• Physics 9702 Doubts | Help Page 24

Question 129: [Waves > Stationary waves]
Basic principle of note production in a horn is to set up stationary wave in an air column.

For any note produced by horn, a node is formed at the mouthpiece and an antinode is formed at bell. Frequency of lowest note is 75 Hz.
What are the frequencies of next two higher notes for this air column?

Reference: Past Exam Paper – June 2011 Paper 11 Q25 & June 2014 Paper 11 Q27



Solution 129:

Answer: D.

As stated from the question, for any note produced by the horn, a node is formed at the mouthpiece and an antinode is formed at the bell.

The lowest note consists of only a node at the mouthpiece and an anti-node at the bell, forming ¼ of a wavelength. The frequency of this lowest note is 75Hz.

The first higher note consists of a node at the mouthpiece, followed by an anti-node, then a node and finally an anti-node at the ball, forming ¾ of a wavelength.

So, its frequency is 3 times that of the lowest note (that is, for the same distance {the length of the horn}, the wave formed is 3 times greater {3/4 is 3(1/4) and ¼ of the wavelength is formed in the lowest note}). Frequency = 3(75) = 225Hz.

Similarly, the second higher note would formed 5/4 of a wavelength, with a frequency = 5(75) = 375Hz










Question 130: [Power > Hydroelectric power]
Turbine at hydroelectric power station is situated 30 m below the level of surface of a large lake. Water passes through turbine at a rate of 340 m³ per minute.
Overall efficiency of turbine and generator system is 90%.
What is the output power of power station? (Density of water is 1000 kg m^–3.)
A 0.15 MW                 B 1.5 MW                   C 1.7 MW                   D 90 MW

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q19



Solution 130:
Go to
A turbine at a hydroelectric power station is situated 30 m below the level of the surface of a large lake.






Question 131: [Pressure > Work done > Gas]
Gas is enclosed inside cylinder which is fitted with frictionless piston.

Initially, gas has volume V₁ and is in equilibrium with an external pressure p. Gas is then heated slowly so that it expands, pushing piston back until the volume of the gas has increased to V₂.
How much work is done by gas during this expansion?
A p(V₂ – V₁)               B ½ p(V₂ – V₁)           C p(V₂ + V₁)               D ½ p(V₂ + V₁)

Reference: Past Exam Paper – June 2013 Paper 12 Q18



Solution 131:

Answer: A.

Increase in volume, ΔV = (V₂ – V₁)

Work done by gas during expansion = pΔV = p(V₂ – V₁)

{The above formula for work done on/by gas is given in the formula list}








Question 132: [Waves > Stationary Waves]
Diagram shows two tubes.

Tubes are identical except tube X is closed at its lower end while tube Y is open at its lower end. Both tubes have open upper ends.
Tuning fork placed above tube X causes resonance of air at frequency f. No resonance is found at any lower frequency than f with tube X.
Which tuning fork will produce resonance when placed just above tube Y?
A a fork of frequency f / 2
B a fork of frequency 2f / 3
C a fork of frequency 3f / 2
D a fork of frequency 2f

Reference: Past Exam Paper – November 2010 Paper 12 Q25



Solution 132:
Answer: D.
For tube X, a node is present at lower (closed) end while for tube Y, an antinode is present at the lower (open) end. For both tubes, antinodes are present at the upper open end.

For closed tube X, resonant frequency are at f_c = nv / 4L
where v: speed of wave, L: length of tube and n = 1, 3, 5, …

For open tube Y, resonant frequency are at f_o = nv / 2L
where n = 1, 2, 3, …

{Note that the above formulae may be derived by studying the wave formed at each frequency. As explained in Question 129 at Physics 9702 Doubts | Help Page 24 (above), at the lowest frequency, a quarter of a wavelength will be formed in the length L of tube X. So, L = λ / 4 giving λ = 4L. Speed v = fλ . Frequency f = v / λ = v / 4L.  A similar reasoning may be used to find the frequency for an open tube at the lowest frequency. The same is done at higher frequencies. The general formulae are given above.}

So, the resonant frequency for the open tube Y, f_o = 2f_c for n =1.










Question 133: [Matter > Elastic Deformation]
Spring is stretched over range within which elastic deformation occurs. Its spring constant is 3.0 N cm^–1.
Which row, for stated applied force, gives the correct extension and strain energy?

Reference: Past Exam Paper – June 2013 Paper 13 Q 21



Solution 133:

Answer: D.

Hooke’s law: Force F = ke      where k is the spring constant and e is the extension.

Spring constant, k = 3.0Ncm^-1 = 300Nm^-1  

Extension = F / k

Strain energy = ½ Fe

Consider A:

Force = 3.0N and extension = 3 / 300 = 0.01m

Strain energy = 0.5 (3.0) (0.01) = 0.015J = 15mJ                   [A is incorrect]

Consider B:

Force = 6.0N and extension = 6 / 300 = 0.02m

Strain energy = 0.5 (6.0) (0.02) = 0.06J = 60mJ                     [B is incorrect]

Consider C:

Force = 12.0N and extension = 12 / 300 = 0.04m                  [C is incorrect]

Consider D:

Force = 24.0N and extension = 24 / 300 = 0.08m

Strain energy = 0.5 (24.0) (0.08) = 0.96J = 960mJ

Question 134: [Forces > Moment]
Diving board of length 5.0 m is hinged at one end and supported 2.0 m from this end by spring of spring constant 10 kN m^–1. Child of mass 40 kg stands at far end of the board.

What is the extra compression of the spring caused by child standing on the end of board?
A 1.0 cm                     B 1.6 cm                      C 9.8 cm                      D 16 cm

Reference: Past Exam Paper – November 2013 Paper 13 Q15



Solution 134:

Answer: C.

Both the child and the spring cause a moment, at a distance from the hinge, on the board.

For equilibrium of the board (for the board to stay horizontal), the torque of the child (anticlockwise moment produced by child) should be equal to the torque of the spring (clockwise moment produced by spring).

Let the force by the spring = F

(40 x 9.81) x 5.0 = F x 2.0

Spring force, F = 981N

Hooke’s law: F = ke

Spring constant, k = 10kNm^-1 = 10000Nm^-1

Extension = F / k = 981 / 10000 = 0.0981m = 9.8cm








Question 135: [Electric Field > Non-uniform electric field]
Diagram shows non-uniform electric field near a positively charged and a negatively charged sphere.
Four electrons, A, B, C and D, are shown at different positions in field.
On which electron is direction of the force on the electron shown correctly?

Reference: Past Exam Paper – November 2011 Paper 12 Q31



Solution 135:
Go to
The diagram shows a non-uniform electric field near a positively charged and a negatively charged sphere.

Question 136: [Electric Field > Electric Potential energy]
Positive charge experiences force F when placed at point X in a uniform electric field.
Charge is then moved from point X to point Y.
Distances r and s are shown on diagram.

What is the change in potential energy of charge?
A decreases by Fs
B increases by Fs
C decreases by Fr
D increases by Fr

Reference: Past Exam Paper – June 2008 Paper 1 Q16 & November 2013 Paper 13 Q18



Solution 136:

Answer: A.

Electric field direction is (away) from positive to negative. The positive charge loses potential energy when moving in the direction of the electric field.

(This is similar to the gravitational potential energy. The gravitational field is usually downward, but a mass would have a higher potential energy when its height is greater – height is in opposite direction to the direction of the gravitational field)

Work done = Force x distance moved in direction of force

Since the distance moved in the direction of the electric field is s, the potential energy decreases by Fs.

Question 137: [Kinematics > Projectile motion]
Bullet is fired horizontally with speed v from rifle. For short time t after leaving rifle, the only force affecting its motion is gravity. Acceleration of free fall is g.
Which expression gives the value of the ratio of the horizontal distance travelled in time t to the vertical distance travelled in time t?
A vt / g                        B v / gt                                    C 2vt / g                      D 2v / gt

Reference: Past Exam Paper – June 2011 Paper 12 Q 6



Solution 137:

Answer: D.

The acceleration of free fall acts (vertically) downwards. It does not affect the horizontal motion.

Horizontal distance travelled in time t, s = vt

Since the bullet is fired horizontally, it’s vertical component of velocity is initially zero.  

Vertical distance travelled in time t, s = ut + ½ at² = 0 + ½ gt² = ½ gt²

Ratio = vt / (½gt²) = 2v / gt