# 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 x

_{0}= 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√(x

_{0}^{2}– x^{2}) = 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 = πr

^{2}
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}= 4πr^{2}= 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.

thanks!

ReplyDeleteThanks 🙏

ReplyDeleteBut the still got more problem's

ReplyDelete