# Physics 9702 Doubts | Help Page 183

__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}= kr^{2},
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 × (πr

^{3}ρ_{f }g)
(ii)

EITHER Resultant downward force =
4/3 × [πr

^{3}(ρ_{s}– ρ_{f}) g]
OR Resultant downward force = 4/3 × [πr

^{3}(ρ_{s}– ρ_{f}) g] – viscous force**(b)**

6πrηv

_{t}= 4/3 × [πr^{3}(ρ_{s}– ρ_{f}) g]
{Simplifying the equation,
we obtain ηv

_{t}= 2/9 × [r^{2}(ρ_{s}– ρ_{f}) g]
v

_{t}= 2/(9η) × [r^{2}(ρ_{s}– ρ_{f}) g]
So, v

_{t}depends on r^{2}. 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}= kr^{2}
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.}

160

^{2}+ 70^{2}= Y^{2}
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

12

^{2}= 9^{2}+ 15^{2}- 2(9)(15)cosθ_{1}
θ

_{1}= cos^{-1}(162 / 270) = 53.13^{o}
Similarly, 9

^{2}= 15^{2}+ 12^{2}– 2(15)(12)cosθ_{2}giving θ_{2}= 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

T

_{1 }sin(53.13) + T_{2 }sin(36.87) = 100
The horizontal components of the
tensions should equate since the weight does not have a horizontal component.

T

_{1 }sin(53.13) = T_{2 }sin(36.87)
Solving the 2 equations
simultaneously gives the tension in the 9m string, T

_{1}= 80N and the tension in the 12m string, T_{2}= 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)

H

_{0}is not known with any certainty
The mass of matter in the Universe is
not known

The extent of the Universe is
unknown

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ReplyDeleteFor 04/M/J/04 Q,3(c)(ii), see solution 900 at

Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-184.html

For 04/M/J/06 Q.4(c),, see solution 903 at

http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-184.html