Physics 9702 Doubts | Help Page 80
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 = IP / IQ
= RQ / RP
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, RQ
= 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, RP
= 2ρL/(π(0.5d)2) = 8 x(ρL/πd2)
Ratio = IP / IQ
= RQ / RP = (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 / RT
= 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
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