Tuesday, March 10, 2015

Physics 9702 Doubts | Help Page 81

  • Physics 9702 Doubts | Help Page 81



Question 426: [Current of Electricity]
Light-dependent resistor (LDR) is connected in series with resistor R and a battery.

Resistance of LDR is equal to the resistance of R when no light falls on the LDR.
When light intensity falling on the LDR increases, which statement is correct?
A The current in R decreases.
B The current in the LDR decreases.
C The p.d. across R decreases.
D The p.d. across the LDR decreases.

Reference: Past Exam Paper – June 2012 Paper 12 Q36



Solution 426:
Answer: D.
The resistance of the LDR decreases as the light intensity increases.

Since the LDR and resistor R are connected in series, the current in both of them are the same (same in whole circuit).

Ohm’s law: V = IR
Since the resistance of the LDR decreases, the potential difference V across it also decreases.  

As the battery has a fixed e.m.f., this causes the p.d. across R to increase. [C is incorrect]

A decrease in the resistance of the LDR causes the overall resistance RT in the circuit to decrease. From Ohm’s law, the current I in the circuit is given by I = V / RT. So, the current increases. This current is the same current that flows through the LDR and R. [A and B are incorrect]










Question 427: [Dynamics > Momentum]
Diagram shows masses and velocities of two trolleys about to collide.

After impact they move off together.
What is total kinetic energy of the trolleys after the collision?
A 1.3 J                                     B 12 J                         C 18 J                          D 19 J

Reference: Past Exam Paper – June 2007 Paper 1 Q12



Solution 427:
Answer: B.
For any collision in a closed system, the law of conservation of momentum applies, i.e. the sum of momentum before collision is equal to the sum of momentum after collision.

Momentum = mv

Total momentum before collision = 2(4) + 4(1) = 12Ns

After collision, both trolleys move together (with the same speed v). So, the total mass = 2 + 4 = 6kg. The sum of momentum after collision should be equal to the momentum of collision.

(2+4)v = 12
Speed v = 12 / 6 = 2ms-1

Kinetic Energy = ½ mv2 = ½ (6)(22) = 12J











Question 428: [Electric field]
Diagram shows electric field near a point charge and two electrons X and Y.

Which row describes forces acting on X and Y?
direction of force                    magnitude of force on X
A         radially inwards                      less than force on Y
B         radially inwards                      greater than force on Y
C         radially outwards                    less than force on Y
D         radially outwards                    greater than force on Y

Reference: Past Exam Paper – November 2008 Paper 1 Q29



Solution 428:
Answer: B.
The direction of electric field lines is given from a positive charge to a negative charge. So, the electric force on an electron (which is negatively charged) would be in the opposite direction, towards the positive charge. So, the direction of the electric force is radially inwards.

The electric force F = Q1Q2 / 4πϵor2
So, the electric field is inversely proportional to the square of the separation r between the 2 charges Q1 and Q2. Therefore, the magnitude of the electron force on X is greater than the electric force on Y.









Question 429: [Waves]
A circular bowl of diameter 400mm contains water at rest. If its side is tapped gently, a completely circular pulse can be produced on the surface of the water which travels inwards with a speed of 250mms-1. The radius of the pulse and its direction of travel, 1 second after the pulse is produced are
A zero, stationary
B 50mm, outwards
C 50mm, inwards
D 150mm, outwards
E 150mm, inwards

Reference: Past Exam Paper – N77 / II / 12



 Solution 429:
Answer: B.
Radius of circular bowl = 400 /2 = 200mm

Speed of pulse = 250mms-1
In 1 second, the distance travelled by the pulse would be
Distance travelled = Speed x Time = 250 x 1 = 250mm

However, the pulses formed on the surface of the water would reflect each other at the centre of the circular bowl, where they meet. So, any specific pulse would move (from the side of the bowl) to the centre of the bowl (a distance of 200mm from the side of the bowl), and then it gets reflected back (now moving towards the side of the bowl).

In 1 second, a pulse moves a total distance of 250mm. For the first distance of 200mm, it moves inwards (towards the centre). Thus, for the remaining 50mm, the pulse is moving outwards, at a radius of 50mm.










Question 430: [Medical Physics > X-ray]
(a) Quality of image produced using X-rays depends on sharpness and contrast. State what is meant by, and briefly explain the causes of,
(i) sharpness
(ii) contrast      

(b) Parallel beam of X-ray photons is produced by X-ray tube with 80 keV across it. Beam has its intensity reduced to one half of its original value when it passes through a thickness of 1.0 mm of copper.
(i) Describe energies of the X-ray photons in the beam.
(ii) Determine linear absorption coefficient μ of the X-ray photons in copper.
(iii) Suggest, with reason, the effect on linear absorption coefficient if the beam is comprised of 100 keV photons.

Reference: Past Exam Paper – November 2002 Paper 6 Q7



Solution 430:
(a)
(i)
Sharpness is the clear distinction between boundaries.
Example (cause): parallel X-ray beam / point source

(ii)
Contrast is the (large) differences in blackening of different regions (allow changes in colour)
Example (cause): differences in attenuation coefficient

(b)
(i)
The maximum {possible} energy of a photon {here} is 80keV. {For energies}Below this energy of 80keV, there is a continuous spectrum {a continuous distribution of energies} with sharp peaks {superimposed on the distribution}.

(ii)
Intensity I = I0exp(-μx)
{Intensity is halved when thickness x = 1.0mm. So, I/I0 = 1/2.}
½ = exp(-μ)
Linear absorption coefficient μ = 0.693mm-1

(iii) The X-rays are now more penetrating. So, linear absorption coefficient μ is smaller.




4 comments:

  1. number 36, 9702/11/m/j/12..why is it b ?

    ReplyDelete
    Replies
    1. See solution 762 at
      http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-154.html

      Delete
  2. 11) Two equal masses travel towards each other on a frictionless air track at speeds of 60 cm s–1 and
    40 cm s–1. They stick together on impact.
    60cms–1 40cms–1
    What is the speed of the masses after impact?

    A 10 cm s–1 B 20 cm s–1 C 40 cm s–1 D 50 cm s–1

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-169.html

      Delete

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