• Physics 9702 Doubts | Help Page 128

Question 643: [Matter]

(a) With reference to arrangement of atoms, distinguish between metals, polymers and amorphous solids.

(b) On Fig.1, sketch variation with extension x of force F to distinguish between a metal and a polymer.

Reference: Past Exam Paper – November 2012 Paper 22 Q3

Solution 643:

(a)

Metals: regular / repeated / ordered arrangement / pattern / lattice or long range order (of atoms / molecules / ions)

Polymers: tangled chains (of atoms / molecules) or long chains (of atoms / molecules / ions)

Amorphous solids: disordered / irregular arrangement or short range order (of atoms / molecules / ions)

(b)

Metal: straight line or straight line then curving with less positive gradient 

Polymer: curve with decreasing gradient with steep increasing gradient at end

Question 644: [Nuclear Physics]

(a) Radioactive decay of some nuclei gives rise to the emission of α-particles.

State

(i) what is meant by α-particle

(ii) two properties of α-particles

(b) One possible nuclear reaction involves bombardment of a stationary nitrogen-14 nucleus by an α-particle to from oxygen-17 and another particle.

(i) Complete nuclear equation for this reaction

¹⁴₇N     +          ^…_…α     - - - >   ¹⁷₈O                 +          ^…_…p

(ii) The total mass-energy of nitrogen-14 nucleus and the α-particle less than that of the particles resulting from the reaction. This mass-energy difference = 1.1MeV.

1. Suggest how it is possible for mass-energy to be conserved in this reaction

2. Calculate speed of an α-particle having kinetic energy of 1.1MeV

Reference: Past Exam Paper – June 2010 Paper 22 Q7

Solution 644:

(a)

(i) An α-particle EITHER is a helium nucleus            OR contains 2 protons and 2 neutrons

(ii) Choose any 2:

Its range is a few cm in air / sheet of thin paper

It has a speed of up to 0.1c

It causes dense ionization in air

It is positively charged or deflected in magnetic or electric fields

(b)

(i) ¹⁴₇N +          ⁴₂α        - - - >   ¹⁷₈O                 +          ¹₁p

(ii)

1. Initially, the α-particle must have some kinetic energy

2.

{First, convert the energy given in MeV into joules.}

In Joules, 1.1MeV = 1.1 x (1.6x10^-13) = 1.76x10^-13J

Kinetic energy, E_k = ½ mv²

{The α-particle is a helium nucleus – it consists of 2 protons and 2 neutrons. That is, it contains 4 nucleons.

Unified atomic mass constant, u = 1.66×10^-27 kg.

So, the mass of the α-particle is 4u.}

1.76×10^-13 = ½ (4 × 1.66×10^-27) × v²  

Speed v = 7.3×10⁶ ms^-1

Question 645: [Matter]

Ice at temperature of 0 °C is a rare example of a solid that floats on its liquid form, in this case water, when they are both at the same temperature.

What is the explanation for this?

A The average speed of the molecules in the ice is greater than the average speed of the molecules in the water.

B The average speed of the molecules in the water is greater than the average speed of the molecules in the ice.

C The mean separation of the molecules in the ice is greater than the mean separation of the molecules in the water.

D The mean separation of the molecules in the water is greater than the mean separation of the molecules in the ice.

Reference: Past Exam Paper – June 2013 Paper 13 Q17

Solution 645:

Answer: C.

At the same temperature, the molecules have the same kinetic energy in both ice and water. So, the average speed of the molecules in both of them are the same. [A and B are incorrect]

Floating depends on the density of the substances present. If a solid has a lower density than the liquid in which it is put, then the solid will float.

Density = Mass / Volume

Since ice floats on water, it means that its density is less than water. For the same mass of both substances, the volume of ice should be greater than that of water. This explains why the density of ice is lower than that of water.

For greater volume, the mean separation of the molecules in the ice should is greater than the mean separation of the molecules in the water. [D is incorrect]

Question 646: [Alternating Current]

(a) Explain, in terms of heating effect, what is meant by root-mean-square (r.m.s.) value of an alternating current.

(b) State relation between the peak current I₀ and the r.m.s. current I_rms of a sinusoidally-varying current.

(c) Value of a direct current and peak value of a sinusoidal alternating current are equal.

(i) Determine the ratio

(power dissipation in a resistor of resistance R by the direct current) / (power dissipation in the resistor of resistance R by the alternating current)

(ii) State one advantage and one disadvantage of use of alternating rather than direct current in the home.

(d) A current I varies with time t as shown in Fig.1.

For this varying current, state

(i) peak value,

(ii) r.m.s. value.

Reference: Past Exam Paper – June 2004 Paper 4 Q5

 

Solution 646:

(a) The root-mean-square (r.m.s.) value of an alternating current is the (value of the) direct current that dissipates (heat) energy at the same rate (in a resistor).

(b) √2 I_rms = I₀

(c)

(i)

Power dissipated ∝ I²                         or Power P = I²R         or P = VI

{Direct current: P_dc = I₀²R

Alternating current: P_ac = I_rms²R = (I₀/√2)²R = I₀²R / 2}

Ratio (= P_dc / P_ac) = 2.0

(ii)

Advantage: e.g. It is easy to change the voltage

Disadvantage: e.g.

The cables require greater insulation

Rectification – with some justification

(d)

(i) Peak value = 3.0 A

(ii) r.m.s. value = 3.0 A

{As explained in the first part, the root-mean-square (r.m.s.) value of an alternating current is the (value of the) direct current that dissipates (heat) energy at the same rate (in a resistor).

In any general current-time graph, to find the r.m.s value, we take the area under the graph for a specific time interval and divide this area by that time interval.

By since the graph shown is periodic, the time interval to be consider is actually the period. From the graph, the period is 2(0.5) = 1.0ms.

The total area for 1 period is given by the area of 2 rectangles. Total area = (3×0.5) + (3×0.5) = 3 A ms. The time interval is the period which is 1.0ms.

So, I_rms = (3 A ms) / (1ms) = 3A.

This is obvious for a rectangular wave – the r.m.s. value is equal to the peak value (without actually having to go through the calculations I did).}