• 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