Wednesday, January 21, 2015

Physics 9702 Doubts | Help Page 48

  • Physics 9702 Doubts | Help Page 48



Question 284: [Pressure]
Diagram shows a flask connected to a U-tube containing liquid. Flask contains air at atmospheric pressure.

Flask is now gently heated and the liquid level in right-hand side of the U-tube rises through a distance h. Density of the liquid is ρ.
What is the increase in pressure of the heated air in the flask?
A hρ                B ½ hρg                      C hρg              D 2hρg

Reference: Past Exam Paper – June 2009 Paper 1 Q18



Solution 284:
Answer: D.
The heating causes the liquid in the right-hand side to rise through a distance h from its original position.

But this also causes the liquid in the left-hand side to sink by a distance h from its original position.

So, the difference in height of the liquid is h + h = 2h.

Increase in pressure = (Δh)ρg = 2hρg 









Question 285: [Kinematics > Linear motion > Graph]
Student investigates speed of a trolley as it rolls down a slope, as illustrated.

Speed v of trolley is measured using a speed sensor for different values of the time t that the trolley has moved from rest down the slope.
Fig shows variation with t of v.

(a) Use Fig to determine acceleration of the trolley at the point on the graph where t = 0.80 s.

(b)
(i) State whether acceleration is increasing or decreasing for values of t greater than 0.6 s. Justify your answer by reference to Fig.
(ii) Suggest an explanation for this change in acceleration.

(c) Name feature of Fig that indicates the presence of
(i) random error
(ii) systematic error

Reference: Past Exam Paper – November 2006 Paper 2 Q2



Solution 285:
(a)
{The acceleration can be found by calculating the gradient of the tangent at time t = 0.80s. Marks are usually awarded for the following}
Use of a tangent (anywhere), not at a single point
The tangent is drawn at the correct position (at time t = 0.80s)
Acceleration (= gradient) = 1.7 ± 0.1

(b)
(i) Since the slope (gradient of tangent of graph) is decreasing, the acceleration is decreasing

(ii) E.g.
{Since the trolley is accelerating downwards, its speed increases. Air resistance increases with speed. So, the resultant force on the trolley decreases, so the acceleration decreases.}
Air resistance increases (with speed)
{If the slope of the ramp decreases [that is, the ramp becomes closer to being horizontal], the component of the acceleration of the gravity contribution to the acceleration of the trolley decreases. So, the acceleration of the trolley would decrease if the slope of the ramp decreases.}
The (angle of) the slope of the ramp decreases
{Friction is usually taken to be constant for a specific path. Friction would only cause the resultant force to be lower [the value is still constant], it does not cause the resultant force to decrease to time. A changing resultant force causes the resultant acceleration to change but a constant resultant force results in a constant acceleration. If the resultant force is big [and constant], the resultant acceleration is also big and if the resultant force is small [and constant], the resultant force is also small.}

(c)
(i)
{The ‘scatter of the points’ is not enough. It should be related to the line (to their distribution about the curve). Since not all the points lie on the line, random error is present.}
Random error is indicated by the scatter of the points about the line

(ii)
{Since the line does not go through the origin – it has an intercept, it indicates that a systematic error is present.}
The intercept / line does not go through the origin








Question 286: [Matter > Solids]
What’s the difference between crystalline, polymeric and amorphous materials?



Solution 286:
Solids can be classified according to the ways atoms are arranged.

Crystalline solids are solids where the atoms are arranged in fixed repetitive manner over a large distance.
Examples of crystalline solids are sodium chloride crystals, quartz, copper and silver.
In a crystal, the network of fixed structures which consists of individual cells is known as crystal lattice.

Rubber is an example of a polymer. It consists of very long chain molecules. The chains of molecules are all tangled up. The chains of molecules are linked to each other by transverse strings which are actual sulphur molecules.
Another example of a polymer is plastic or polythene.

Amorphous materials do not possess repeated structures over a long range as in crystalline. In amorphous materials, only short range structures occur..
An example of an amorphous solid is glass.
An amorphous solid is formed when a liquid is cooled rather rapidly, i.e. supercooled. The rapid cooling process does not give the atoms sufficient time to arrange themselves in an orderly manner. The atoms are bonded in short range structure. Since the material is cooled, the movement of the atoms is reduced and this explains the fixed shape of glass.
When glass is heated, it becomes soft and then starts to melt. The melting point of glass is not fixed. This is a special characteristic of amorphous solids.

Source: “Pacific Physics A Level” Volume 1, by POH LIONG YONG, pg 345 - 347








Question 287: [Measurements > Micrometer]
Diameter of a cylindrical metal rod is measured using micrometer screw gauge.
Diagram below shows an enlargement of the scale on micrometer screw gauge when taking the measurement.

What is cross-sectional area of the rod?
A 3.81 mm2                 B 11.4 mm2                 C 22.8 mm2                 D 45.6 mm2

Reference: Past Exam Paper – June 2012 Paper 12 Q4



Solution 287:
Answer: B.
Main scale reading = 3.5mm.
Thimble scale reading = 0.31mm
Diameter of rod = 3.5 + 0.31 = 3.81mm

Cross-sectional area = πr2 = π(d/2)2 = π(3.81/2)2 = 11.4mm2








Question 288: [Matter > Energy]
A hard ball and soft ball, with equal masses and volumes, are thrown at glass window. Balls hit the window at the same speed. Suggest why hard ball is more likely than the soft ball to break the glass window.

Reference: Past Exam Paper – June 2005 Paper 2 Q4(c)



Solution 288:
Note that the mass, density and kinetic energy of both balls are all similar.
If it is hard, the ball does not deform (much) and
EITHER (all) its kinetic energy is converted to strain energy {kinetic energy of the balls is converted into strain energy strain energy in the ball which affects the strain energy within the glass only}. If the ball is soft, the kinetic energy becomes strain energy of {both} the ball and the window {during collision, the soft ball compresses. Thus, some of its kinetic energy becomes its strain energy}.

OR The impulse {Impulse = Force x time = mv} for the hard ball takes place over a shorter time {the time of collision is smaller. This affects the rate of change of momentum, and thus the force on the glass}. So, there is a larger force {Force F = mv / time. Since time is small, force is big} / greater stress {on the glass}.










Question 289: [Magnetism]
(a) Define the tesla.

(b) Large horseshoe magnet produces uniform magnetic field of flux density B between its poles. Outside region of the poles, the flux density is zero.
Magnet is placed on a top-pan balance and stiff wire XY is situated between its poles, as shown.

Wire XY is horizontal and normal to magnetic field. Length of wire between the poles is 4.4 cm.
A direct current of magnitude 2.6 A is passed through wire in the direction from X to Y.
Reading on the top-pan balance increases by 2.3 g.
(i) State and explain the polarity of pole P of the magnet.
(ii) Calculate flux density between the poles.

(c) Direct current in (b) is now replaced by a very low frequency sinusoidal current of r.m.s. value 2.6 A.
Calculate variation in the reading of the top-pan balance.

Reference: Past Exam Paper – June 2009 Paper 4 Q6



Solution 289:
(a) The tesla is the unit of magnetic flux density / magnetic field strength for which a (uniform) field is normal to a wire carrying a current of 1A, giving a force (per unit length) of 1Nm-1.

(b)
(i)
{Since the reading on the balance increases,}The force on the magnet / balance is downwards. (So by Newton’s third law,) The force on the wire is upwards.
{Force in the wire is upward. Current in the wire is from X to Y. Using Fleming’s left hand rule [thumb – force, forefinger – magnetic field and middle finger – current], the magnetic field is from pole P towards the other pole. The direction of magnetic field is draw from North pole to South pole. So, pole P is a north P.}
Therefore, pole P is a north pole.

(ii)
Upward force F = BIL and Downward force = mg (g missing, then 0/3 in (ii))
{From Newton’ third law, the magnitude of the upward force on the wire should be equal to the magnitude of the downward force on the magnet. Upward force = BIL. The downward force causes an increase of 2.3g in mass, so the downward force is (2.3 x g) where g here is the acceleration due to gravity.}
(2.3x10-3) x 9.8 = B x 2.6 x (4.4x10-2)
Flux density B = 0.20T

(c)
{r.m.s value of current, Irms = 2.6A. Peak value of current, I0 = Irms x 2.
Since F = BIL, the force is proportional to the current [B and L are constant for the magnet and wire] and from Newton’s third law, BIL = mg. So, the current is proportional to the mass. So, the increase in mass is 2.3 x √2.}
Reading for the maximum current = 2.3 x √2
{The current is now sinusoidal. So, it can be positive and negative. The maximum positive current is + Irms x 2, causing the mass to increase by (2.3 x √2). The maximum negative current is – Irms x 2, [the direction of current is now opposite, so the force on the magnet is now upward – the mass decreases], causing the mass to decrease by (2.3 x √2). So, the total variation in the mass is 2 times (2.3 x √2). That is, total variation = 2 x 2.3 x √2.}
Total variation = 2 x 2.3 x √2 = 6.5g




No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 48