# Physics 9702 Doubts | Help Page 113

__Question 573: [Forces > Equilibrium]__Diagram shows a child’s balancing game.

Wooden rod is uniform and all the rings are of equal mass. Two rings are hung on peg 13 and one on peg 1.

On which hook must a fourth ring be hung in order to balance the rod?

A 2 B 3 C 5 D 6

**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q15 & Paper 13 Q13*

__Solution 573:__**Answer: C.**

Let the weight of 1 ring be W.

For the rod to balance
(equilibrium),

Clockwise moment = Anticlockwise
moment

The 2 rings on peg 13 causes a
clockwise moment.

Clockwise moment = 2[(13 – 8)xW] =
10W

Anticlockwise moment = (8 – 1)W + (8
– y)W

where y is the position of hook
where fourth ring must be hung.

For equilibrium,

(8 – 1)W + (8 – y)W = 10W

7W + (8 – y)W = 10W

(8 – y)W = 3W

8 – y = 3 giving y = 8 – 3 = 5

__Question 574: [Electric field]__**(a)**An Î±-particle and a proton are at rest a distance 20 Î¼m apart in a vacuum, as illustrated in Fig.1.

(i)
State Coulomb’s law.

(ii)
The Î±-particle and proton may be considered to be point charges.

Calculate
electric force between the Î±-particle and the proton.

**(b)**

(i)
Define electric field strength.

(ii)
A point P is distance x from Î±-particle along the line joining the Î±-particle
to the proton (see Fig.1). Variation with distance x of electric field strength
E

_{Î±}due to the Î±-particle alone is shown in Fig.2.
Variation
with distance x of the electric field strength E

_{P}due to the proton alone is also shown in Fig.2.
1.
Explain why the two separate electric fields have opposite signs.

2.
On Fig.2, sketch variation with x of the combined electric field due to the Î±-particle
and the proton for values of x from 4 Î¼m to 16 Î¼m.

**Reference:**

*Past Exam Paper – November 2013 Paper 41 & 42 Q4*

__Solution 574:__**(a)**

(i)
Coulomb’s law states that the (electric) force between 2

__point charges__is proportional to the product of the 2 charges and inversely proportional to the square of their separation
(ii)

Distance
between Î±-particle and proton, r = 20Î¼m = 20 x 10

^{-6}m
Charge
of Î±-particle (contains 2 protons) = 2 x (1.6x10

^{-19}) C
Charge
of a proton, q = 1.6x10

^{-19}C
Ïµ

_{0}= 8.85 x 10^{-12}Fm^{-1}
Electric
force, F = Qq/4Ï€Ïµ

_{0}r^{2}
F
= 2 x (1.6x10

^{-19})^{2}/{4Ï€ x 8.85 x 10^{-12}x (20 x 10^{-6})^{2}} = 1.15 x 10^{-18}N
{Q is the charge of the Î±-particle which contains 2 protons and thus
has charged +2e = 2 x (1.6x10

^{-19}) C. q is the charge of the proton which is +e = (1.6x10^{-19}) C. The product Qq is 2 x (1.6x10^{-19})^{2}}**(b)**

(i) Electric field strength is defined as the
force per unit charge acting on EITHER a stationary charge OR a positive charge.

(ii)

1.

Choose
any 2:

Electric
field is a vector quantity

Electric
fields are in opposite directions

The
2 charges repel one another

2.
In the graph, the line should always be between the given lines, crossing the
x-axis between 11.0Î¼m and 12.3Î¼m. The curve should have a reasonable shape.

{The value of electric field strength at any value of x is the sum
of the corresponding values of E

_{Î±}and E_{p}at that point – this value would be less than E_{Î±}and greater than E_{p}since E_{p}is negative.}
[A typical drawing is shown. This can be done in more properly on
paper.]

__Question 575: [Dynamics > Momentum]__Diagram shows a cannon ball fired from a cannon.

Mass of the cannon is 1000 kg and mass of the cannon ball is 10 kg.

Recoil velocity of the cannon is 5 m s

^{–1}horizontally.

What is horizontal velocity of the cannon ball?

A 200 m s

^{–1}B 500 m s

^{–1}C 2000 m s

^{–1}D 5000 m s

^{–1}

**Reference:**

*Past Exam Paper – November 2006 Paper 1 Q12*

__Solution 575:__**Answer: B.**

Before firing the cannon ball, the
cannon + cannon ball are stationary. So, there initial momentum is zero.

As the cannon fires the ball, it
moves backwards. This is the recoil velocity which is 5ms

^{-1}horizontally.From the conservation of momentum, the magnitude of the momentum of the cannon ball (moving forwards) is equal to the magnitude of the momentum of the cannon (moving backwards).

Let the velocity of the cannon ball be v.

10v = 1000 (5)

Speed v = 5000 / 10 = 500 ms

^{-1}

__Question 576:__

__[Electromagnetism > Hall Probe]__**(a)**State relation between magnetic flux density B and magnetic flux Î¦, explaining any other symbols you use.

**(b)**Large horseshoe magnet has uniform magnetic field between its poles. Magnetic field is zero outside the space between poles.

A small Hall probe moved at constant
speed along line XY that is midway between, and parallel to, the faces of poles
of magnet, as shown in Fig.1.

An e.m.f. is produced by Hall probe
when it is in the magnetic field.

Angle between plane of the probe and
direction of magnetic field is not varied.

On axes of Fig.2, sketch graph to
show variation with time t of the e.m.f. V

_{H}produced by Hall probe**(c)**

(i) State Faraday’s law of
electromagnetic induction.

(ii) Hall probe in (b) is replaced
by small flat coil of wire. Coil is moved at constant speed along line XY.
Plane of coil is parallel to the faces of the poles of magnet.

On axes of Fig.3, sketch graph to
show variation with time t of e.m.f. E induced in the coil

**Reference:**

*Past Exam Paper – November 2012 Paper 43 Q5*

__Solution 576:__**(a)**

EITHER

Î¦ = BAsinÎ¸ where A is the area
(through which flux passes) and Î¸ is the angle between magnetic flux density B
and (the plane of) A

OR

Î¦ = BA where A is the area normal to
the magnetic flux density B

**(b)**For the graph, V

_{H}is constant and non-zero between the poles and zero outside. There is a sharp increase / decrease at the ends of the magnet

{V

_{H}is constant between the poles because the area through which flux passes is constant as the angle between the plane of the Hall probe and the direction of magnetic field is not varied. There is a sharp increase is due to the sudden cutting of the flux by the area of the plane of the Hall probe.}**(c)**

(i) Faraday’s law of electromagnetic
induction states that the (induced) e.m.f. is proportional to the rate of
change of (magnetic) flux (linkage)

(ii) For the graph, there is a short
pulse on entering and on leaving the region between the poles. The pulses have
approximately the same shape but opposite polarities. The e.m.f. is zero
between the poles and outside.

{As the coil enters the
field, the sharp increase in flux produces a pulse of e.m.f., which acts in
such a way as to oppose the change of flux producing it (Lenz’s law). The same
applies when the coil is leaving the magnetic field.

Since coil is moved a
constant speed, once the whole coil has entered the field completely, the rate of
change of flux is then zero (there is a flux, but it is not changing – so, the
rate of change of flux is zero), causing the e.m.f. to be zero (from Faraday’s
law).

The pulses have opposite
polarities because as the coil enters the field, the change is an increase in
flux and when it leaves the field, the change is a decrease in flux.}

__Question 577: [Forces > Equilibrium]__
A cylinder of weight W is placed on
smooth slope. Contact force of the slope on the cylinder is R. A thread is
attached to surface of the cylinder. The other end of the thread is fixed.

Which diagram shows the cylinder in
equilibrium?

**Reference:**

*Past Exam Paper – November 2014 Paper 13 Q13*

__Solution 577:__**Answer: C.**

For equilibrium, the resultant force
and the resultant torque on the system should be zero.

The resultant force can be
identified by considering the components of the forces involved. Since the
forces are not drawn to scale, we can assume that the forces act in such a way
that the resultant is zero in this case.

Additionally, for equilibrium, the
resultant torque should be zero.

Torque or moment = Force x
perpendicular distance of line of action of the force to the pivot

Without doing any calculations, the
easiest way to know if the resultant torque is zero is to make sure that all
the force act on the same point – this causes the ‘perpendicular distance of
line of action of the force to the pivot’ to be zero. Thus, the resultant
torque would be zero. [Only C corresponds to this]

loveeee youuuu for helping so many students.

ReplyDeletegreat work,keep it up.

ReplyDeleteawesome work

ReplyDeleteSolution 576:

ReplyDelete(b)

How is there an emf in between the poles? Can't it be zero because the flux is constant? Please explain.

the hall probe is moving, so it is always cutting the field.

DeleteThanks it was very helpful.

ReplyDelete