Thursday, January 22, 2015

Physics 9702 Doubts | Help Page 49

  • Physics 9702 Doubts | Help Page 49



Question 290: [Matter > Pressure]
(a)
(i) Define the terms
1. tensile stress
2. tensile strain
3. the Young modulus:

(ii) Suggest why Young modulus is not used to describe deformation of a liquid or a gas\
(b) Change ΔV in the volume V of some water when the pressure on the water increases by Δp is given by expression
Δp = 2.2x109(ΔV / V)
where Δp is measured in pascal.
In many applications, water assumed to be incompressible. By reference to expression, justify this assumption:

(c) Normal atmospheric pressure is 1.01 × 105 Pa. Divers in water of density 1.08 × 103 kg m–3 frequently use approximation that every 10 m increase in depth of water is equivalent to 1 atmosphere increase in pressure. Determine the percentage error in this approximation.

Reference: Past Exam Paper – June 2008 Paper 2 Q4



Solution 290:
(a)
(i)
1. Tensile stress is defined as the force per (cross-sectional) area
2. Tensile strain is defined as the ratio of extension to the original length
3. The Young modulus is defined as the ratio of stress to strain.

(ii) EITHER Fluids cannot be deformed in one direction / cannot be stretched OR Fluids can only have a volume change OR Fluids have no fixed shape

(b)
{It should be noted that the assumption that the water is incompressible implies that the change in volume should be very small (nearly zero). From the equation given, since the value of the constant (2.2x109) is very large, it means that a very large change in pressure would be required to cause a significant change in volume (if the water was compressible). So, it is justified to assume that the water is incompressible.}
EITHER Unless Δp is very large OR 2.2x109 is a large number
ΔV is very small OR ΔV / V is very small, (so ‘incompressible’)

(c)
Change in pressure, Δp = hρg
1.01x105 = h (1.08x103) (9.81)
Correct depth, h = 9.53m
Δh / h = 0.47 / 10        OR 0.47 / 9.53
Percentage error = 4.7%          OR 4.9%         OR 5%








Question 291: [Forces > Equilibrium]
Picture on a wall is supported by a wire looped over a nail.

Mass of the picture is 4.2 kg.
What is tension in the supporting wire?
A 5.0 N                       B 23 N                        C 49 N                        D 97 N

Reference: Past Exam Paper – November 2012 Paper 12 Q18



Solution 291:
Answer: C.
The tensions are along the wire and towards the nail.
For the picture not to fall, the sum of forces acting on it should be zero. The force of gravity acts downwards. The horizontal components of the tensions in the wire cancel each other.

So, the vertical components of the tensions should be equal to the weight.
2Tsin(25) = 4.2 x 9.81
Tension T = 48.7 = 49N








Question 292: [Gases > Pressure]
Why is it wrong to say that the ideal gas exerts pressure in its container due to the weight of the molecules exerting a force on the walls of the container?



Solution 292:
Weight is a specific force. It is due to the force of gravity acting on a mass and acts downwards (towards the Earth – assuming we are on Earth). The explanation stated in the question is incorrect. Even if we assume it was correct, the mass of molecules are very small, so the pressure would be almost negligible.

The correct explanation is as follows.
A gas consists of many molecules in constant random free motion, colliding elastically with each other and the walls of the container. The pressure of the gas is due to the collision of the gas molecules with the walls of the container.
Source: “Pacific Physics A Level,” Volume 1, by POH LIONG YONG, pg 445









Question 293: [Matter > Elasticity]
Sample of metal is subjected to a force which increases to maximum value and then decreases back to zero. A force-extension graph for sample is shown.

When sample contracts it follows the same force-extension curve as when it was being stretched.
What is the behaviour of the metal between X and Y?
For June 2008 Paper 1 Q22:
A both elastic and plastic
B elastic but not plastic
C plastic but not elastic
D not elastic and not plastic

For June 2014 Paper 12 Q19:
A both elastic and plastic
B not elastic and not plastic
C plastic but not elastic
D elastic but not plastic

Reference: Past Exam Paper – June 2008 Paper 1 Q22 & June 2014 Paper 12 Q19

Solution 293:
June 2008 Paper 1 Q22 – Answer: B.
June 2014 Paper 12 Q19 – Answer: D.

In the graph, point X is the limit of proportionality and point Y is the elastic limit. Beyond the elastic limit, the metal would have a plastic behaviour, but between extension = 0 to point Y, the metal has an elastic behaviour. [B is correct]

{For more information about the limit of proportionality and the elastic limit, check Solution 279 at Physics Doubt | Help Page 47 - http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-47.html}

As long as point X is the limit of proportionality and point Y is the elastic limit (as in the graph given), the metal will not have a plastic behaviour. The graph given cannot display plastic behaviour.
Plastic behaviour can be shown beyond point Y. In such a case, when the force is zero, the extension will have a non-zero value. Consider a point Z beyond point Y. When the sample contracts the graph would be a straight line between point Z and the non-zero value of extension when the force is zero.









Question 294: [Forces > Hooke’s law]
A wire is stretched by a force F which causes an extension l. The energy stored in the wire is ½ Fl only if
A the extension of the wire is proportional to the force applied
B the weight of the wire is negligible
C the wire is not stretched beyond its elastic limit
D the acceleration of free fall is constant
E the cross-sectional area of the wire remains constant

Reference: Past Exam Paper – N76 / II / 40



Solution 294:
Answer: A.
From a graph of force F against extension l of the wire, the energy stored in the wire is given by the area.

Hooke’ law states that the extension of the wire is proportional to the force applied, provided the elastic limit has not been reached. So, for a graph obeying Hooke’s law, the area under the graph is given by ½ Fl given the graph is a straight line.

Note that at some point before the elastic limit, there a point known as the limit of proportionality beyond which the extension of the wire is no longer proportional to the force applied, but the wire still behaves elastically. Therefore, just before the elastic limit, the graph is not a straight line and so, the area under the graph is NOT ½ Fl.
{For more information about the limit of proportionality and the elastic limit, check Solution 279 at Physics Doubt | Help Page 47 - http://physics-ref.blogspot.com/2015/01/physics-9702-doubts-help-page-47.html}





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