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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 6.00x10²⁴kg and 7.40x10²²kg 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 / x² = GM_M / (D – x)²

(6.0x10²⁴) / (7.4x10²²) = x² / (60R_E – x)²

{(6.0x10²⁴) / (7.4x10²²) = 81.08 ≈ 81

81 = x² / (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 / y²    

M_E / M_M = (D – y)² / y²

M_E / M_M = (6.0x10²⁴) / (7.4x10²²) = 81.08 ≈ 81

81 = (D – y)² / y²

Take square root on both sides of the equation

9 = (60R_E – y) / y

9y = (60R_E – y)

10y = 60R_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 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, k_eff = k₁ + k₂ + ….

In series: effective spring constant, 1/k_eff = 1/k₁ + 1/k₂ + ….

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 / 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.