• Physics 9702 Doubts | Help Page 57

Question 328: [Pressure]
Diagram shows two liquids, labeled P and Q, which do not mix. The liquids are in equilibrium in an open U-tube.

What is the ratio of density of P / density of Q?
A ½                             B 2/3                           C 3/2                           D 2

Reference: Past Exam Paper – November 2004 Paper 1 Q20 & November 2014 Paper 13 Q22



Solution 328:
Answer: A.
Since the liquids are in equilibrium in the open U-tube, the pressure at which the liquids meet should be the same. That is, at the position where they are in contact (without mixing), the pressure that liquid P exerts at that point and the same as the pressure exerted by liquid Q at the point.

That position is at a height of x from the bottom of the U-tube.

The pressure that a liquid would exert at that point depends on the length of that liquid above that point.

Pressure due to liquid P = (2x) (ρ_P) g
Pressure due to liquid Q = (x) (ρ_Q) g

Since the pressures are the same,
(2x) (ρ_P) g = (x) (ρ_Q) g
ρ_P / ρ_Q = xg / 2xg = 1 / 2









Question 329: [Nuclear Physics]

(a) One isotope of gold is represented as ¹⁹⁷₇₉Au. State number of neutrons in one nucleus of this isotope.

(b) In an α-particle scattering experiment, an α-particle approaches isolated gold nucleus, as illustrated in Fig.

Complete Fig to show path of the α-particle as it passes by, and moves away from, the gold nucleus.

(c) α-particle in (b) is replaced by one having greater initial kinetic energy. State what change, if any, will occur in final deviation of the α-particle.

Reference: Past Exam Paper – November 2001 Paper 2 Q8



Solution 329:

(a) Number of neutrons = (197 – 79 =) 118

(b) The path should have a correct shape in the correct position relative to the nucleus.

(c) There will be a smaller deviation.

Question 330: [Work, Energy, Power]

Sphere has volume V and is made of metal of density ρ.

(a) Write down an expression for mass m of the sphere in terms of V and ρ.

(b) Sphere is immersed in a liquid. Explain the apparent loss in weight of the sphere.

(c) Sphere in (b) has mass 2.0 x 10^–3 kg. When sphere is released, it eventually falls in the liquid with a constant speed of 6.0 cms^–1.

(i) For this sphere travelling at constant speed, calculate

1. its kinetic energy

2. its rate of loss of gravitational potential energy

(ii) Suggest why it is possible for sphere to have constant kinetic energy whilst losing potential energy at a steady rate.

Reference: Past Exam Paper – November 2001 Paper 2 Q4



Solution 330:

(a) Mass m = ρV

(b) The pressure of the liquid depends on the depth. The bottom of the sphere has a greater pressure on it than the top of the sphere. So, the resultant force or pressure is upwards.

{The resultant force or pressure referred above is the resultant of the 2 pressures (one at top of sphere and one at the bottom) mentioned above. There is a loss in apparent weight whenever the resultant downward force is less than the weight of the object. So, the difference in pressure causes an upward force. This in turn causes the resultant downward force to decrease. As stated in part (c), when the sphere is released, it eventually falls in the liquid.}

(c)

(i)

1. Kinetic energy = ½ mv² = 0.5 (2x10^-3) (6x10^-2)² = 3.6x10^-6 J

2.

Potential energy = mgh

{Rate of loss of potential energy = mgh / t = mgv}

Rate of loss of potential energy = mgv = (2x10^-3) (9.8) (6x10^-2) = 1.2x10^-3 Js^-1

(ii) The potential energy of the sphere is given / lost to the liquid to overcome the drag forces or to produce eddies or friction etc.

Question 331: [Kinematics > Linear motion]

A falling stone strikes soft ground at speed u and suffers a constant deceleration until is stops. Which one of the following graphs best represents the variation of the stone’s speed v, with distance s, measured downwards, from the surface of the ground?

Reference: Past Exam Paper – N81 / II / 6

Solution 331:

Answer: A.

The stone hits the soft ground with speed u. The stone then sinks in the soft ground, with a constant deceleration until it stops.

Distance s is measured downwards, from the surface of the ground.

Deceleration is the decrease in speed with time. So, in a speed-time graph, a straight line would be seen. However, the choices present speed-distance graphs. The required graph is not a straight line (as will be explained). [C is incorrect]

Consider the following example. The initial speed is 10ms^-1 and the constant deceleration is 2ms^-2. That is, after each second, the speed decreases by 2ms^-1.

After 1^st second, speed changes from 10ms^-1 to 8ms^-1. [The fraction of speed that has changed is {decrease in speed / original speed} 2 / 10 = 0.2]

After 2^nd second, speed changes from 8ms^-1 to 6ms^-1. [Fraction of speed that has changed is 2 / 8 = 0.25]

After 3^rd second, speed changes from 6ms^-1 to 4ms^-1. [Fraction of speed that has changed is 2 / 6 = 0.33]

As can be seen, as time increase, the fraction change in speed becomes greater, even though the deceleration is constant.

But we are interested in the change in speed with distance. Obviously, as time increases, the distance s also increases. Therefore, initially, when s is smaller, the fraction change in speed is smaller, and when s is large, the fraction change in speed is bigger. This is represented by the gradient of the speed-distance graph.

The gradient of the speed-distance graph changes and this change is an increase – that is, the gradient of the graph increases. [D and E incorrect]

But the way that the gradient increases should also be considered. In the example given, consider the 1^st and 2^nd seconds: the increase is 0.25 – 0.2 = 0.05. Now, consider the 2^nd and 3^rd seconds: the increase in 0.33 – 0.25 = 0.08. That is, the increase in gradient should not be uniform. The increase in gradient becomes more significant as s increases. [B is incorrect since the increase in gradient increases.]