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Question 382: [Electric field]
An isolated, solid metal sphere of radius R is given an electric charge. Which one of the following best represents the way in which the density of charge varies with distance r from the centre of the sphere?

Reference: Past Exam Paper – J83 / II / 22



Solution 382:
Answer: A.
The charges are found at the surface of the sphere. So, anywhere inside the sphere, it is assumed that there is no charge. Thus, the charge density before distance r = radius R is zero.

For points inside the sphere, the electric field intensity E is zero.

At the surface (distance r = radius R) of the sphere, all charges are found. So, the charge density is maximum at distance r = R and zero before and after r = R.











Question 383: [Waves > Graphs]
Graph shows the shape at a particular instant of part of transverse wave travelling along a string.

Which statement about the motion of points in string is correct?
A The speed at point P is a maximum.
B The displacement at point Q is always zero.
C The energy at point R is entirely kinetic.
D The acceleration at point S is a maximum.

Reference: Past Exam Paper – June 2007 Paper 1 Q23



Solution 383:
Answer: D.
The wave is NOT stationary but, as stated in the question, the graph only represents the shape of the wave at a particular instant. So, the displacements of all the points on the string would change with time. [B is incorrect]

As the transverse wave travels along the string, the particles (points) in the string vibrate up and down.  So, the motion of the points is periodic (the motion repeats itself). The particles undergo simple harmonic motion.

In a simple harmonic motion, the acceleration is maximum when the displacement is maximum (at amplitude). [D is correct]

The speed at the maximum displacement (at P, R and S) is zero since the particles are momentarily at rest there (compare it with the simple harmonic motion of a pendulum). [A is incorrect] Thus, at these points, kinetic energy is zero (since speed is zero). [C is incorrect]

At the rest position (at point Q), the displacement is zero but the speed is maximum.










Question 384: [Matter > Ideal Gases]

(a) State what is meant by ideal gas.

(b) Two cylinders A and B are connected by tube of negligible volume, as shown.

Initially, tap T is closed. Cylinders contain an ideal gas at different pressures.

(i) Cylinder A has constant volume of 2.5 × 10³ cm³ and contains gas at pressure 3.4 × 10⁵ Pa and temperature 300 K.

Show that cylinder A contains 0.34 mol of gas.

(ii) Cylinder B has constant volume of 1.6 × 10³ cm³ and contains 0.20 mol of gas. When tap T is opened, pressure of the gas in both cylinders is 3.9 × 10⁵ Pa. No thermal energy enters or leaves the gas.

Determine the final temperature of gas.

(c) By reference to work done and change in internal energy, suggest why temperature of the gas in cylinder A has changed.

Reference: Past Exam Paper – June 2013 Paper 41 Q2



Solution 384:

(a) An ideal gas is one that obeys the equation pV = constant          OR pV = nRT

where p is the pressure of the gas, V is the volume of the gas and T is the temperature of the gas (p, V and T explained)

at all values of p, V and T / fixed mass / n is constant

(b)

(i)

{pV = nRT} {1 cm³ = 1×10^-6 m³}

(3.4 × 10⁵) (2.5 × 10³ cm³) = n (8.31) (300)

Number of moles, n = 0.34 mol

(ii)

For the total mass / amount of gas

{Both cylinders contain the same ideal gas and the pressure of the gas in both of them when the tap is opened is p = 3.9 × 10⁵ Pa. After the tap is opened, both cylinders may be considered as a single system with the stated pressure.

The total volume of gas is the sum of volumes of gases in both cylinders (= 2.5 × 10³ cm³ + 1.6 × 10³ cm³ = (2.5+1.6) × 10³ cm³ = (2.5+1.6) × 10³ × 10^-6 m³)

The total amount of gas changes when the tap is opened. The total number of moles, n = 0.34 + 0.20}

{pV = nRT}

(3.9 × 10⁵) ([2.5+1.6] × 10³) × 10^-6 = (0.34+0.20) (8.31) T

Temperature T = 360K

(c) When the tap is opened, gas passes (from cylinder B) to cylinder A {since the pressure in cylinder A has increased, it means that there is more collisions of the gas molecules with the wall of cylinder A. So, the total number of molecules should have increased}. So, work is done on the gas in cylinder A (and no heating, i.e. without the system being heated – no energy is being provided to the system). So, the internal energy and hence the temperature increase.

Question 385: [Waves > Double slit experiment]
Diagrams show the arrangement of apparatus for a Young’s slits experiment and also pattern formed on the screen with a ruler placed next to it.

What is wavelength of the light?
A 4.8 × 10^–7 m            B 5.4 × 10^–7 m             C 3.2 × 10^–6 m             D 3.4 × 10^–6 m

Reference: Past Exam Paper – June 2011 Paper 12 Q29



Solution 385:

Answer: B.

Formula for Young’s slit experiment:             x = λD /a

where

x = fringe separation,

λ = wavelength,

D = distance between slits and screen and

a = slit separation.

Fringe separation, x = 31.5 – 28.5 = 3mm (the separation between any 2 similar (dark or bright) fringe could also be have been considered. Here the separation between the 2 top dark fringes is considered.)

Wavelength λ = xa / D = (3x10^-3)(0.9x10^-3) / 5 = 5.4x10^-7m