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Question 896: [Options > Physics of Fluids]

A small sphere of radius r and density ρ_s is falling with speed v in fluid of density ρ_f and viscosity η.

The flow of fluid around the sphere is streamline.

(a) Write down expressions for

(i) upthrust on the sphere,

(ii) resultant downward force on the sphere.

(b) Drag force F on a sphere in streamline flow is given by the expression

F = 6πrηv.

Show that the terminal speed v_t of the sphere is given by

v_t = kr²,

where k is a constant.

(c) A student determines the viscosity of oil by measuring the terminal speed of a steel sphere as it falls through the oil contained in a wide vertical tube. Suggest

(i) how it can be checked that sphere is falling at terminal speed when measurements are taken,

(ii) why tube should have a diameter at least ten times that of the steel sphere.

Reference: Past Exam Paper – November 2003 Paper 6 Q6

Solution 896: [Options > Physics of Fluids]

(a)

(i) Upthrust = 4/3 × (πr³ρ_f g)

(ii)

EITHER Resultant downward force = 4/3 × [πr³ (ρ_s – ρ_f) g]

OR Resultant downward force = 4/3 × [πr³ (ρ_s – ρ_f) g] – viscous force

(b)

6πrηv_t = 4/3 × [πr³ (ρ_s – ρ_f) g]

{Simplifying the equation, we obtain ηv_t = 2/9 × [r² (ρ_s – ρ_f) g]

v_t = 2/(9η) × [r² (ρ_s – ρ_f) g]

So, v_t depends on r². The other terms on the right-hand side of the equation are all constant. So, they can be represented by the constant k.}

Hence, v_t = kr²

constant k discussed { k = 2/(9η) × [(ρ_s – ρ_f) g] }

(c)

(i) Examples:

Find the speed near the ‘top’ and near the ‘bottom’ of the tube

By using equally spaced markers (or other detail)

(ii) The oil flowing past the wall of the tube would cause extra drag.

Question 897: [Measurement > Vectors]

(a) Force is a vector quantity. State three other vector quantities.

(b) Three coplanar forces X, Y and Z act on an object, as shown in Fig.1.

Force Z is vertical and X is horizontal. The force Y is at an angle θ to the horizontal. The force Z is kept constant at 70 N.

In an experiment, magnitude of force X is varied. The magnitude and direction of force Y are adjusted so that the object remains in equilibrium.

Fig.2 shows the variation of the magnitude of force Y with the magnitude of force X.

(i) Use Fig.2 to estimate the magnitude of Y for X = 0.

(ii) State and explain the value of θ for X = 0.

(iii) Magnitude of X is increased to 160 N. Use resolution of forces to calculate the value of

1. angle θ,

2. the magnitude of force Y.

(c) The angle θ decreases as X increases. Explain why the object cannot be in equilibrium for θ = 0.

Reference: Past Exam Paper – November 2014 Paper 23 Q3

Solution 897:

(a) Displacement / velocity / acceleration / momentum / etc.

(b)

(i) Y = 70 N [allow 71 N as +½ small square on graph]

(ii) Angle θ = 90°

{X = 0 means that there is no X vector. So, for equilibrium, vector Y should not have any horizontal component.}

(For equilibrium) the direction of Y must be opposite to Z

OR using Y sinθ = Z, hence sin θ = 70/70 = 1, θ = 90°

(iii)

1.

{Horizontal component of Y = Y cosθ. This should be equal to X, for equilibrium. Considering equilibrium vertically, Y sinθ = 70.}

Y cosθ = 160 and Y sinθ = 70

{Diving the 2^nd equation by the 1^st one gives (sin / cos = tan)}

tan θ = 70 / 160           hence angle θ = 23.6° (24°)

2.

EITHER

{Using either of these 2 equations: Y cosθ = 160 and Y sinθ = 70, we can calculate Y since θ is now known.}

Y = 160 / cos23.6° OR 70 / sin23.6°

Y = 174.6 or 175 or 170 N

OR

{Using Pythagoras’ theorem, where Y is the hypotenuse of its perpendicular components which are equal to the vectors X and Z respectively.}

160² + 70² = Y²

Y = 174.6 or 175 or 170 N

(c) (Equilibrium is not possible as) there is no vertical component from Y to balance Z.

Question 898: [Measurement > Vectors]

A load of 10 kg is hung using two strings 9 m and 12 m long tied to two points at the same level 15 m apart. Find the tension in each string?

Reference: ???

Solution 898:

Taking acceleration of free fall, g = 10ms^-2

Weight of the load = mg = 10 x 10 = 100N

The angles can be found using the cosine rule

12² = 9² + 15² - 2(9)(15)cosθ₁

 θ₁ = cos^-1(162 / 270) = 53.13^o

Similarly, 9² = 15² + 12² – 2(15)(12)cosθ₂  giving θ₂ = 36.87^o

(It can also be determined that the angle at the load is 53.13^o + 36.87^o = 90^o.)

For equilibrium,

The sum of the vertical components of the tensions should equal the weight

T₁ sin(53.13) + T₂ sin(36.87) = 100

The horizontal components of the tensions should equate since the weight does not have a horizontal component.

T₁ sin(53.13) = T₂ sin(36.87)

Solving the 2 equations simultaneously gives the tension in the 9m string, T₁ = 80N and the tension in the 12m string, T₂ = 60N

Question 899: [Options > Astrophysics and Cosmology]

The Universe may be described as ‘open’, ‘flat’ or ‘closed’.

(a) State clearly the factor on which the ultimate fate of the Universe depends.

(b) Fig.1 illustrates the variation with time of the extent of a ‘flat’ Universe.

(i) On Fig.1, draw a line to show the variation with time of the extent of a closed Universe.

(ii) Suggest three reasons why the ultimate fate of the Universe is not known.

Reference: Past Exam Paper – June 2006 Paper 6 Q2

Solution 899:

(a) The (mean) density of matter in the Universe.

(b)

(i) The line is a symmetrical curve below given line touching given line at ‘present time’.

(ii)

H₀ is not known with any certainty

The mass of matter in the Universe is not known

The extent of the Universe is unknown