Shi Qi 4 Mar 2018 - 2

Question

In two widely-separated planetary systems whose suns have masses S₁ and S₂, planet P₁ of mass M₁ and planet P₂ of mass M₂ are observed to have circular orbits of equal radii respectively. If P₁ completes and orbit in half the time by P₂, it may be deduced that

A) S₁ = S₂ and M₁ = 0.25 M₂

B) S₁ = 4 S₂ only

C) S₁ = 4 S₂ and M₁ = M₂

D) S₁ = 0.25 S₂ only

E) S₁ = 0.25 S₂ and M₁ = M₂

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Answer

For the system with S₁ and P₁, the gravitational force on P₁ provides the centripetal force.

Similarly for the system with S₂ and P₂,

Since the period of P₁ is half of P₁,

Since the periods do not depend on the masses of the planets, we can only deduce that S₁ = 4 S₂.

Hence, answer is B.

Shi Qi 4 Mar 2018 - 1

Question

At a point outside the Earth and a distance of x from its centre, the Earth’s gravitational field strength is 5 N kg^-1. At the Earth’s surface, the field strength is about 10 N kg^-1. Which of the following gives an approximate value for the radius of the Earth?

A) x/5

B) x/2√2

C) x√2

D) x/√2

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Answer

At point x, the gravitational field strength is given by the formula:

At the surface of the Earth, the gravitational field strength is given by the formula:

Since g_s ≈ 2 g_x,

Hence, the answer is D.

Shi Qi 20 Feb 2018

Question

A wire that obeys Hooke’s Law of length x₁ when it is in equilibrium under tension T₁. It’s length becomes x₂ when the tension is increased to T₂. What is the extra energy stored in the wire as a result of this process?

A) 1/4 ( T2 + T1 ) ( x2 - x1 )

B) 1/4 ( T2 + T1 ) ( x2 + x1 )

C) 1/2 ( T2 + T1 ) ( x2 - x1 )

D) 1/2 ( T2 + T1 ) ( x2 + x1 )

E) ( T2 - T1 ) ( x2 - x1 )

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Answer

The extra energy is stored as elastic potential energy.

The graph of tension against total length is shown below. The extra elastic potential  energy is given by the shaded area.

Answer is C.

Shi Qi 2 Oct 2017

Question

A metal nugget floats in between some water and mercury. The densities of the metal nugget, mercury and water are 7900 kg m^-3, 13600 kg m^-3 and 1000 kg m^-3 respectively.

What is the ratio of the volume of the nugget submerged in water to that of mercury?

A) 0.541
B) 0.826
C) 0.924
D) 1.21

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Answer

Since the nugget is suspended in the liquids, the weight of the nugget is equal to the upthrust.

The nugget displaces both water and mercury, hence, the upthrust is due to the both the weight of the water displaced and the weight of the mercury displaced.

The volume of the nugget is the sum of the volume of water displaced and the volume of mercury displaced.

The answer is B.

Shi Qi 22 Sep 2017

Question

Two containers of volume 4.0m³ and 6.0m³ contain an ideal gas at pressures of 100Pa and 50Pa respectively. Their temperatures are equal. They are joined by a tube of negligible volume. The gas flows from one container to the other with no change in temperature. The final pressure will be

A) 70 Pa
B) 75 Pa
C) 80 Pa
D) 150 Pa

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Answer

Before the valve is opened, we first find an expression for the initial number of moles of gas in each container:

Now, the total initial number of moles of gas can be found:

After the valve is opened, the pressure in each container is now the same and the temperature remains the same.

The new expression for the final number of moles in each container is found again:

Now, the total final number of moles of gas can be found:

Since the total number of moles remains the same,

Answer is A.