Physics 9702 Doubts | Help Page 194
Question 940: [Current
of Electricity > Resistance]
Two copper wires X and Y have same
volume. Wire Y is four times as long as wire X.
What is the ratio resistance of wire
Y / resistance of wire X ?
A 4 B
8 C 16 D 64
Reference: Past Exam Paper – June 2008 Paper 1 Q34
Solution 940:
Answer: C.
The volume V of both wires are the
same. Both are copper wires, so the resistivity ρ
of both are the same.
Volume V = AL where A is the cross-sectional area and L is the length
(Cross-sectional area, A = V / L)
Let the cross-sectional area of wire
X be A and its length be L.
Resistance of wire X = ρL / A (= R)
The length of wire Y is 4L. Since
the volume of both wires are the same,
Cross-sectional area of wire Y = V /
4L = 0.25A
Resistance of wire Y = ρ (4L) /
0.25A = (4 / 0.25) × (ρL / A) = 16R
Ratio = resistance of wire Y /
resistance of wire X = 16R / R = 16
This question required great care.
One small slip and a factor of 2 can be lost or the reciprocal of the answer can
be given.
Question 941: [Work,
Energy and Power]
What is the average power output of
a laser that can deliver 0.20 J of energy in 10 ns?
A 2 nW B 20 mW C
200 kW D 20 MW
Reference: Past Exam Paper – June 2014 Paper 11 Q14
Solution 941:
Answer:
D.
Time
= 10ns = 10 × 10-9 s
Power
= energy / time = 0.20 / (10×10-9) = 2×107 W = 20 MW
1MW = 1 × 106 W
Power = 2×107
W = 20 MW
Question 942: [Kinematics
> Graph]
At time t = 0, a body moves from
rest with constant acceleration in a straight line. At time t, body is distance
s from its rest position.
A graph is drawn of s against t2,
as shown.
Which statement describes the
acceleration of the body?
A It is equal to half the value of
the gradient of the graph.
B It is equal to the value of the
gradient of the graph.
C It is equal to twice the value of
the gradient of the graph.
D It is equal to the reciprocal of
the gradient of the graph.
Reference: Past Exam Paper – June 2013 Paper 13 Q8
Solution 942:
Answer: C.
Equation for uniformly accelerated
motion: s = ut + ½ at2
For an object starting from rest (initial
speed, u = 0),
Distance s = ½ at2
So, s / t2 = ½ a
All the choices provided deal with
the gradient of the graph. So, we need to relate the above equation with the gradient.
On the graph s / t2 is
the gradient so, to find the acceleration, twice the gradient is needed (twice
the gradient = 2 × ½ a = a).
Question 943: [Vectors]
Two forces, each of 10 N, act at a
point P as shown in diagram. The angle between the directions of the forces is
120°.
What is the magnitude of the
resultant force?
A 5N B 10N C
17N D 20N
Reference: Past Exam Paper – June 2003 Paper 1 Q3
Solution 943:
Go toTwo forces, each of 10 N, act at a point P as shown in the diagram. The angle between the directions of the forces is 120°.
thank u that was so useful I seriously didn't know how to solve them
ReplyDeleteQuestion 943: never thought of it that way... haha, i used cos rule, it was wrong wkwk
ReplyDeleteSame lol
Deletealways makes me happy when i see solutions to the mcq's being solved haha
ReplyDeletewhy cant we use cos rule in Q 943?
ReplyDeleteyou can (if you have the available data)
Delete