# 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.**

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 1

^{st}wire: E = 3 (stress) / 1.5

For 2

^{nd}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 1

^{st}wire: E = 1 (stress) / 1

For 2

^{nd}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 F

_{C}.

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 = F

_{C}

Since the tension is stated to be zero at the highest point, centripetal force, F

_{C}= mg

Centripetal force, F

_{C}= mv

^{2}/ r = mg

Speed, v

_{h}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 v

_{l}.

From the law of conservation of energy,

KE at lowest position = KE at highest position + GPE at highest position

½ mv

_{l}

^{2}= ½ mv

_{h}

^{2}+ mg(2r) [but v

_{h}= √(gr)]

0.5 v

_{l}

^{2}= 0.5(gr) + 2gr

Speed, v

_{l}at lowest position = √(5gr)

Now, consider the forces at the lowest position

T – mg = mv

_{l}

^{2}/ 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 kx

^{2}= 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

Assalamualaikum!

ReplyDeletein 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.

Wslm.

DeleteThis 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

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

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

DeleteJazak Allah! can i get help with these questions?

ReplyDeleteMay june 2003

19, 20, 21 and 36

MayJune 2004

10 and 11

November 2013

28 and 22

For May june 2003 Q19, see question 737 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-149.html

November 2013 which variant?

For MayJune 2004 Q10, see solution 738 at

http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-149.html

For ayJune 2004 Q11, see solution 218 at

http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-35.html

For May june 2003 Q20, see solution 742 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-150.html

please help with the remaining questions ASAP in addition to these:

Deletes12/11/ question 15

w05/q33

s13/11/38

For w05/q33, see solution 839 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-167.html

salam can I also get help with these MCQs ASAP?

ReplyDeletew04: 19

s05: 15

w05: 33

For w04: 19, see solution 748 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-151.html

For s05: 15, see solution 786 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-158.html