# 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 Î±-particles.

State

(i) what is meant by Î±-particle

(ii) two properties of Î±-particles

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

(i) Complete nuclear equation for
this reaction

^{14}

_{7}N +

^{…}

_{…}Î± - - - >

^{17}

_{8}O +

^{…}

_{…}p

(ii) The total mass-energy of
nitrogen-14 nucleus and the Î±-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 Î±-particle
having kinetic energy of 1.1MeV

**Reference:**

*Past Exam Paper – June 2010 Paper 22 Q7*

__Solution 644:__**(a)**

(i) An Î±-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)

^{14}_{7}N +^{4}_{2}Î± - - - >^{17}_{8}O +^{1}_{1}p
(ii)

1. Initially, the Î±-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^{-13}J
Kinetic energy, E

_{k}= ½ mv^{2}
{The Î±-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
Î±-particle is 4u.}

1.76×10

^{-13}= ½ (4 × 1.66×10^{-27}) × v^{2}
Speed v = 7.3×10

^{6 }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 I

_{0}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}= I

_{0}

**(c)**

(i)

Power dissipated ∝ I

^{2}or Power P = I^{2}R or P = VI
{Direct current: P

_{dc}= I_{0}^{2}R
Alternating current: P

_{ac}= I_{rms}^{2}R = (I_{0}/√2)^{2}R = I_{0}^{2}R / 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).}

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