Saturday, June 6, 2015

Physics 9702 Doubts | Help Page 164

  • Physics 9702 Doubts | Help Page 164



Question 818: [Work, Energy and Power]
A body travelling with speed of 20 m s–1 has kinetic energy Ek.
If speed of the body is increased to 80 m s–1, what is its new kinetic energy?
A 4Ek                          B 8Ek                           C 12Ek                                     D 16Ek

Reference: Past Exam Paper – November 2013 Paper 13 Q18



Solution 818:
Answer: D.
Initial Kinetic energy = ½ mv2 = ½ m(20)2 = Ek

Final Kinetic energy = ½ mv2 = ½ m(80)2

We need to write the final kinetic energy in terms of Ek which is itself equal to ½ m(20)2.
Final Kinetic energy = ½ m (4×20)2 = 16 × [½ m(20)2] = 16 Ek











Question 819: [Dynamics > Newton’s laws of motion]
Water is pumped through hose-pipe at a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with speed of 20 m s–1.
Which force is required from a person holding the hose-pipe to prevent it moving backwards?
A 30 N                        B 270 N                      C 1800 N                    D 10 800 N

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q8



Solution 819:
Answer: A.
From Newton’s 2nd law,
Force = rate of change of momentum = Δp / t

The emerging water has some momentum and hence exerts a force on the person. From Newton’s 3rd law, the person should exert a force similar in magnitude but in the opposite direction to prevent the hose-pipe from moving backwards.

Force = rate of change of momentum = Δp / t

In this case, the water is emerging a constant speed, but the water is being pumped at a specific rate.
Δp / t = (Δm)v / t = (Δm / t) v
Δm / t = rate of water being pumped = 90kg / min
1 min = 60s corresponds to 90kg of water
1s corresponds to 90 / 60 = 1.5kg

Rate of water being pumped = Δm / t = 1.5kg s-1

Force = Δp / t = (Δm / t) v = 1.5 × 20 = 30N











Question 820: [Radioactivity]
The first artificial radioactive substance was made by bombarding aluminium, 2713Al, with α-particles. This produced an unstable isotope of phosphorus, 3015P.
What was the by-product of this reaction?
A an α-particle
B a β-particle
C a neutron
D a proton

Reference: Past Exam Paper – June 2014 Paper 12 Q40



Solution 820:
Answer: C.
In a reaction, the mass number and proton number should be conserved.
2713Al               +          42α        - - - >   3015P                +          ZAX

27 + 4 = 30 + Z           giving Z = 1
13 + 2 = 15 + A          giving A = 0

10X is a neutron (10 n).











Question 821: [Electric field]
In diagram, the shaded area represents a uniform electric field directed away from the observer (at right-angles into the plane of the paper).

A horizontal beam of electrons enters the field, travelling from left to right.
In which direction is this beam deflected by the field?
A upwards (in the plane of the paper)
B downwards (in the plane of the paper)
C away from the observer
D towards the observer

Reference: Past Exam Paper – June 2010 Paper 11 Q29 & Paper 12 Q27 & Paper 13 Q26



Solution 821:
Answer: D.
Electric field is drawn from positive to negative.

Since the electric field is said to be into the plane of the paper, the negative is in the paper while positive is out of the paper.

An electron, which is negatively charged, is attracted to the positive. So the deflection of the electron is towards observer (the field is directed away from the observer, so the position of the positive charge is at the position of the observer).










Question 822: [Current of Electricity]
What describes the electric potential difference between two points in wire that carries a current?
A the force required to move a unit positive charge between the points
B the ratio of the energy dissipated between the points to the current
C the ratio of the power dissipated between the points to the current
D the ratio of the power dissipated between the points to the charge moved

Reference: Past Exam Paper – November 2004 Paper 1 Q31 & June 2011 Paper 12 Q32



Solution 822:
Answer: C.
Power P dissipated = VI
p.d. V = P / I
The electric potential difference between two points in wire that carries a current is described as the ratio of the power dissipated between the points to the current. [C is correct]

Current and charge are different quantities. [D is incorrect]
Power dissipated and energy dissipated are different quantities.  [B is incorrect]









Question 823: [Pressure]
The Mariana Trench in the Pacific Ocean has a depth of about 10 km.
Assuming that sea water is incompressible and has density of about 1020 kg m–3, what would be the approximate pressure at that depth?
A 105 Pa                      B 106 Pa                      C 107 Pa                      D 108 Pa

Reference: Past Exam Paper – June 2014 Paper 13 Q22



Solution 823:
Answer: D.
Pressure = hρg = 10000 ×1020 × 9.81 = 1×108 Pa










Question 824: [Matter > Deformation > Young modulus]
A lift is supported by two steel cables, each of length 10 m and diameter 0.5 cm.
The lift drops 1 mm when a man of mass 80 kg steps into the lift.
What is the best estimate of value of the Young modulus of the steel?
A 2 × 1010 N m–2
B 4 × 1010 N m–2       
C 2 × 1011 N m–2
D 4 × 1011 N m–2

Reference: Past Exam Paper – November 2012 Paper 13 Q25



Solution 824:
Answer: C.
Young modulus = Stress / Strain
Stress = Force F / Area A
Strain = extension e / Original length L

Young modulus = (F/A) / (e/L) = FL / Ae

The lift is supported by 2 wires which are in parallel to each other. Both wires are made of steel, so they have the same spring constant, k.
For wires in parallel,
Effective spring constant, keff = k + k = 2k

Hooke’s law: F = keff e
Weight of the person is approximately = mg = 80 × 10 = 800N
Effective spring constant, keff = F / e = 800 / 0.001 = 8.0×105 Nm-1

Effective spring constant, keff = 2k = 8.0×105 Nm-1
Spring constant of 1 steel wire = 8.0×105 / 2 = 4.0×105 Nm-1


For only 1 wire,
Young modulus = (F/A) / (e/L) = FL / Ae = (F/e) (L/A)
Spring constant, k = F / e = 4.0×105 Nm-1
Young modulus = (F/e) (L/A) = kL / A = (4.0×105) × 10 / (π(0.005)2/4) = 2 × 1011 N m–2


OR Simply,
Young modulus = (F/A) / (e/L) = FL / Ae

Since there are 2 wires, the total cross-sectional area = 2[πd2 / 4] = πd2 / 2
Young modulus = 800 × 10 / [(π(0.005)2 / 2) × 0.001] = 2 × 1011 N m–2



4 comments:

  1. In solution 821 how can I know the uniform electric field is into the plane of the paper?

    ReplyDelete
    Replies
    1. I suppose it would be mentioned in the question and indicated by a X.

      Away from paper is indicated by a circle containing a dot

      Delete
  2. Solution 820: answer is C
    Where Z=1 and A=0, the particle is not a proton but it's a neutron.

    ReplyDelete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 164