# 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}) = 2×10^{7 }W = 20 MW
1MW = 1 × 10

^{6}W
Power = 2×10

^{7 }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 t

^{2}, 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 at

^{2}
For an object starting from rest (initial
speed, u = 0),

Distance s = ½ at

^{2}
So, s / t

^{2}= 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 / t

^{2}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:__**Answer: B.**

Since the angle on the right of the
diagonal vector is 120° with the horizontal, the angle on the LEFT of the
diagonal vector is 180° – 120° = 60°with the horizontal.

Consider the resultant vector.

Since the 2 vectors given are of the
same magnitude, we can assume that the triangle formed when the resultant is
formed will at least be an isosceles triangle – that is, the angle that these 2
vectors make with the resultant vector will have the same value. Let this angle
be x.

Sum of angles in a triangle is 180°.

60° + x + x = 180°

Angle x = (180° – 60°) / 2 = 60°

Since all the angles are equal in the
triangle, it is actually an equilateral triangle. So, the sides will have the
same magnitude.

Thus, resultant vector = 10N

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

ReplyDeletealways makes me happy when i see solutions to the mcq's being solved haha

ReplyDelete