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Question 472: [Stationary waves]

A string fixed at both ends and of length L is plucked at its midpoint and emits its fundamental note of f₁. When the string is plucked at a different point, the first overtone frequency f₂ is also produced. Which one of the following correctly gives both f₂ / f₁ and v, where v is the speed of transverse waves in the string?

            f₂ / f₁                 v

A         2                      f₁L

B         0.5                   f₁L

C         2                      f₂L

D         0.5                   f₂L

E          2                      f₂L / 2

Reference: Past Exam Paper – J81 / II / 13

Solution 472:

Answer: C.

The higher overtone frequencies are always integral multiple of the fundamental frequency, that is, the overtone frequencies can be 2f₁, 3f₁, 4f₁, ….

The first overtone frequency f₂ = 2f₁

So, the ration f₂ / f₁ = 2f₁ / f₁ = 2        [B and D are incorrect]

Consider the fundamental frequency.

Since the string is fixed at both ends, nodes will be found at these ends. The fundamental frequency consist of only these nodes. It can be seen that half a wave will be formed at this frequency. That is, when the frequency = f₁, the length of the string, L = λ / 2.

Wavelength λ = 2L

Speed v = f λ = f₁(2L) = 2f₁L

Since f₂ / f₁ = 2,

Fundamental frequency f₁ = f₂ / 2

Speed v = 2 (f₂/2) L = f₂L

At the first overtone frequency, a complete wave in formed in the string.

Question 473: [Dynamics > Momentum]

Isolated system consists of two bodies on which no external forces act. The two bodies collide with each other and stick together on impact.

Which row correctly compares total kinetic energy and total momentum of the bodies before and after the collision?

Reference: Past Exam Paper – November 2013 Paper 13 Q11

Solution 473:

Answer: B.

For any isolated system, the law of conservation of momentum applies, that is, the total momentum before collision is equal to the total momentum after collision. [A and C are incorrect]

Kinetic energy = ½ mv²

Let the speeds of the two bodies before collision be v₁ and v₂. Let the speed of the 2 bodies after collision be v_f.

From the conservation of linear momentum,

mv₁ + mv₂ = (2m)v_f

Speed v_f = (v₁ + v₂) / 2

Total kinetic energy before collision = ½ mv₁² + ½ mv₂²

Total kinetic energy after collision = ½ (2m) [(v₁ + v₂) / 2]²

Therefore, the kinetic energy after collision is different from that before collision.

Question 474: [Current of Electricity > Potential difference]

Six resistors, each of resistance 5 Ω, are connected to 2 V cell of negligible internal resistance.

What is potential difference between terminals X and Y?

A 2 / 3 V                     B 8 / 9 V                     C 4 / 3 V                     D 2 V

Reference: Past Exam Paper – November 2002 Paper 1 Q36

Solution 474:

Answer: A.

To obtain the potential difference between terminals X and Y, we first need to know the values of the potential at X and Y.

The positive terminal of the cell can be taken to be at a potential of 2V and the negative terminal to be at a potential of 0V.

It can be seen that the circuit consists of 2 loops, with a junction at the top and one at the bottom. Since there are no components connected between the upper junction and the positive terminal of the cell, it can be concluded that the upper junction at the same potential of 2V. Similarly, the lower junction is at a potential of 0V.

Thus, in any loop, going from up and down, the potential decreases across each components until it is zero at the lower junction. Current flows from the positive terminal of the cell to the negative one. So, following the direction of current flow, the potential decreases after going across a component.

From Kirchhoff’s law, the sum of p.d. in any loop is equal to the e.m.f in the circuit.

Consider the first loop (where terminal X is connected).

From the potential divider equation,

p.d. across the upper resistor = [5 / (5+5+5)] x 2 = 2/3 V

Note that this is the potential DIFFERENCE across the upper resistor. It is also the p.d. across each resistor since they all have the same resistance.

Since the upper junction is at a potential of 2V,

Potential at terminal X = 2 – 2/3 = 4/3 V

Consider the second loop (where terminal Y is connected).

From the upper junction to terminal Y, there are 2 resistors and the p.d. across each is 2/3V.

Potential at terminal Y = 2 – 2 (2/3) = 2/3 V

Potential difference between X and Y = 4/3 – 2/3 = 2/3 V

Question 475: [Measurement > Uncertainty]

Steel wire is stretched in an experiment to determine Young modulus for steel.

Uncertainties in the measurements are given below.

measurement               uncertainty

load on wire                ±2%

length of wire              ±0.2%

diameter of wire          ±1.5%

extension                     ±1%

What is the percentage uncertainty in Young modulus?

A 1.3%                        B 1.8%                        C 4.7%                        D 6.2%

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q4

Solution 475:

Answer: D.

Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae

where F is the load on the wire, L is the length of wire, A is the cross-sectional area of the wire and e is the extension

Cross-sectional area of wire = πd² / 4

where d is the diameter of the wire

Young modulus, E = 4FL / πd²e

(ΔE / E) x 100% = [(ΔF / F) + (ΔL / L) + 2(Δd / d) + (Δe / e)] x 100%

Percentage uncertainty, (ΔE / E) x 100% = 2 + 0.2 + 2(1.5) + 1 = 6.2%

.

Question 476: [Work done]

Graph is a force-extension graph for a wire that is being stretched.

How much work needs to be done by tensile force, to two significant figures, to cause extension of 7.0 mm?

A 0.088 J                     B 0.12 J                       C 0.53 J                       D 120 J

Reference: Past Exam Paper – June 2012 Paper 12 Q24

Solution 476:

Answer: B.

The work done by the tensile force is given by the area under the force-extension graph from extension = 0mm to 7.0mm.

Since the work done is given in joules, we need to convert the unit of extension from mm to m.

The area can be approximated by the area of a triangle formed by the x-axis, point (0, 0) and point (3.5, 21) + area of a trapezium formed as shown.

Area of triangle = ½ (3.5x10^-3) (21) = 0.03675J

At extension = 7mm, force = 25N.

Area of trapezium = ½ (sum of parallel sides) (height)

Area of trapezium = ½ (21 + 25) ([7.0 – 3.5]x10^-3) = 0.0805J

Total work done = 0.03675 + 0.0805 = 0.11725 = 0.12J

Question 477: [Pressure]

The Mariana Trench in Pacific Ocean has a depth of about 10 km.

Assuming that sea water is incompressible and has density of about 1020 kg m^–3, what would be the approximate pressure at that depth?

A 10⁵ Pa                      B 10⁶ Pa                      C 10⁷ Pa                      D 10⁸ Pa

Reference: Past Exam Paper – November 2010 Paper 11 Q19

Solution 477:

Answer: D.

Depth h = 10km = 10 000 m

Density ρ = 1020 kg m^–3

Assume that the acceleration of free fall, g = 10 ms^-2.

Pressure = hρg = (10 000) (1020) (10) = 1.02x10⁸ Pa