# Physics 9702 Doubts | Help Page 200

__Question 968: [Current of Electricity > Power]__
A 100 Ω resistor conducts a current
with changing direction and magnitude, as shown.

What is the mean power dissipated in
the resistor?

A 100 W B 150 W C
250 W D 400 W

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q36*

__Solution 968:__**Answer: C.**

First, consider the period of the
alternating current.

Look at the period just next to the
y-axis. The period of the signal consists of a first section where the current
I = 2A, followed by another section where I = -1A.

Power dissipated in a resistor = I

^{2}R
When current I = 2A (1

^{st}half of the cycle), Power dissipated = (2)^{2}× 100 = 400W
When current I = -1A (2

^{nd}half of the cycle), Power dissipated = (-1)^{2}× 100 = 100W
So, mean power = (400+100) / 2 =
250W

__Question 969: [Kinematics > Graph]__
The diagram shows a velocity-time
graph.

What is the displacement during the
last 2 seconds of the motion?

A 6 m B 12 m C
18 m D 24 m

**Reference:**

*Past Exam Paper – November 2009 Paper 11 Q7 & Paper 12 Q6*

__Solution 969:__**Answer: C.**

In a velocity-time graph, the
displacement is given by the area under the graph.

Displacement = area under graph for
last 2s = area of trapezium

Displacement = ½ × sum
of parallel sides × width = ½ × (6+12) × (2) = 18m

__Question 970: [Radioactivity]__
Uranium-236 (

^{236}_{92}U) and Uranium-237 (^{239}_{72}U) are both radioactive.
Uranium-236 is an α-emitter and
Uranium-237 is a β-emitter.

**(a)**Distinguish between an α-particle and a β-particle.

**(b)**The grid of Fig. 7.1 shows some proton numbers Z on the x-axis and the number N of neutrons in the nucleus on the y-axis.

The α-decay of Uranium-236 (

^{236}_{92}U) is represented on the grid. This decay produces a nucleus of thorium (Th).
(i) Write down the nuclear equation
for this α-decay.

(ii) On Fig. 7.1, mark the position
for a nucleus of

1. Uranium-237 (mark this position
with the letter U),

2. Neptunium, the nucleus produced
by the β-decay of Uranium-237 (mark this position with the letters Np).

**Reference:**

*Past Exam Paper – June 2008 Paper 2 Q7*

__Solution 970:__**(a)**

α-particle: either helium nucleus OR contains 2 protons + 2 neutrons OR

^{4}_{2}He
β-particle: either electron OR

^{0}_{-1}e
Choose any 2 of the
following:

The speed of α speed is less than
that of β.

α-particles have discrete values of
speed/energy while β-particles have a continuous spectrum.

EITHER The ionising power of α-particles
is much greater than that of β-particles OR
The range of or α-particles is much less than that of β-particles.

**(b)**

(i)

^{236}_{92}U →^{232}_{90}Th +^{4}_{2}He
{Since it is an alpha
decay, a helium nucleus must be produced. We can then calculate the
corresponding proton and nucleon number of the other product.}

(ii)

1. correct position for U at Z = 92,
N = 145

{Since it is still
uranium, the proton number would still be 92. Thus, the neutron number = 237 –
92 = 145}

2. correct position for Np relative
to U i.e. Z + 1 and N – 1

{

^{237}_{92}U →^{237}_{93}Np +^{0}_{-1}e
Proton number = 93.
Neutron number = 237 – 93 = 144}

__Question 971: [Nuclear Physics]__
Each of the nuclei below is
accelerated from rest through the same potential difference.

Which one completes the acceleration
with the

**lowest**speed?
A

^{1}_{1}H B^{4}_{2}He C^{7}_{3}Li D^{9}_{4}Be**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q40*

__Solution 971:__**Answer: C.**

A force is needed to cause an
acceleration.

Electric force F = Eq = (V/d) q = Vq
/ d

This should be equal to F = ma (the
resultant force on the nuclei.).

So, ma = Vq/d giving

**a = Vq / md**
Since V and d are constant for all
particles, the acceleration a depends on

**q/m.**
Hence, the acceleration would depend
on the charge and mass of the particle. So, the q/m ratio is important. This
ratio is also called the

**specific charge**.
The particle that completes the
acceleration with lowest speed is that one having the lowest value of this
ratio.

The mass is the nucleon number while
the charge is taken to be equal to the proton number (since we are dealing with
nuclei, and not the atoms).

For

^{1}_{1}H, ratio = 1/1 = 1
For

^{4}_{2}He, ratio = 2/4 = 0.5
For

^{7}_{3}Li, ratio = 3/7 ≈ 0.429
For

^{9}_{4}Be, ratio = 4/9 ≈ 0.44
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.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

41/M/J/10 Q.6(a),Q.7(a)

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

41/M/J/11 Q.8(a)

For 04/M/J/09 Q.11(b)(iii), see solution 973 at

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