FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Wednesday, March 11, 2015

Physics 9702 Doubts | Help Page 82

  • Physics 9702 Doubts | Help Page 82



Question 431: [Pressure]
Bore-hole of depth 2000 m contains both oil and water as shown. Pressure due to the liquids at bottom of the bore-hole is 17.5 MPa. Density of the oil is 830 kg m–3 and density of the water is 1000 kg m–3.

What is depth x of the oil?
A 907 m                      B 1000 m                    C 1090 m                    D 1270 m

Reference: Past Exam Paper – June 2013 Paper 11 Q21



Solution 431:
Answer: D.
The pressure at the bottom of the bore-hole (= 17.5MPa) is the sum of pressures exerted by water and oil on the bottom.

Pressure at a point = hρg
where h is the length of the column of the liquid above that point.

For the column of water,
Pressure exerted by water = (2000 – x) (1000) (9.81)
Where x is the length of the column of oil

For the column of oil,
Pressure exerted by oil = x (830) (9.81)

The sum of pressures is 17.5MPa.
(2000 – x) (1000) (9.81) + x (830) (9.81) = 17.5x106
(1.96x107) – 9810x + 8142.3x = 17.5x106
Length x = 1260m










Question 432: [Electric field]
Diagram shows insulating rod with equal and opposite point charges at each end. Electric field of strength E acts on rod in a downwards direction.

Which row is correct?
resultant force                         resultant torque
A         zero                             clockwise
B         downwards                 clockwise
C         zero                             anti-clockwise
D         downwards                 anti-clockwise

Reference: Past Exam Paper – November 2011 Paper 11 Q32 & Paper 13 Q31



Solution 432:
Answer: C.
The electric force acting on a charge Q is given by F = EQ.

The direction of the electric field is from positive to negative – that is, it shows the direction of the electric force acting on a positive charge. So, the direction of the electric force on the charge +Q is in the same direction as the electric field E (downwards). The electric force on the charge –Q is in the opposite direction (upwards).

Resultant force = E(+Q) + E(-Q) = 0

Both forces cause an anti-clockwise moment at the ends of the rod. So, there is a resultant anti-clockwise torque.










Question 433: [Measurement > Uncertainty]
Quantity X varies with temperature θ as shown.

θ is determined from corresponding values of X by using this graph.
X is measured with percentage uncertainty of ±1 % of its value at all temperatures.
Which statement about the uncertainty in θ is correct?
A The percentage uncertainty in θ is least near 0 °C.
B The percentage uncertainty in θ is least near 100 °C.
C The actual uncertainty in θ is least near 0 °C.
D The actual uncertainty in θ is least near 100 °C.

Reference: Past Exam Paper – November 2012 Paper 11 Q6



Solution 433:
Answer: C.
Percentage uncertainty in X is given by (ΔX / X) x 100%.

The graph obtained is a straight line.
The general equation for a straight line is: y = mx + c (do not confuse the x-axis with the X quantity in the question)

So, the equation of the graph shown in the question is
X = mθ + c = mθ         (since the y-intercept c is zero)

Considering the percentage uncertainties,
(ΔX / X) x 100% = (Δθ / θ) x 100%

From the above equation, the percentage uncertainty in θ is equal to the percentage uncertainty in X, which is itself a constant. So, the percentage uncertainty in θ is also constant and equal to 1%. [A and B are incorrect]

So, (Δθ / θ) x 100% = 1%
Δθ / θ = 0.01

The actual uncertainty Δθ = 0.01 (θ)
Thus, when θ is smallest, the actual uncertainty is also smallest.










Question 434: [Current of Electricity > Potential difference]
Cell of e.m.f. 2.0 V and negligible internal resistance is connected to network of resistors shown.

V1 is potential difference between S and P. V2 is potential difference between S and Q.
What is the value of V1 – V2?
A +0.50 V                   B +0.20 V                   C –0.20 V                   D –0.50 V

Reference: Past Exam Paper – June 2007 Paper 1 Q33



Solution 434:
Answer: C.
S is connected to the negative terminal of the cell, so it is at a potential of 0V.

From Kirchhoff’s law, the sum of p.d. in each loop is equal to the e.m.f. of the battery.

Consider the loop containing point P.
Since the 2 resistors have the same resistance, the p.d. across each of them will be equal. So, the potential difference between point P and S is 1.0V and since point S is at a potential of 0V, point P should be at a potential of 1.0V. V­1 = 1.0V.

Consider the loop containing point Q.
From the potential divider equation, the p.d. across the 3.0kΩ resistor is
p.d. V2 = [3 / (2+3)] x 2 = 6 / 5 = 1.2V

Thus, V1 – V­2 = 1.0 – 1.2 = -0.2V




2 comments:

  1. Assalamualaikum, I have difficulties understanding solution 434, why does the pd between p and s equals to 1?

    ReplyDelete
    Replies
    1. Wslm.
      The sum of p.d.`s in the middle loop is 2.0V.
      Potential divider equation: V1 = [R1 / (R1+R2)] × E
      (p.d. across P and S,) V1 = [5 / (5+5)] × 2 = 1.0 V

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 82