Wednesday, November 13, 2019

An object shaped as a hemisphere rests with its flat surface on a table. The object has radius r and density ρ.


Question 33
An object shaped as a hemisphere rests with its flat surface on a table. The object has radius r and density ρ.

The volume of a sphere is 4/3 π r3.

Which average pressure does the object exert on the table?
A 1/3 ρr2                      B 1/3 ρr2g                   C 2/3 ρr                      D 2/3 ρrg





Reference: Past Exam Paper – March 2019 Paper 12 Q14





Solution:
Answer: D.

The object is shaped as a hemisphere (half a sphere).


Pressure = Force / Area
Here, the force is the weight of the object and the area is the cross-sectional area of the object which is in contact with the table.


Volume of object = ½ × volume of sphere
Volume of object = ½ × 4/3 πr3 = 2/3 πr3

Density = mass / volume

Mass of object = ρV = 2/3 πρr3
Weight of object = mg = 2/3 πρr3g


The area in contact with the table has the shape of a circle of radius r.
Area = πr2


Pressure = Force (or weight) / Area
Pressure = (2/3 πρr3g) / (πr2)
Pressure = 2/3 ρrg

Tuesday, November 12, 2019

A stone is thrown vertically upwards from a point that is 12 m above the sea. It then falls into the sea below after 3.4 s.


Question 32
A stone is thrown vertically upwards from a point that is 12 m above the sea. It then falls into the sea below after 3.4 s.

Air resistance is negligible.

At which speed was the stone released when it was thrown?
A 3.5 m s-1                        B 6.6 m s-1                        C 13 m s-1                        D 20 m s-1





Reference: Past Exam Paper – March 2019 Paper 12 Q7





Solution:
Answer: C.

The motion of the stone consists of both an upward and a downward movement. So, we need to be careful with the signs of the vector quantities involved.

Let the upward direction be positive. Any quantity pointing downwards would be negative.

Initial velocity = u       (???)

Take the point at which the stone is released to be the origin.
The sea is at a distance of 12 m below the origin. So, the displacement would be negative.
Displacement s = – 12 m

Acceleration a = – 9.8 m s-2 (this is also negative as it is downwards)

Time t = 3.4 s


Consider the equation of uniformly accelerated motion:
s = ut + ½ at2
- 12 = 3.4u + (½ × -9.8 × 3.42)
- 12 = 3.4u = 3.4u – 56.644
3.4u = -12 + 56.644
u = 44.644 / 3.4

Initial speed u = 13.1 m s-1   

Monday, November 11, 2019

Some gas in a cylinder is supplied with thermal energy q.


Question 34
Some gas in a cylinder is supplied with thermal energy q.

The gas does useful work in expanding at constant pressure p from volume V0 to volume VF, as shown.

Which expression gives the efficiency of this change?






Reference: Past Exam Paper – June 2016 Paper 11 Q16





Solution:
Answer: C.

The gas is supplied with thermal energy q.
(Input) thermal energy = q


Work done by expanding gas = pΔV
Work done by expanding gas = p × (Vfinal – Vinitial) = p (VF – VO)

This is the output energy.
Output energy = p (VF – VO)


Efficiency = output energy / input energy
Efficiency = p (VF – VO) / q
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