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Friday, September 4, 2015

Physics 9702 Doubts | Help Page 194

  • 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 to
Two 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°.



6 comments:

  1. thank u that was so useful I seriously didn't know how to solve them

    ReplyDelete
  2. Question 943: never thought of it that way... haha, i used cos rule, it was wrong wkwk

    ReplyDelete
  3. always makes me happy when i see solutions to the mcq's being solved haha

    ReplyDelete
  4. why cant we use cos rule in Q 943?

    ReplyDelete

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