MCQ 9702/12/O/N/10


Uncertainties Problems:

1. x is to be determined from this equation: x= P – Q

P is measured as 1.27 ± 0.02 m

Q is measured as 0.83 ± 0.01 m

What is the percentage uncertainty in x?

How to solve:

P(1.27) – Q(0.83) = x = 0.44

0.02 + 0.01 = 0.03 (maximum uncertainty)

Percentage uncertainty is   0.03 ⁄x × 100 =  6.(81) = 6.82%

 

Uniform acceleration Problems

1. A moving body undergoes uniform acceleration while travelling in a straight line between pts. X, Y and Z. The distance XY ≡ YZ = 40 m. The time to travel from X to Y is 12 sec and from Y to Z is 6 sec.

What’s the acceleration of the body?

How to solve:

You have the four equations of motions. Select the one which is best to use in this situation. That is s = ut + ½at²

Make a system of equations.

1. For t = 12,  40 = 12u + ½a (12)²

2. For t = 6,  40 = 6v +  ½a (6)²;        v here is the speed it has at Y.

v= u + at = u + 12a ⇒ For t = 6,  40 = 6( u + 12a) +  ½a (6)²;

We must find a, therefore we extract u from the first equation.

u = (40 – 72a) / 12

and replace it in equation number two.

40 = 6[ (40-72a)/12 + 12a] +  ½a (6)²

this will give a = 0.37 m/s/s

 

Deforming Solids

1. Two wires P and Q are made of the same material.

Wire P is initially twice the diameter and twice the length of the wire Q. The same force is applied to each wire, causing them to extend elastically.

What is the ratio of extension in P to extension in Q?

The theory: 

  • Twice the length is twice the extension
  • Twice the radius is half the extension

Therefore: Extension in P = 2 / ( 2 x 2 )

ratio is  2/4 ⇒ 1/2.

Superposition of Waves

1. A double slit experiment using light of wavelength 600 nm, results in fringes being produced on a screen. The fringe separation is found to be 1.0 mm

When the distance between the double slits and the viewing screen is increased by 2.0 m, the fringe separation increases to 3.0 mm.

What is the separation of the double slits?

The theory:

For the double slit experiment we have Λ=ax/D

Attention: These buggers have set a trap so re-read the question carefully paying attention to the bold words.

How to solve:

Make a system of equations:

1. 600× 10^(-9) = 0.001x / D

2. 600× 10^(-9) = 0.003x/ D + 2

Solve for D in equation 1 and replace it in equation 2

0.003x =  600 × 10^(-9) [ (0.001x + 2 × 600× 10^(-9)) /  600× 10^(-9)]

solve fore x ⇒ x = [2×600×10^(-9)]/ 0.002 = 0.6×10^(-3)m = 0.6 mm

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