# Physics 9702 Doubts | Help Page 216

__Question 1032: [Matter > Deformation]__
The graph shows the relationship between stress and strain for three
wires of the same linear dimensions but made from different materials.

Which statements are correct?

1 The extension of P is approximately twice that of Q for the same
stress.

2 The ratio of the Young modulus for P to that of Q is approximately
two.

3 For strain less than 0.1, R obeys Hooke’s law.

A 1, 2 and 3 B 1 and
3 only C 2 and 3
only D 2 only

**Reference:**

*Past Exam Paper – November 2011 Paper 12 Q21*

__Solution 1032:__**Answer: C.**

Stress = Force / Area and Strain = extension / original
length

Statement 1: [incorrect]

Consider an equal amount of stress
on wire P and Q (draw a horizontal line on the graph). Wire Q has a greater
strain than wire P. This means that the extension of wire Q is greater than
that of P. The statement is wrong.

It should have been: “

*The extension of*”.**Q**is approximately twice that of**P**for the same stress
Statement 2: [correct]

Young modulus E = stress / strain

For the same stress, the strain of Q
is approximately twice that of P, thus its Young modulus is half that of P.

Ratio of E

_{P}/ E_{Q}= E / 0.5E = 2
Statement 3: [correct]

Hooke’s law: F = ke

Since stress is proportional to the
force and strain is proportional to the extension, it can be concluded that, for
Hooke’s law, stress is proportional to strain. In the graph, it is a straight line
passing through the origin. So, R obeys Hooke’s law for strain less than 0.1.

__Question 1033:__

__[Electromagnetism]__**(a)**A solenoid is connected in series with a resistor, as shown in Fig. 7.1.

As the magnet is being moved into the solenoid, thermal energy is
transferred in the resistor.

Use laws of electromagnetic induction to explain the origin of this
thermal energy.

**(b)**Explain why the alternating current in the primary coil of a transformer is not in phase with the alternating e.m.f. induced in the secondary coil.

**Reference:**

*Past Exam Paper – June 2015 Paper 42 Q7*

__Solution 1033:__**(a)**The moving magnet gives rise to/causes/induces an e.m.f./current in solenoid/coil. This (induced current) creates a field/flux in the solenoid that opposes the (motion of) magnet. Work is done/energy is needed to move the magnet (into the solenoid). The (induced) current gives heating effect (in resistor) which comes from the work done.

**(b)**The current in the primary coil give rise to a (magnetic) flux/field. The (magnetic) flux / field (in the core) is in phase with the current (in primary coil). This (magnetic) flux threads/links/cuts the secondary coil inducing an e.m.f. in the secondary coil. The e.m.f. induced is proportional to the

__rate__of change/cutting of flux/field, so it is not in phase {with the alternating current in the primary coil itself}.

__Question 1034: [Resistance]__
A copper wire is to be replaced by
an aluminium alloy wire of the same length and resistance.

Copper has half the resistivity of
the alloy.

What is the ratio diameter of alloy wire / diameter of
copper wire ?

A √2 B 2 C
2√2 D 4

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q31*

__Solution 1034:__**Answer: A.**

Resistance of a wire = ρL /
A

Copper wire: Resistance R = ρ

_{c}L / A
Cross-sectional A = ρ

_{c}L / R
Since A proportional to the diameter

^{2}(A = πd_{c}^{2 }/ 4),
d

_{c}^{2}= 4ρ_{c}L / R [since we are dealing with a ratio, the π may be eliminated as it is a constant]
Diameter of copper wire, d

_{c}= 2 √(ρ_{c}L/R)
Aluminium wire: Resistance R = (2ρ

_{c})L / A
Cross-sectional A = 2ρ

_{c}L / R
d

_{A}^{2}= 8ρ_{c}L / R
Diameter of aluminium wire, d

_{A}= 2√(2ρ_{c}L/R) = 2√2 √(ρ_{c}L/R)
So, ratio = d

_{A}/ d_{c}= √2

__Question 1035: [Dynamics > Air resistance]__
A body is released from rest and falls vertically in air of constant
density.

Which statement about the motion of the falling body is correct?

A As it accelerates, its weight decreases so that its acceleration
decreases until it travels with constant velocity.

B It accelerates initially at 9.8 m s

^{–2}but the drag force increases so its acceleration decreases.
C Its velocity increases at a constant rate until its velocity becomes
constant.

D The drag force of the air increases continually and eventually the
velocity decreases.

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q7*

__Solution 1035:__**Answer: B.**

As the body is released, it falls
under the effect of gravity – with an initial acceleration of 9.8 m s

^{–2}.
However, the drag force of the air
causes the resultant force on the body to decrease, and thus the acceleration
decreases.

The weight of a body is constant in
one particular location. [A is incorrect]

‘Velocity increases at a constant rate’ means a constant acceleration, which is incorrect. [C is incorrect]

The drag force increases with an
increase in velocity of the body. At one point, the weight would be equal to
the drag force and the resultant force on the body becomes zero. From this
point, the body no longer accelerates – that is, it travels with a constant
velocity. Consequently, the drag force remains constant. [D is incorrect]

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.10(c)

04/M/J/08 Q.5(b),Q.9(b)

41/O/N/09 Q.10

51/M/J/10 Q.2(d)

For 41/O/N/09 Q.10, see solution 1038 at

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can you do q34 the way you did the other resistance questions example when it says the length is twice so you write resistance as 2R since R=resistivity X L/A.HOPE U GET WHAT IAM SAYING

ReplyDeleteyes, but note that R does not only depend on the length

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