# 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 g

_{s}.
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
6.00x10

^{24}kg and 7.40x10^{22}kg respectively.
Radius of the Earth is R

_{E}and separation of the centres of Earth and Moon is 60R_{E}, 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 R

_{E}, 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, 1.0g

_{s}), (2R, g_{s}/4), (3R, g_{s}/9) and (4R, g_{s}/16)
Graph needs to have correct
curvature

from (R, 1.0g

_{s}) & 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 (=60R

_{E}) be the distance between Earth and Moon,
{Distance of the point
from the Moon is (D – x)}

GM

_{E}/ x^{2}= GM_{M}/ (D – x)^{2}
(6.0x10

^{24}) / (7.4x10^{22}) = x^{2}/ (60R_{E}– x)^{2}
{(6.0x10

^{24}) / (7.4x10^{22}) = 81.08 ≈ 81
81 = x

^{2}/ (60R_{E}– x)^{2}
Take square root on both
sides of the equation,

9 = x / (60R

_{E}– x)
9 (60R

_{E}– x) = x
10x = 540R

_{E}giving x = 54R_{E}}
Distance x from the Earth = 54R

_{E}
{Alternatively, let y be
the distance of the point from the moon. The distance of the point of the
centre of Earth = D – y = 60R

_{E}– y
GM

_{E}/ (D – y)^{2}= GM_{M}/ y^{2}
M

_{E}/ M_{M}= (D – y)^{2}/ y^{2}
M

_{E}/ M_{M}= (6.0x10^{24}) / (7.4x10^{22}) = 81.08 ≈ 81
81 = (D – y)

^{2}/ y^{2}
Take square root on both
sides of the equation

9 = (60R

_{E}– y) / y
9y = (60R

10y = 60R_{E}– y)_{E}giving y = 6R

_{E}

(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

g

_{E}and g_{M}are in opposite directions
A correct curvature (by eye) and g

_{E}> g_{M}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 CV

^{2}OR Energy = ½ QV__and__Q = CV
Energy = ½ (4700x10

^{-6}) (18^{2}– 12^{2}) = 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, k

_{eff}= k_{1}+ k_{2}+ ….
In series: effective spring
constant, 1/k

_{eff}= 1/k_{1}+ 1/k_{2}+ ….
Therefore, when springs are attached
in parallel, the effective spring constant is greater and hence, the extension
e (= F / k

_{eff}) is smaller. Also, for springs attached in series, k_{eff}is smaller and hence the extension is larger.
Let the spring constant of each
spring be k.

For arrangement A,

1/k

_{eff}= 1/k + 1/k = 2/k
Extension, e = F / k

_{eff}= 2 (2/k) = 4 / k
For arrangement B,

1/k

_{eff}= 1/k + 1/k + 1/k = 3/k
Extension, e = F / k

_{eff}= 1 (3/k) = 3 / k
For arrangement C,

k

_{eff}= k + k = 2k
Extension, e = F / k

_{eff}= 6 / 2k = 3 / k
For arrangement D,

k

_{eff}= k + k + k = 3k
Extension, e = F / k

_{eff}= 8 / 3k
Therefore, the extension is larger
for arrangement A.

__Question 507: [Units]__
Drag coefficient C

_{d}is a number with no units. It is used to compare drag on different cars at different speeds. It is given by the equation
C

_{d}= 2F / ρv^{n}A
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 = kgm

^{-2}
Units of density, ρ = kgm

^{-3}
Units of cross-sectional area, A = m

^{2}
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 = kgm

^{-2}
Denominator: Units of ρv

^{n}A = (kgm^{-3}) (ms^{-1})^{n}(m^{2})
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)

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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.

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