• 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 × 10³ cm³ intervals,

(ii) mark a possible position for volume of 1.0 × 10⁵ cm³.

(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 (²₁H) 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^-14m.

(a) Use energy considerations to show that initial speed v of deuterium nuclei must be approximately 2.5x10⁶ms^-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 mv². 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 mv²).

Electric potential energy = q₁q₂ / 4πϵ_or where q₁ and q₂ 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). q₁q₂ = q²}

2(½ mv²) = q² / 4πϵ_or              

2[½ (2 x 1.67x10^-27) v²] = (1.6x10^-19)² / [4π (8.85x10^-12) (1.1x10^-14)]]

Initial speed, v = 2.5x10⁶ms^-1

(b)

pV = (1/3)Nm<c²>      and pV = NkT

½ m<c²> = (3/2) kT

½ (2 x 1.67x10^-27) x (2.5x10⁶)² = (3/2) (1.38x10^-23) T

Temperature, T = 5x10⁸K

(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]