# Physics 9702 Doubts | Help Page 242

__Question 1121:__

__[Gravitation]__

**(a)**By reference to the definition of gravitational potential, explain why gravitational potential is a negative quantity. [2]

**(b)**Two stars A and B have their surfaces separated by a distance of 1.4 × 10

^{12}m, as illustrated in Fig. 1.1.

Fig. 1.1

Point P lies on the line joining the centres of the two stars. The
distance x of point P from the surface of star A may be varied.

The variation with distance x of the gravitational potential φ at point
P is shown in Fig. 1.2.

Fig. 1.2

A rock of mass 180 kg moves along the line joining the centres of the
two stars, from star A towards star B.

(i) Use data from Fig. 1.2 to calculate the change in kinetic energy of
the rock when it moves from the point where x = 0.1 × 10

^{12}m to the point where x = 1.2 × 10^{12}m.
State whether this change is an increase or a decrease. [3]

(ii) At a point where x = 0.1 × 10

^{12}m, the speed of the rock is v.
Determine the minimum speed v such that the rock reaches the point where
x = 1.2 × 10

^{12}m. [3]**Reference:**

*Past Exam Paper – June 2016 Paper 43 Q1*

__Solution 1121:__**(a)**

The (gravitational) potential at infinity is defined as/is zero. The (gravitational)
force is

__attractive__so work got out/is done as the object moves from infinity.
(so the potential is negative)

**(b)**

(i)

ΔE = mΔφ

{Change in KE = Change in PE. If PE is lost, then KE
has been gained and vice versa. Since in this case, PE has been lost, it means
that KE has been gained.

At x = 1.2 × 10

^{12}m, φ = – 14 × 10^{8}J kg^{-1}
At x = 0.1 × 10

^{12}m, φ = – 10 × 10^{8}J kg^{-1}
Δφ = final φ – initial φ = – 14 × 10

^{8}– (– 10 × 10^{8}) = – 4 × 10^{8}J kg^{-1}
The calculations below are considering only the
magnitude since KE is gained.}

ΔE = 180 × (14 – 10) × 10

^{8}
ΔE = 7.2 × 10

^{10}J
The change is an increase {in KE}.

(ii)

{The rock needs to reach only the point where the field
is zero. In the graph, this is where the change in gravitational potential is
zero. Value of φ = – 4.4 × 108 J kg

^{-1}. The rock needs only to reach this point. The rock would be attracted by the field of the other planet afterwards.}
Energy required = 180 × (10 – 4.4) × 10

^{8}OR Energy per unit mass = (10 – 4.4) × 10^{8}
½ × 180 × v

^{2}= 180 × (10 – 4.4) × 10^{8}OR ½ × v^{2}= (10 – 4.4) × 10^{8}
Minimum speed v = 3.3 × 10

^{4}m s^{–1}

__Question 1122: [Measurement > Accuracy and Precision]__
Four students each made a series of
measurements of the acceleration of free fall g. The table shows the results
obtained.

Which set of results could be
described as precise but not accurate?

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q5*

__Solution 1122:__**Answer: D.**

A precise set of results have values
that are very close to each other. An accurate set of results is one in which
the mean is close to the actual value (g = 9.81) of the quantity under
consideration.

Choice A: Mean = (9.81 + 9.79 + 9.84
+ 9.83) / 4 = 9.8175

Choice B: Mean = (9.81 + 10.12 +
9.89 + 8.94) / 4 = 9.69

Choice C: Mean = (9.45 + 9.21 + 8.99
+ 8.76) / 4 = 9.1025

Choice
D: Mean = (8.45 + 8.46 + 8.50 + 8.41) / 4 = 8.455 [precise but not accurate]

__Question 1123: [Current Electricity > Internal resistance]__
A battery of electromotive force
(e.m.f.) 12 V and internal resistance r is connected in series to two resistors,
each of constant resistance X, as shown in Fig. 5.1.

Fig. 5.1

The current Ι

_{1}supplied by the battery is 1.2 A.
The same battery is now connected to
the same two resistors in parallel, as shown in Fig. 5.2.

Fig. 5.2

The current Ι

_{2}supplied by the battery is 3.0 A.**(a)**(i) Show that the combined resistance of the two resistors, each of resistance X, is four times greater in Fig. 5.1 than in Fig. 5.2. [2]

(ii) Explain why Ι

_{2}is not four times greater than Ι_{1}. [2]
(iii) Using Kirchhoff’s second law,
state equations, in terms of e.m.f., current, X and r, for

1. the circuit of Fig. 5.1,

2. the circuit of Fig. 5.2.

[2]

(iv) Use the equations in (iii) to
calculate the resistance X. [1]

**(b)**Calculate the ratio

power transformed in one resistor of
resistance X in Fig. 5.1 / power transformed in one resistor of resistance X in
Fig. 5.2

[2]

**(c)**The resistors in Fig. 5.1 and Fig. 5.2 are replaced by identical 12 V filament lamps.

Explain why the resistance of each
lamp, when connected in series, is not the same as the resistance of each lamp
when connected in parallel. [2]

**Reference:**

*Past Exam Paper – November 2014 Paper 22 Q5*

__Solution 1123:__**(a)**(i)

in series 2X

__or__in parallel X / 2
{In series, combined
resistance R

_{S}= X + X = 2X
In parallel, combined
resistance R

_{P}= [1/X + 1/X]^{-1}= [2/X]^{-1}= X/2}
other relationship given

__and__4× greater in series (than in parallel)
{R

_{S}/ R_{P}= 2X / 0.5X = 4}
(ii) Due to the internal resistance,
the total resistance for the series circuit is not four times greater than the resistance
for the parallel circuit

(iii)

1.

{e.m.f = sum of p.d.}

E = I

_{1}(2X + r)**or**12 = 1.2(2X + r)
2. E = I

_{2}(X/2 + r)**or**12 = 3.0(X/2 + r)
(iv)

{From 12 = 1.2(2X + r),} 2X + r = 10 and {From 12 = 3.0(X/2 + r),} X/2 + r = 4

{Subtracting the 2
equations,}

X = 4.0 Ω

**(b)**

P = I

^{2}R or V^{2}/ R or VI
ratio = [(1.2)

^{2}× 4] / [(1.5)^{2}× 4] = 0.64**(c)**The resistance (of a lamp) changes with V (or I).

Can u plz explain solution 1123 c part in detail

ReplyDeleteThe I-V graph of a filament lamp shows that its resistance increases as the p.d. across it increases.

DeleteIn a simple series circuit of 2 similar resistors for example, the e.m.f. E in the circuit is divided equally between the resistors.

p.d. across each resistor = E/2

In a simple parallel circuit, the p.d. across each resistor (each branch) is the same as the e.m.f.

p.d across each resistor = E

Thus, in a parallel circuit, the e.m.f. is greater and so, will be the resistance.