{6}m. Its mass of 6.4 × 10^{23}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:__**(a)**

Gravitational field strength, g = GM
/ R

^{2}
Gravitational field strength, g = (6.67
× 10

^{–11}) (6.4 × 10^{23}) / (3.4 × 10^{6})^{2}= 3.7 Nkg^{-1}**(b)**

Change in gravitational potential
energy, ΔE

_{P}= mgΔh
because the vertical height Δh ≪ R
(or 1800 m ≪ 3.4 × 10

^{6}m) the gravitational field strength g is constant
Change in gravitational potential
energy, ΔE

_{P}= 2.4 × 3.7 × 1800 = 1.6 × 10^{4}J**(c)**

Gravitational potential energy = (–)
GMm / x

{x is the separation
between the rock and the centre of the planet (since the planet is considered
to be a point mass, all its mass is concentrated at the centre).}

{The GPE is all transferred
to kinetic energy. So, considering the magnitudes, ½ mv

^{2}= GMm / x, giving v^{2}= 2GM / x.}
v

^{2}= 2GM / x
{The rock is at a distance
3.5D from the surface of the planet. The surface is itself at a distance of D/2
= 0.5D form the centre of the planet. So, x = 3.5D + 0.5D = 4D}

Separation x = 4D = 4 × (6.8 × 10

^{6}) m
v

^{2}= 2 × (6.67 × 10^{–11}) × (6.4 × 10^{23}) / (4 × 6.8 × 10^{6}) = 3.14 × 10^{6}
Speed v = 1.8 × 10

^{3}m s^{–1}

__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 / 4πx

^{2}

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

^{-1}m

^{-2}.

An intensity of 0.1 Jm

^{-2}s

^{-1}is produced by N photons, each of energy 3.2x10

^{-19}J. {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}) = 3.125x10

^{17}= 3x10

^{17}m

^{-2}s

^{-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

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