Physics 9702 Doubts | Help Page 157
Question 777: [Dynamics
> Equilibrium]
Diagram shows a crane supporting a
load L.
A mass provides a balancing load W.
Position of the load is such that the system is perfectly balanced with Wx =
Ly. The ground provides a reaction force R. Distance x does not change.
If the load is moved further out so
that distance y increases and the crane does not topple, which statement is
correct?
Reference: Past Exam Paper – June 2012 Paper 11 Q15 & Paper 13 Q13
Solution 777:
Answer: D.
For equilibrium, the net torque on
the system should be zero. That is, the clockwise moment should be equal to the
anticlockwise moment.
The pivot in this system is in the
middle of the vertical bar, at the top, where it crosses with the horizontal
bar.
Wx = Ly
It is said that the distance x does
not change.
Distance y has been increased. This
increases the clockwise moment caused by load L. So we need a force that would
cause an extra anticlockwise moment in order to achieve equilibrium again.
When the reaction force R moves to
the right, it causes an anticlockwise moment about the pivot, thus achieving
equilibrium again.
If the reaction force R moves to the
lift, this could further increase the clockwise moment about the pivot. [C is incorrect]
Question 778: [Dynamics]
Brick weighing 20 N rests on an
inclined plane. Weight of the brick has a component of 10 N parallel with the
plane. The brick also experiences a frictional force of 4 N.
What is the acceleration of brick
down the plane? Assume that the acceleration of free fall g is equal to 10 m s–2.
A 0.3 m s–2 B 0.8 m s–2 C 3.0 m s–2 D 8.0 m s–2
Reference: Past Exam Paper – June 2010 Paper 11 Q11 & Paper 12 Q14
& Paper 13 Q8
Solution 778:
Answer: C.
Weight = mg
Mass of brick, m = W / g = 20N /
10ms-2 = 2kg
Resultant force F on brick = ma
Resultant force on brick = 10 – 4 N =
6N
Acceleration a = F / m = 6 / 2 = 3ms-2
Question 779: [Current
of Electricity > Resistance]
For June 2004 Paper 1 Q31:
Two wires made of same material and
of the same length are connected in parallel to the same voltage supply. Wire P
has a diameter of 2 mm. Wire Q has a diameter of 1 mm.
For November 2008 Paper 1 Q31 &
June 2014 Paper 13 Q33:
Two wires P and Q made of same
material and of the same length are connected in parallel to the same voltage
supply. Wire P has diameter 2 mm and wire Q has diameter 1 mm.
What is the ratio current in P /
current in Q?
A ¼ B ½ C
2 D 4
Reference: Past Exam Paper – June 2004 Paper 1 Q31 & November 2008
Paper 1 Q31 & June 2014 Paper 13 Q33
Solution 779:
Answer: D.
Ohm’s law: V = IR
Current I = V / R
Resistance R of a wire = ρL / A
For the same material of the same
length, resistivity ρ and length L are the same.
Cross-sectional area A = π(d/2)2
= πd2/4
Resistance R is inversely
proportional to area A (and thus to d2).
But current I is inversely
proportional to resistance R, so current I is directly proportional to d2.
Diameter of wire P = 2mm and
diameter of wire Q is 1mm.
Ratio of current in P to current in
Q = 22 / 12 = 4 / 1
Question 780: [Current
of Electricity > Charge]
Potential difference across a
component in a circuit is 2.0 V.
How many electrons must flow through
this component in order for it to be supplied with 4.8 J of energy?
A 2.6 × 1018 B 1.5 × 1019 C 3.0 × 1019 D 6.0 × 1019
Reference: Past Exam Paper – June 2014 Paper 11 Q35
Solution 780:
Answer: B.
Energy, E = Charge × p.d.
Energy, E = Q × V
Charge Q = E / V = 4.8 / 2.0 = 2.4C
Total charge Q = ne
where e is the charge of an electron
and n is the number of electrons.
Number of electrons, n = Q / e = 2.4
/ (1.6×10-19)
= 1.5×1019
Question 781: [Current
of Electricity]
A student found two unmarked
resistors. To determine resistance of the resistors, the circuit below was set
up. The resistors were connected in turn between P and Q, noting the current
readings. Voltage readings were noted without the resistors and with each
resistor in turn.
Results were entered into a spreadsheet
as shown.
The student forgot to enter column
headings.
Which order of the headings would be
correct?
Reference: Past Exam Paper – November 2014 Paper 13 Q34
Solution 781:
Answer: D.
The e.m.f. of the supply in the
circuit is constant and independent of the other components in the circuit.
Without the resistors in the circuit, the voltmeter will read the e.m.f. of the
supply. Since the e.m.f is constant, the first column (which is the only column
with constant values) would be values of the e.m.f. [B
and C are incorrect]
Ohm’s law: V = IR
The supply may have some internal
resistance r.
So, from Kirchhoff’s law,
p.d. across resistor + voltage drop
due to internal resistance = e.m.f. of supply
p.d. across resistor = IR
Voltage drop due to internal
resistance = Ir
IR + Ir = e.m.f.
I (R + r) = e.m.f.
To distinguish between the current I
and resistance R columns, one method is to consider that the voltage drop due
to internal resistance (= Ir) must be larger when the current is greater, so
the first row of data is for a greater current than the second row.
The second column, which contain
values close to 1.5, must be the p.d. V across the resistor. The values in this
column increases (from 1.3 to 1.4) when the values of the last column increases
(from 46 to 100). So, these must be the V and I columns respectively.
This can be verified as follows. From
Ohm’s law, resistance R = V / I. The headings indicates that current is
measured in mA – so we should multiply the value of the current column by a
factor of 10-3. So, the ratio of the V column (2nd
column) to that of the I column (3rd column) should give the value of
the resistance (4th column).
Additionally, it can be seen that
when the resistance R is greater, the current is smaller
Note that usually, when filling
column, as in this example, the first column would be the values are remain
constant (if any – here it is the e.m.f.) and the last column is a quantity
that needs to be calculated (here the resistance). The middle columns usually
contain the measured quantities (here, current and p.d.).
