Monday, January 26, 2015

Physics 9702 Doubts | Help Page 52

  • Physics 9702 Doubts | Help Page 52

Question 304: [Forces > Hooke’s law]
A sample is placed in a tensile testing machine. It is extended by known amounts and the tension is measured.

What is the work done on the sample when it is given a total extension of 9mm?
A 0.31J                        B 0.36J                        C 0.43J                        D 0.72J

Reference: Past Exam Paper – N83 / II / 40 & J99 / I / 23

Solution 304:
Answer: C.
Work done on sample is given by the area under the graph. In this case, the area can be divided by that of a triangle (from extension – 0 to 5mm) and that of a parallelogram (from extension = 5 to 9 mm)

Area of triangle = ½ (60) (0.005) = 0.150J
Area of parallelogram = ½ (sum of parallel sides) (height)
Area of parallelogram = 0.5 (60 + 80) (0.009 – 0.005) = 0.280J

Total area = 0.150 + 0.280 = 0.430J

Question 305: [Matter > Young modulus]
Wire consists of a 3.0 m length of metal X joined to a 1.0 m length of metal Y.
Cross-sectional area of the wire is uniform.

A load hung from wire causes metal X to stretch by 1.5 mm and metal Y to stretch by 1.0 mm.
Same load is then hung from a second wire of the same cross-section, consisting of 1.0 m of metal X and 3.0 m of metal Y.
What is total extension of this second wire?
A 2.5 mm                    B 3.5 mm                    C 4.8 mm                    D 5.0 mm

Reference: Past Exam Paper – November 2010 Paper 11 Q22 & 13 Q21

Solution 305:
Answer: B.
For a material, the Young modulus is constant. Since the load and the cross-sectional area are the same, the stress (= Force / Area) is also the same.

Young modulus, E = stress / strain = stress / (e/L) = L (stress) / e
where L is the original length and e is the extension

Consider metal X.
For 1st wire: E = 3 (stress) / 1.5
For 2nd wire: E = 1 (stress) / e

Since Young modulus for a particular material is the same,
3 (stress) / 1.5 = stress / e
Extension, e of metal X = 1.5 / 3 = 0.5 mm

Consider metal Y.
For 1st wire: E = 1 (stress) / 1
For 2nd wire: E = 3 (stress) / e

Since Young modulus for a particular material is the same,
Stress = 3 (stress) / e
Extension, e of metal Y = 3.0 mm

Total extension = 0.5 + 3.0 = 3.5 mm

Question 306: [Matter > Polymer]
Which graph represents force-extension relationship of a rubber band that is stretched almost to its breaking point?

Reference: Past Exam Paper – June 2010 Paper 12 Q20 & Paper 13 Q19 & June 2013 Paper 13 Q20

Solution 306:
Answer: A.
Rubber is a polymer. When the elastic band is stretched, the long chain molecules are fully aligned so making any further stretching is almost impossible.

This is represented in the graph A: after the section with a straight line, to extend the rubber band by even a small amount, the force required is much higher [when compared with the ‘straight line’ section]. That is, the gradient at the later part is much higher than the gradient at the earlier part.

Graph B suggest that after the rubber band is stretched [after the straight line], less force is required to cause a small extension.

Graph C is that of a metal obeying Hooke’s law.

Graph D suggests that after the rubber is stretched, no additional force is required to extend the rubber.
[B, C and D are incorrect]

Question 307: [Circular motion]
An object of mass 0.20 kg tied to a string is made to move in a vertical circle. When the object is at the highest point, the tension in the string is zero. Determine the tension in the string when the object is at the lowest point.

Solution 307:
Since the object is moving in circular motion, there is a resultant centripetal acceleration.

The tension is in the string is always towards the centre of the circle and the force of gravity acts downwards (towards the surface of the Earth). The centripetal force is also always towards the centre of the circle.

Let the tension in the string be T and let the centripetal force be FC.
Weight of object, W = mg (= 0.20 x 9.81 = 1.962 N)

At the highest point, the tension and the centripetal force also act downwards.
T + mg = FC
Since the tension is stated to be zero at the highest point, centripetal force, FC = mg

Centripetal force, FC = mv2 / r = mg
Speed, vh of object at highest position = √(gr)

Note that the speed of the object will not be constant. This is because at the highest position, the object also has gravitational potential energy (this varies depending on the height of the object).

At the lowest position, the difference in height of the object when compared to the highest position is 2r where r is the radius of the circle formed. Therefore, the gravitational potential energy of the object at the highest position is greater than that at the lowest position by mg(2r). This additional potential energy will contribute to the kinetic energy of the object at the lowest position.

Let the velocity of the object at the lowest position be vl.

From the law of conservation of energy,
KE at lowest position = KE at highest position + GPE at highest position
½ mvl2 = ½ mvh2 + mg(2r)                  [but vh = √(gr)]
0.5 vl2 = 0.5(gr) + 2gr
Speed, vl at lowest position = √(5gr)

Now, consider the forces at the lowest position
T – mg = mvl2 / r
Tension T = mg + m (5gr) / r = mg + 5mg = 6mg
Tension T = 6 (0.2) (9.81) = 11.8 N

Question 308: [Forces > Hooke’s law]
A catapult consists of 2 strands of rubber, each of unstretched length 0.20 m and each of which stretches by 0.1 m under a tension of 50 N. A stone of mass 0.060 kg is projected vertically upwards from the catapult after each strand has been extended to a total of 0.35 m. What is the energy stored in the stretched catapult? Find the maximum height attained by the stone. (Assume that the rubber obeys Hooke’s law and that air resistance is negligible.) [Acceleration of free fall, g = 10 ms-2.]

Reference: Past Exam Paper – J76 / I / 11

Solution 308:
Hooke’s law: F = kx
For each strand, spring constant k = F / x = 50 / 0.1 = 500 Nm-1

Extension of each strand due to stone = 0.35 – 0.20 = 0.15 m
Energy stored in 1 rubber strand = ½ kx2 = 0.5 (500) (0.15)2 = 5.625 J
Total energy stored = 2 (5.625) = 11.25 J

At the maximum height, the energy of the stone is entirely gravitational potential energy.
mgh = 11.25
0.060 (10) h = 11.25
Maximum height, h = 18.75 = 18.8 m


  1. Assalamualaikum!
    in question 305, why cant the answer be 2.5 mm agn cz the metals are the same and the load is the same? if we do it according to F=kX then we will agn get 2.5 mm, why is this wrong? Plz help ASAP.

    1. Wslm.
      This is a system of 2 wires, we cannot use the simple equation of Hooke's law which is for one wire.
      If the wires were of the same material, we could have calculated the effective spring constant and use it in Hooke's law, but here, even the wires are of different metals.

      + The lengths are different.

      Here, we should use the more fundamental relation relating the Young modulus to stress and strain

    2. so for these type of questions is it always necessary to use youngs modulus?

    3. Yeah. But if it's a simple case, then Hooke's law might suffice.

  2. Jazak Allah! can i get help with these questions?
    May june 2003
    19, 20, 21 and 36
    MayJune 2004
    10 and 11
    November 2013
    28 and 22

    1. For May june 2003 Q19, see question 737 at

      November 2013 which variant?

      For MayJune 2004 Q10, see solution 738 at

      For ayJune 2004 Q11, see solution 218 at

    2. For May june 2003 Q20, see solution 742 at

    3. please help with the remaining questions ASAP in addition to these:
      s12/11/ question 15

    4. For w05/q33, see solution 839 at

  3. salam can I also get help with these MCQs ASAP?
    w04: 19
    s05: 15
    w05: 33

    1. For w04: 19, see solution 748 at

    2. For s05: 15, see solution 786 at


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