Friday, April 3, 2015

Physics 9702 Doubts | Help Page 103

  • Physics 9702 Doubts | Help Page 103



Question 528: [Work, Energy and Power]
Object is thrown into the air.
Which graph shows how potential energy Ep of the object varies with height h above the ground?

Reference: Past Exam Paper – November 2008 Paper 1 Q16



Solution 528:
Answer: A.
The height above the ground is h. So, as h increases, the position of the object above the ground is higher.

Potential energy Ep = mgh

Thus, as h increases, the potential energy Ep also increases and the increase is linear since the power of h is 1 (Ep = mgh1).










Question 529: [Waves > Intensity]
Light wave of amplitude A is incident normally on surface of area S. Power per unit area reaching the surface is P.
Amplitude of light wave is increased to 2A. Light is then focussed on to a smaller area 1/3 S.
What is the power per unit area on this smaller area?
A 36P                          B 18P                          C 12P                          D 6P

Reference: Past Exam Paper – June 2013 Paper 11 Q24



Solution 529:
Answer: C.
Power per unit area = Power / Area = Intensity
When the surface area is S and the amplitude is A, the power per unit area (intensity) = P

Now, consider the first change.
The amplitude of the light wave is doubled. Since the intensity is directly proportional to the square of amplitude, doubling the amplitude causes the intensity to be increased by a factor of (22 =) 4.

Consider the second change.
The surface area is now 1/3 S. But intensity is inversely proportional to the area. So, decreasing the surface area by a factor of 3 causes the intensity to increase by a factor of 3.

So, the first change causes P to increase by a factor of 4 and the second change causes it to increase by a factor of 3.

Thus, the new power per unit area (= intensity) is now (4 × 3) P = 12P











Question 530: [Vectors]
A glider is descending at constant speed at angle of 15° to horizontal. Diagram shows directions of the lift L, air resistance R and weight W acting on the glider.

Which vector triangle could represent the forces acting on the glider?

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q10



Solution 530:
Answer: D.
Since the glider is descending at constant speed, the net acceleration and thus the net force is zero. So, the sum of the lift L and air resistance R should be equal to the weight W. In such a case, the vertical and horizontal components of the forces are all balanced by each other.

Forces are vectors and so have both magnitudes and directions. The direction of air resistance R is at 15° to (above) the horizontal.

Only choice D correspond to the above descriptions.










Question 531: [Matter > Young Modulus]
Two wires P and Q are made from same material.
Wire P is initially twice diameter and twice length of wire Q. Same force, applied to each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
A ½                             B 1                              C 2                              D 4

Reference: Past Exam Paper – November 2010 Paper 12 Q21



Solution 531:
Answer: A.
Since wires P and Q are made up of the same material, they have the same Young modulus.

Young modulus, E = stress / strain = (F/A) / (e/l) = Fl / Ae
Cross-sectional area A = πr2 = π(d/2)2

Wire P has twice the diameter and twice the length of wire Q.

For wire P: Length = 2l and diameter = 2d
Young modulus Ep = F(2l) / (2d/2)2ep = 2Fl / d2ep

For wire Q: Length = l and diameter = d
Young modulus Eq = Fl / (d/2)2eq = 4Fl / d2eq

Since Ep = Eq               (Young modulus of both P and Q are the same)
1/ep = 2/eq
Ratio ep / eq = ½


No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 103