# 9702 June 2014 Paper 23 Worked Solutions | A-Level Physics

## Paper 23

__Question 1__**(a)**

Underline all base quantities in
following list::

ampere charge

__current____mass__second__temperature__weight**(b)**

Potential energy E

_{P}stored in stretched wire is given by E_{P}= ½Cσ^{2}V where C is a constant, σ is strain, V is volume of wire. SІ base units of C:
Base units of σ: no units

Base units of V: m

^{3}
Base units of E

_{p}: kgm^{2}s^{-2}
Base units of C: kgm

^{2}s^{-2}x m^{-3}= kgm^{-1}s^{-2}

__Question 2__**{Detailed explanations for this question is available as Solution 1016 at Physics 9702 Doubts | Help Page 212 -**

*http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-212.html*}

__Question 3__**{Detailed explanations for this question is available as Solution 608 at Physics 9702 Doubts | Help Page 120 -**

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-120.html*}

__Question 4__
Spring hangs vertically from point
P, as shown. Mass M is attached to lower end of spring. Reading x from metre
rule is taken, as shown. Fig shows relationship between x and M.

**(a)**

How apparatus in Fig may be used to
determine load on spring at elastic limit:

Small masses can be added to cause
the extension. Then, the masses hare removed to see if the spring returns to
its original length. This is repeated for larger masses and the maximum mass
for which, when the load is removed, the spring does return to its original
length, is noted.

**(b)**

Explain whether Fig suggests that
spring obeys Hooke’s law:

Hooke’s law requires the force to be
proportional to the extension. The graph shows a straight line, hence the
spring obeys Hooke’s law.

**(c)**

Use Fig to determine spring
constant, in Nm

^{–1}, of spring:
Spring constant, k (= gradient) =
force / extension

k = (0.42x9.81) / [(30 – 21.2)x10

^{-2}] = 47 (46.8) Nm^{-1}

__Question 5__**(a)**

Why terminal potential difference
(p.d.) of cell with internal resistance may be less than electromotive force
(e.m.f.) of cell:

There are lost volts / energy used
within the cell / internal resistance when the cell supplies a current

**(b)**

Battery of e.m.f. 4.5 V and internal
resistance r is connected in series with resistor of resistance 6.0 Ω, as
shown. Current I in circuit is 0.65 A.

(i)

Internal resistance r of battery:

e.m.f. E = I (R + r)

4.5 = 0.65 (6.0 + r)

Internal resistance r = 0.92Ω

(ii)

Terminal p.d. of battery:

V = IR and current I = 0.65A

Terminal p.d. = 0.65 x 6 = 3.9V

(iii)

Power dissipated in resistor:

Power dissipated, P = V

^{2}/R or P = I^{2}R or P = IV
P = (3.9)

^{2}/6 = 2.5W
(iv)

Efficiency of battery:

Efficiency of battery = power out /
power in

Efficiency of battery = I

^{2}R / I^{2}(R + r) = R / (R + r) = 6.0 / (6.0 + 0.92) = 0.87**(c)**

Second resistor of resistance 20 Ω
connected in parallel with 6.0 Ω resistor in Fig. Explain qualitatively change
in heating effect within battery:

The (circuit) resistance decreases.
So, the current increases, causing more heating effect.

__Question 6__**Detailed explanations for this question is available as Solution 533 at Physics 9702 Doubts | Help Page 104 -**}

*http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-104.html*
Can you explain Q6d(ii). Why are we multiplying wavelength with 5/4

ReplyDeleteDetailed updated

Deletein 3(b2) isn't the loss in PE = KE - PE because the whole PE doesn't get converted to KE in 2.5s please help!

ReplyDeleteDetails added.

DeleteFrom the conservation of energy, KE cannot be greater than PE here. + there is a resistive force. Work needs to be done against it. and even if there was no resistive force, the KE would still not be greater than PE.

The equation you gave gives a negative energy, which is wrong.

See the details added

could you please explain why, in question 2(b) (i), 22 is taken as the vertical velocity and not the resultant velocity

ReplyDeleteThe explanation has been updated

DeleteCould you please explain why the Phase difference between P and Q is zero? question 6(c)

ReplyDeleteThis is from the properties of points within a loop of a stationary wave. At the next instant, both points would be moving down. They won’t move right or left since the wave is stationary. So the 2 points are in phase (phase difference = 0).

Delete