# 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 = 2.2x10

^{9}(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 × 10

^{5}Pa. Divers in water of density 1.08 × 10

^{3}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 (2.2x10

^{9}) 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
2.2x10

^{9}is a large number
ΔV is very small OR ΔV / V is very
small, (so ‘incompressible’)

**(c)**

Change in pressure, Δp = hρg

1.01x10

^{5}= h (1.08x10^{3}) (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 F

*l*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 F

*l*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 F

*l*.

{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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