Physics 9702 Doubts | Help Page 98
Question 504: [Gravitation]
(a) Define gravitational field strength
(b) Isolated star has radius R. Mass of the star may be considered to
be point mass at centre of the star.
Gravitational field strength at
surface of the star is gs.
On Fig, sketch a graph to show
variation of gravitational field strength of star with distance from its
centre. You should consider distances in range R to 4R
(c) Earth and Moon may be considered to be spheres that are isolated in
space with their masses concentrated at their centres.
Masses of the Earth and the Moon are
6.00x1024kg and 7.40x1022kg respectively.
Radius of the Earth is RE
and separation of the centres of Earth and Moon is 60RE, as
illustrated in Fig.
(i) Explain why there is a point
between Earth and Moon at which the gravitational field strength is zero
(ii) Determine the distance, in
terms of RE, from the centre of Earth at which gravitational field
strength is zero
(iii) On axes of Fig, sketch graph
to show the variation of gravitational field strength with position between
surface of Earth and surface of Moon
Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q1
Solution 504:
(a) Gravitational field strength is defined as gravitational force per
unit mass.
(b)
Inverse-square
relationship. So, points are (R, 1.0gs), (2R, gs/4), (3R,
gs/9) and (4R, gs/16)
Graph needs to have correct
curvature
from (R, 1.0gs) & at
least 1 other correct point
(c)
(i) The gravitational fields of the
Earth and the Moon are in opposite directions.
EITHER The resultant field is found
by the subtraction of the field strength. OR any other sensible comment.
So, there is a point where the
gravitational field strength is zero.
(ii)
Let the point be at a distance x
from Earth and let D (=60RE) be the distance between Earth and Moon,
{Distance of the point
from the Moon is (D – x)}
GME / x2 = GMM
/ (D – x)2
(6.0x1024) / (7.4x1022)
= x2 / (60RE – x)2
{(6.0x1024) /
(7.4x1022) = 81.08 ≈ 81
81 = x2 / (60RE
– x)2
Take square root on both
sides of the equation,
9 = x / (60RE –
x)
9 (60RE – x) =
x
10x = 540RE
giving x = 54RE}
Distance x from the Earth = 54RE
{Alternatively, let y be
the distance of the point from the moon. The distance of the point of the
centre of Earth = D – y = 60RE – y
GME / (D – y)2
= GMM / y2
ME / MM
= (D – y)2 / y2
ME / MM
= (6.0x1024) / (7.4x1022) = 81.08 ≈ 81
81 = (D – y)2 /
y2
Take square root on both
sides of the equation
9 = (60RE – y)
/ y
9y = (60RE – y)
10y = 60RE giving y = 6RE
(iii) On axes of Fig, sketch graph
to show the variation of gravitational field strength with position between
surface of Earth and surface of Moon
For the graph:
g = 0 at least 2/3 the distance to
Moon
gE and gM are
in opposite directions
A correct curvature (by eye) and gE
> gM at the surface
Question 505: [Current of Electricity > Capacitors]
(a) State 2 functions of capacitors in electrical circuits
(b) 3 capacitors, each marked ‘30μF, 6V max’, arranged as shown in Fig.
Determine, for arrangement shown in
Fig,
(i) total capacitance
(ii) Maximum potential difference
that can safely be applied between points A and B
(c) Capacitor of capacitance 4700μF
is charged to potential difference of 18V. It is then partially discharged
through a resistor. Potential difference reduced to 12V. Calculate energy
dissipated in resistor the during discharge
Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q5
Solution 505:
(a) Choose any 2:
For ‘storage of charge’ / storage of
charge
For blocking of direct current
For producing of electrical oscillations
For smoothing
(b)
(i)
Capacitance of parallel combination (= 30 + 30) = 60μF
Total capacitance (= [1/60 + 1/30]-1) = 20μF
(ii)
The p.d. across the parallel
combination = ½ x p.d. across the single capacitor
Capacitance C = Q / V
where Q is the charge. Current I = Q/t. In a parallel combination, the current
splits at the junction. Since both capacitors in parallel are the same, the current
splits equally. So, the charge is halved.
For a capacitor to still
have the capacitance C, the p.d. across it should also be halved.
Since the maximum p.d. is
6V across a single capacitor, it should be 3V across the parallel combination.
So, the maximum is (6 + 3 =) 9V
(c)
EITHER Energy = ½ CV2 OR Energy = ½ QV and Q = CV
Energy = ½ (4700x10-6)
(182 – 122) = 0.42J
Question 506: [Matter
> Elasticity]
A number of identical springs, each
having same spring constant, are joined in four arrangements. Different load is
applied to each arrangement.
Which arrangement has the largest
extension?
Reference: Past Exam Paper – June 2013 Paper 11 Q23
Solution 506:
Answer: A.
Hooke’s law: F = ke
Extension, e = F / k
Since all the arrangements consist
of more than 1 spring, we need to find the effective spring constant in each
arrangement.
For springs in series and in
parallel, the following formulae for the effective spring constant apply:
In parallel: effective spring
constant, keff = k1 + k2 + ….
In series: effective spring
constant, 1/keff = 1/k1 + 1/k2 + ….
Therefore, when springs are attached
in parallel, the effective spring constant is greater and hence, the extension
e (= F / keff) is smaller. Also, for springs attached in series, keff
is smaller and hence the extension is larger.
Let the spring constant of each
spring be k.
For arrangement A,
1/keff = 1/k + 1/k = 2/k
Extension, e = F / keff =
2 (2/k) = 4 / k
For arrangement B,
1/keff = 1/k + 1/k + 1/k
= 3/k
Extension, e = F / keff =
1 (3/k) = 3 / k
For arrangement C,
keff = k + k = 2k
Extension, e = F / keff =
6 / 2k = 3 / k
For arrangement D,
keff = k + k + k = 3k
Extension, e = F / keff =
8 / 3k
Therefore, the extension is larger
for arrangement A.
Question 507: [Units]
Drag coefficient Cd is a
number with no units. It is used to compare drag on different cars at different
speeds. It is given by the equation
Cd = 2F / ρvnA
where F is drag force on the car, ρ
is density of the air, A is cross-sectional area of the car and v is the speed
of the car.
What is the value of n?
A 1 B
2 C
3 D 4
Reference: Past Exam Paper – November 2013 Paper 11
& 12 Q3
Solution 507:
Answer: B.
Consider the units of the different quantities
in terms of their base units.
Units of Force, F = kg ms-2
Units of density, ρ = kgm-3
Units of cross-sectional area, A = m2
Units of speed, v = ms-1
The drag coefficient has no units.
For the equation to be homogeneous and hence correct, the overall unit on the
right-hand side of the equation should be the same as on the left-hand side –
that it the LHS should have no overall units. Therefore, the units of the
numerator should be the same as the unit of the denominator.
Numerator: Units of Force, F = kg ms-2
Denominator: Units of ρvnA
= (kgm-3) (ms-1)n (m2)
For the units of the denominator to
be the same as that of the numerator, n should be 2.
In 42/O/N/10 Can you show the sketch of the graph for Q.1(c)(iii)
ReplyDeleteAdded
DeleteYou are a life saver Thanks.
ReplyDeleteThere are so many answers I did not understand from markingscheme, and you explain it so well I understood everything. Will definitely aim for A* now in physics
could you please explain Q.507 in much better way especially the last part. I couldn't understand. Please. I really need to know.
ReplyDeleteI tried again. I succeeded. Thank you.
ReplyDeleteCan you solve 9702/11/M/J/16 question number 20
ReplyDeletego to
Deletehttps://physics-ref.blogspot.com/2018/09/a-number-of-identical-springs-are.html
can the gravitational field of earth be in positive y axis and the gfe of moon in negative?
ReplyDeleteyes, if we consider the left direction to be positive.
DeleteIt is correct as long as they are of opposite sign.
F=ma
ReplyDeleteso units of force is kg.m/s^2
you mistakenly didnt mention second squared with unit of force
without incorporating this,
n will not be proved equal to 2
thanks, it has been corrected
Delete