• Physics 9702 Doubts | Help Page 199

Question 963: [Simple harmonic motion]

A tube, closed at one end, has a uniform area of cross-section. The tube contains some sand so that the tube floats upright in a liquid, as shown in Fig.1.

When the tube is at rest, the depth d of immersion of the base of the tube is 16 cm.

The tube is displaced vertically and then released.

The variation with time t of the depth d of the base of the tube is shown in Fig.2.

(a) Use Fig.2 to determine, for the oscillations of the tube,

(i) amplitude,

(ii) period.

(b)

(i) Calculate vertical speed of the tube at a point where the depth d is 16.2 cm.

(ii) State one other depth d where the speed will be equal to that calculated in (i).

(c)

(i) Explain what is meant by damping.

(ii) Liquid in (b) is now cooled so that, although the density is unchanged, there is friction between the liquid and the tube as it oscillates. Having been displaced, the tube completes approximately 10 oscillations before coming to rest.

On Fig.2, draw a line to show variation with time t of depth d for the first 2.5 s of the motion.

Reference: Past Exam Paper – June 2008 Paper 4 Q3

Solution 963:

(a)

(i) Amplitude = 0.5 cm

(ii) Period = 0.8 s

(b)

(i).

Angular speed, ω = 2π / T = 7.85 rad s^–1

{Amplitude x₀ = 0.5cm. Equilibrium position is at 16.0cm. When depth d = 16.2cm, the tube is at a displacement of 16.2 – 16.0 = 0.2cm from its equilibrium position.}

Speed v = ω√(x₀² – x²) = 7.85 × √({0.5 × 10^–2}² – {0.2 × 10^–2}²) = 3.6 cm s^–1

(ii) Depth d = 15.8 cm

{At a displacement equal in magnitude to the above, that is 0.2cm from the equilibrium position.}

(c)

(i) Damping is the (continuous) loss of energy / reduction in amplitude (from the oscillating system) caused by a force acting in opposite direction to the motion / friction / viscous forces

(ii)

The line should have the same period / small increase in period. The line displacement is always less than that on Fig.3.2 (ignore first T/4). The peak get progressively smaller

Question 964: [Current of Electricity]

Two cells of e.m.f. 3.0 V and 1.2 V and negligible internal resistance are connected to resistors of resistance 9.0 Ω and 18 Ω as shown.

What is the value of current I in the 9.0 Ω resistor?

A 0.10 A                     B 0.20 A                     C 0.30 A                     D 0.47 A

Reference: Past Exam Paper – June 2009 Paper 1 Q33

Solution 964:

Answer: B.

Current normally flows from the positive terminal of a cell to its negative one, but since the polarities of the 2 cells are reversed in this case, we need to find the equivalent e.m.f. in the circuit.

Equivalent e.m.f. in circuit = 3.0 – 1.2 = 1.8 V

The resistors are connected in parallel with the batteries, so the p.d. across each resistor would be equal to the e.m.f. in the circuit.

Ohm’s law: V = IR

Current I = 1.8 / 9.0 = 0.20 A

Question 965: [Magnetic field]

(a) Describe what is meant by a magnetic field.

(b) A small mass is placed in a field of force that is either electric or magnetic or gravitational.

State nature of the field of force when the mass is

(i) charged and the force is opposite to direction of the field,

(ii) uncharged and the force is in direction of the field

(iii) charged and there is a force only when the mass is moving

(iv) charged and there is no force on the mass when it is stationary or moving in a particular direction

Reference: Past Exam Paper – November 2008 Paper 4 Q8

Solution 965:

(a) A magnetic field is a region (of space) / area where a force is experienced by a current-carrying conductor / moving charge / permanent magnet.

(b)

(i) Electric

(ii) Gravitational

(iii) Magnetic

(iv) Magnetic

{Gravitational force does not depend on the charge (gravitational is eliminated).

In an electric field, as long as the particle is charged, it will experience a force (electric is eliminated).

The direction of the force on a moving charge in a magnetic field is given by Fleming’s left-hand rule. If the charge is not moving, there will be no force. Now, for Fleming’s left-hand rule to apply, the direction of motion should be perpendicular (at right angle) to the magnetic field. If the direction of motion of the charge is in the same direction as the magnetic field, there will be no force on it.}

Question 966: [Current of Electricity > Resistance]

A copper wire is cylindrical and has resistance R.

What will be resistance of a copper wire of twice the length and twice the radius?

A R / 4                         B R / 2                         C R                              D 2R

Reference: Past Exam Paper – November 2013 Paper 13 Q35

Solution 966:

Answer: B.

Resistance R of a wire = ρL / A                    

where L is the length and A is the cross-sectional area.

Cross-sectional area A = πr²

Now consider another wire also made of copper. The resistivity ρ will be the same.

The wire has twice the length (2L instead of L).

The wire has twice the radius (2r instead of r).

New cross-sectional area = π(2r)² = 4πr² = 4A

Resistance of new wire = ρ (2L) / 4A = (ρL/A) / 2 = R / 2

Question 967: [Matter > Deformation]

A piece of copper is drawn into a continuous wire.

What behaviour is the copper exhibiting?

A brittle only

B elastic only

C plastic only

D both brittle and elastic

Reference: Past Exam Paper – June 2007 Paper 1 Q17

Solution 967:

Answer: C.

Glass is a brittle material – that is, it would break if sufficient force is applied. Since the copper is drawn into a continuous wire, it has not been broken. [A and D are incorrect]

Now, an elastic obtained would return to its original shape as soon as the deforming force is removed. Since the copper wire remains in its continuous form, it is not displaying an elastic behaviour here. [B is incorrect]

For the copper to remain in its continuous wire shape, it should be displaying a plastic behaviour only – that is, it does not regain its original shape when the deforming force is removed.