# Physics 9702 Doubts | Help Page 109

__Question 556: [Measurement > Calibration]__
Uncalibrated scale and pointer of a
meter are shown in Fig.

Pointer is shown in the zero
position.

The meter is to be used to indicate
the volume of fuel in tank of a car.

A known volume V of fuel is poured
into the tank and deflection θ of the pointer is noted.

Fig. shows the variation with θ of
V.

**(a)**On Fig.,

(i) calibrate the scale at 20 × 10

^{3}cm^{3}intervals,
(ii) mark a possible position for
volume of 1.0 × 10

^{5}cm^{3}.**(b)**Suggest one advantage of this scale, as compared with a uniform scale, for measuring fuel volumes in tank of the car.

**Reference:**

*Past Exam Paper – June 2007 Paper 2 Q1*

__Solution 556:__**(a)**

(i)

{We need to find the
corresponding deflections for the volumes V in the graph. Then, we measure
these angles of deflection from the position of the pointer shown in the
figure. The one I drew below is not to scale, but you should draw it to scale.}

All the positions should be marked
to within ±5°

The positions are 40°, 70°, 90° and
102°

(ii) The angle is about (allow) 107°
→ 113° (in same diagram above)

**(b)**E.g. The meter is more sensitive at low volumes

__Question 557: [Work, Energy and Power]__
Initially, four identical uniform
blocks, each of mass m and thickness h, are spread on table.

How much work is done on blocks in
stacking them on top of one another?

A 3 mgh B 6 mgh C
8 mgh D 10 mgh

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q17*

__Solution 557:__**Answer: B.**

To stack them all on each other, we
only need to lift 3 of them, with the remaining lowest one already being on the
table.

When on the ground, a block has zero
gravitational potential energy. When at a height x above the ground, the
gravitational potential energy of the block is mgx. So, the work that needs to
be done to lift a block from the ground to a height x is equal to mgx.

So first, we need to lift 1 block by
a height h: Work done is mgh.

Then, we need to lift 1 block by a
height 2h: Work done is mg(2h) = 2mgh

Finally, we need to lift 1 block by
a height 3h: Work done is mg(3h) = 3mgh

Total work done = mgh + 2mgh + 3mgh
= 6mgh

__Question 558: [Kinematics]__
One object moves directly from P to
R.

In a shorter time, second object
moves from P to Q to R.

Which statement about the two
objects is correct for the journey from P to R?

A They have the same average speed.

B They have the same average
velocity.

C They have the same displacement.

D They travel the same distance.

**Reference:**

*Past Exam Paper – November 2013 Paper 11 & 12 Q6*

__Solution 558:__**Answer: C.**

The first object moves directly from
P to R while the second one moves from P to Q to R in a shorter time.

The distance travelled by the second
object is longer than the first one. [D is
incorrect] But since the second object reaches R in a shorter time, its average
speed (= total distance / total time) should be larger. [A is incorrect]

Since both objects starts from P and
ends at R, their displacements are both the same. [C
is correct] Average velocity = total displacement / total time. The time
taken by the second object is less, so its average velocity is greater than the
first object. [B is incorrect]

__Question 559: [Nuclear Physics + Ideal gas]__
Two deuterium (

^{2}_{1}H) nuclei travelling directly towards one another. When their separation is large compared with their diameters, they each have speed v as illustrated in Fig.
Diameter of deuterium nucleus is
1.1x10

^{-14}m.**(a)**Use energy considerations to show that initial speed v of deuterium nuclei must be approximately 2.5x10

^{6}ms

^{-1}in order that they may come into contact. Explain your working.

**(b)**For fusion reaction to occur, deuterium nuclei must come into contact.

Assuming that deuterium behaves as an
ideal gas, deduce a value for temperature of deuterium such that nuclei have an
r.m.s speed equal to the speed calculated in (a).

**(c)**Comment on answer to (b)

**Reference:**

*Past Exam Paper – November 2008 Paper 4 Q5*

__Solution 559:__
.

**(a)**
Change / loss in kinetic energy =
change / gain in electric potential energy

{Kinetic energy of 1
deuterium nuclei = ½ mv

^{2}. Both nuclei are moving with speed v. So, we need to consider the kinetic energies of both nuclei. KE of 2 deuterium nuclei = 2(½ mv^{2}).
Electric potential energy
= q

_{1}q_{2}/ 4πϵ_{o}r where q_{1}and q_{2}are the 2 charges of the particles involved. r is their separation. Since both particles are the same nuclei, they are the same charge (this is represented as q). q_{1}q_{2}= q^{2}}
2(½ mv

^{2}) = q^{2}/ 4πϵ_{o}r
2[½ (2 x 1.67x10

^{-27}) v^{2}] = (1.6x10^{-19})^{2}/ [4π (8.85x10^{-12}) (1.1x10^{-14})]**]**
Initial speed, v = 2.5x10

^{6}ms^{-1}**(b)**

pV = (1/3)Nm<c

^{2}> and pV = NkT
½ m<c

^{2}> = (3/2) kT
½ (2 x 1.67x10

^{-27}) x (2.5x10^{6})^{2}= (3/2) (1.38x10^{-23}) T
Temperature, T = 5x10

^{8}K**(c)**Any sensible comment:

This is a

__very__high temperature
This temperature is found in stars

__Question 560: [Current of Electricity]__
I-V characteristics of two
electrical components P and Q are shown below.

Which statement is correct?

A P is a resistor and Q is a
filament lamp.

B The resistance of Q increases as
the current in it increases.

C At 1.9 A the resistance of Q is
approximately half that of P.

D At 0.5 A the power dissipated in Q
is double that in P.

**Reference:**

*Past Exam Paper – June 2005 Paper 1 Q34*

__Solution 560:__**Answer: D.**

The graph for the I-V
characteristics of a filament lamp is one with a decreasing gradient [unlike the graph Q shown which has an increasing gradient] as
the potential difference V across it increases. This is because as V increases,
the temperature of the lamp increases, causing the resistance to increase. So,
at high value of V, the current I is relatively smaller compared to the current
for small values of V. [A is incorrect]

So, graph Q shown a component whose
resistance decreases with V, unlike the graph for a filament lamp. As V increases,
the current I keeps on getting greater. From Ohm’s law, I = V / R, so the
resistance R should be getting smaller. [B is
incorrect]

At current I = 1.9 A, the 2 graphs
intercept. So, they have the same value of V at that point, and hence the same
resistance. [C is incorrect]

Power dissipated = VI. At current I
= 0.5A, the value of V for graph P = 2.0V while for graph Q, V = 4.0V. So, for
graph Q p.d. V is twice, therefore the power dissipated is double that in P. [D is correct]

in question 559,why is the total change in electric potential energy not 2× q^2/4piEr and gravitational potential energy 2×Gm^2/r since both nuclei gain electrical potential energy and lose gravitational energy and in the second part we are equating this change to the kinetic energies of both the nuclei summed?

ReplyDeleteKE is possessed by the individual nucleus while electric potential energy is the energy between the 2 nuclei. It’s the same energy.

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