# 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)
= 17.5x10

^{6}
(1.96x10

^{7}) – 9810x + 8142.3x = 17.5x10^{6}
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 clockwiseB 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.

V

_{1}is potential difference between S and P. V

_{2}is potential difference between S and Q.

What is the value of V

_{1}– V

_{2}?

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.**

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. V

_{1}= 1.0V.

Consider the loop containing point Q.

From the potential divider equation, the p.d. across the 3.0kΩ resistor is

p.d. V

_{2}= [3 / (2+3)] x 2 = 6 / 5 = 1.2V

Thus, V

_{1}– V

_{2}= 1.0 – 1.2 = -0.2V

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

ReplyDeleteWslm.

DeleteThe 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