Sunday, May 31, 2015

Physics 9702 Doubts | Help Page 158

  • Physics 9702 Doubts | Help Page 158



Question 782: [Current of Electricity]
In circuit shown, the resistance of the thermistor decreases as temperature increases.


Which graph shows variation with Celsius temperature θ of potential difference V between points P and Q?


Reference: Past Exam Paper – November 2013 Paper 13 Q37



Solution 782:
Answer: C.
V is the p.d. across the fixed resistor.

From Kirchhoff’s law, the sum of p.d. across the thermistor and the p.d. across the fixed resistor is equal to the e.m.f. of the d.c. supply.

Ohm’s law: V = IR
As the temperature increases, the resistance of the thermistor decreases, and hence the p.d. across it also decreases since V = IR. Thus, as temperature increases, the p.d. V across the fixed resistor increases (so that the sum of p.d.’s is always equal to the e.m.f.). [A and B are incorrect]

But the resistance of a thermistor does not vary linearly with temperature. So, the increase in the p.d. V with the temperature is not linear.










Question 783: [Measurement > Uncertainty]
Diagram shows an experiment to measure speed of a small ball falling at constant speed through a clear liquid in a glass tube.

There are two marks on the tube. Top mark is positioned at 115 ± 1 mm on the adjacent rule and the lower mark at 385 ± 1 mm. The ball passes the top mark at 1.50 ± 0.02 s and passes the lower mark at 3.50 ± 0.02 s.
Constant speed of the ball is calculated by (385 – 115) / (3.50 – 1.50) = 270 / 2.00 = 135 mms-1.
Which expression calculates fractional uncertainty in the value of this speed?


Reference: Past Exam Paper – June 2011 Paper 11 Q5 & Paper 13 Q4



Solution 783:
Answer: A.
Speed v = Distance s / Time t
Δv / v = (Δs / s) + (Δt / t)

Fractional uncertainty of speed = Δv / v

Distance s = 385 – 115 = 270 mm
Δs = ± (1 + 1) = ± 2 mm

Time t = 3.50 – 1.50 = 2.00s
Δt = ± (0.02 + 0.02) = ± 0.04 s









Question 784: [Kinematics]
Diagram shows a laboratory experiment in which a feather falls from rest in a long evacuated vertical tube of length L.

The feather takes time T to fall from top to the bottom of the tube.
How far will the feather have fallen from the top of the tube in time 0.50 T?
A 0.13 L                      B 0.25 L                      C 0.38 L                      D 0.50 L

Reference: Past Exam Paper – June 2012 Paper 11 Q8 & Paper 13 Q9



Solution 784:
Answer: B.
Since a vacuum, there is no air resistance.

Initial speed, u = 0 since the feather falls from rest.
The acceleration of free fall, g, acts downwards.

Equation for uniformly accelerated motion: s = ut + ½ at2

When distance s = L, the time taken to fall = T
L = 0 + 0.5gT2 = 0.5gT2

When time to fall = 0.5T, let the distance be s
s = 0 + 0.5g(0.5T)2
Distance s = gT2 / 8

From L = 0.5gT2, gT2 = 2L

Distance s = gT2 / 8 = 2L / 8 = L/4 = 0.25L

           









Question 785: [Dynamics > Resultant force]
Diagrams show a metal cube suspended from a spring balance before and during immersion in water.

A reduction in the balance reading occurs as a consequence of the immersion.
Which statement is correct?
A The balance reading will be further reduced if the cube is lowered further into the water.
B The balance reading during immersion corresponds to the upthrust of the water on the cube.
C The forces acting on the vertical sides of the cube contribute to the change in the balance reading.
D The gravitational pull on the cube is unchanged by the immersion.
Reference: Past Exam Paper – November 2010 Paper 12 Q13



Solution 785:
Answer: D.
Before the immersion, the balance reads the weight of the object, which is downwards. The weight of the cube is the gravitational pull on the cube and this is unchanged in a specific location since weight W = mg. [D is correct]

As the cube in immersed, it experiences an upthrust force upwards. This force is due to the difference in pressures acting on the top and bottom of the cube.

Pressure = hρg
Since the cube has a specific size, the difference pressure acting on the cube, as long as it is fully immersed, is unchanged. Thus, the upthrust is unchanged no matter how further the cube is immersed. [A is incorrect]

The forces acting on the vertical sides of the cube contribute are equal in magnitude since they acts at the same depth in the water. So, they cancel each other and are not the cause in the change in the balance reading. [C is incorrect]

Thus, the balance reading is not the upthrust but the weight (which acts downwards) minus the upthrust (which acts upwards). The balance reads the resultant. [B is incorrect]









Question 786: [Work, Energy and Power]
Steel ball is falling at constant speed in oil.
Which graph shows variation with time of the gravitational potential energy Ep and the kinetic energy Ek of the ball?


Reference: Past Exam Paper – June 2005 Paper 1 Q15 & June 2008 Paper 1 Q18



Solution 786:
Answer: B.
Kinetic energy Ek = ½ mv2
Since the speed is constant, the kinetic energy is also constant.

Gravitational potential energy Ep = mgh
Speed = Distance / Time

As the speed is constant, the rate at which the distance (h) decreases is constant. So, the decrease in Ep is also constant.

7 comments:

  1. 9702/12/O/N/10 question number 7

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html

      Delete
  2. Car X is travelling at half the speed of car Y. Car X has twice the mass of car Y.
    Which statement is correct?

    A Car X has half the kinetic energy of car Y.
    B Car X has one quarter of the kinetic energy of car Y.
    C Car X has twice the kinetic energy of car Y.
    D The two cars have the same kinetic energy.

    which statement is correct and why??? please help the real answer is (A)

    ReplyDelete
    Replies
    1. see at
      http://physics-ref.blogspot.com/2016/12/car-x-is-travelling-at-half-speed-of.html

      Delete
  3. i am thankful for this website

    ReplyDelete
  4. in q.786 , why is option A incorrect?

    ReplyDelete
    Replies
    1. Consider the graph for potential energy.

      Gradient of the graph = ΔEp / Δt = mgΔh / Δt
      Gradient = mg (Δh/Δt) since mg is constant

      but Δh/Δt is the speed (which is constant). So, the graph should be constant; that is, a straight line

      Delete

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