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Question 422: [Kinematics > Linear motion]

Stone is dropped from top of a tower of height 40 m. Stone falls from rest and air resistance is negligible.

What time is taken for stone to fall the last 10 m to the ground?

A 0.38 s                       B 1.4 s                         C 2.5 s                         D 2.9 s

Reference: Past Exam Paper – June 2007 Paper 1 Q8

Solution 422:

Answer: A.

Consider the equation for uniformly accelerated motion: s = ut + ½ at²

First, let’s calculate the time t₁ for the stone to fall a distance of 40m to reach the ground.

s = ut + ½ at²

40 = 0 + 0.5(9.81)t₁²

Time t₁ = 2.86s.

Now, let’s calculate the time t₂ for the stone to fall a distance of 30m.

30 = 0 + 1.5(9.81)t₂²

Time t₂ = 2.47s

Thus, the additional time required to fall the last 10m is given by

Time = t₁ – t₂ = 0.38s

Question 423: [Current of Electricity]

Two wires P and Q made of same material are connected to same electrical supply. P has twice the length of Q and one-third of the diameter of Q, as shown.

What is ratio of current in P to current in Q?

A 2 / 3                         B 2 / 9                         C 1 / 6                         D 1 / 18

Reference: Past Exam Paper – June 2013 Paper 12 Q35

Solution 423:

Answer: D.

Ohm’s law: V = IR

Current I = V / R        (the current is inversely proportional to the resistance)

From Kirchhoff’s law, there is the same p.d. across P and Q.

So, Ratio = I_P / I_Q = R_Q / R_P

Resistance R of a wire = ρL/A = ρL/ (π[d/2]²)

For wire Q, the length is L and the diameter is 3d (instead of d in the above equation).

Resistance of wire Q, R_Q = ρL/(π(1.5d)²) = (4/9)x(ρL/πd²)

For wire P, the length is 2L (instead of L in the equation for resistance R of a wire) and the diameter is d.

Resistance of wire P, R_P = 2ρL/(π(0.5d)²) = 8 x(ρL/πd²)

Ratio = I_P / I_Q = R_Q / R_P = (4/9) / 8 = 1 / 18

Question 424: [Current of Electricity > Internal resistance]

Power supply of electromotive force (e.m.f.) 12 V and internal resistance 2.0 Ω is connected in series with 13 Ω resistor.

What is power dissipated in the 13 Ω resistor?

A 8.3 W                      B 9.6 W                       C 10 W                        D 11 W

Reference: Past Exam Paper – June 2013 Paper 11 Q32

Solution 424:

Answer: A.

Power P dissipated in resistor = I²R

Current I in circuit = V / R_T = 12 / (2.0+13) = 0.80A

Power P dissipated in resistor = I²R = (0.8)² (13) = 8.3W  

Question 425: [Forces > Newton’s laws of motion]

Box of mass 8.0 kg rests on horizontal, rough surface. A string attached to box passes over smooth pulley and supports a 2.0 kg mass at its other end.

When box is released, a friction force of 6.0 N acts on it.

What is acceleration of the box?

A 1.4 m s^–2                  B 1.7 m s^–2                  C 2.0 m s^–2                  D 2.5 m s^–2

Reference: Past Exam Paper – November 2008 Paper 1 Q11 & June 2012 Paper 12 Q13 & June 2015 Paper 11 Q13

Solution 425:

Answer: A.

The weight of the 2.0kg mass acts downwards. So, the 2.0kg mass pulls the 8.0kg box to the right with a force (equal to its weight) of

Weight of 2.0kg mass = 2.0 (9.81) = 19.62N

The force of friction opposes motion and thus it acts to the left.

Resultant force on the system = 19.62 – 6.0 = 13.62N = ma.

It is important to note that the system consists of an 8.0kg box AND a 2.0kg mass. So, the total mass in the system is 8.0 + 2.0 = 10.0kg. So, both the box and the mass need to be accelerated.

ma = 13.62N

Acceleration a = 13.62 / 10 = 1.36 ≈ 1.4ms^-2