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Practice each question, then open the answer or worked solution.
In the expression P = E l2 m–5 G–2, E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.
From the problem, P = E l2 m–5 G–2
E is the energy = [ML2T-2]
m is the mass = [M]
L is the angular momentum = [ML2T1]
G is the gravitational constant = [M-1L2T2]
Substituting the values we get, [P] = [M0L0T0]
If the velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.
The principle of homogeneity is used to solve this problem
21-1-13-2Let m = kcahbGc
Solving the above we get,
m = kc1/2h1/2G-1/2 = k√ch/G
Let L = kcahbGc
Solving the above we get,
L = kc-3/2h1/2G1/2 = k√hG/c3
Let T = cahbGc
Solving the above we get,
L = kc-5/2h1/2G1/2 = k√hG/c5
An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that T = k/R √r3/g where k is a dimensionless constant and g is acceleration due to gravity.
From Kepler’s third law, we know that
T2 a3 where T2 is the square of time period of the satellite revolving around a planet and is proportional to the cube of the radius of the orbit r3.
T2 r3
T r3/2
T depends on R and g
T r3/2gaRb
In an experiment to estimate the size of a molecule of oleic acid, 1mL of oleic acid is dissolved in 19mL of alcohol. Then 1mL of this solution is diluted to 20mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water, forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film, we can calculate the thickness of the film, which will give us the size of oleic acid molecule.
Read the passage carefully and answer the following questions:
e) What will be the volume of oleic acid in one drop of this solution?
a) Oleic is dissolved in the alcohol because it does not get dissolved in water.
b) With the help of lycopodium powder one can measure the area over which the oleic acid spreads as lycopodium powder clears the circular area when oleic acid is added.
c) 20mL of oleic acid contains 1mL of oleic acid.
This means that each mL of solution contains 1/20 mL of oleic acid.
1mL of this solution is diluted to 20mL by adding alcohol.
Therefore, each mL of solution prepared, volume of oleic acid = (1/20)(1/20) = 1/400mL
d) To calculate the volume of n drops of this solution of oleic, burette and measuring cylinder can be used.
e) The volume of oleic acid in one drop = 1/400mL
a) How many astronomical units (AU) make 1 parsec?
b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2) degree from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
c) Mars has approximately half of the earth’s diameter. When it is closer to the earth it is at about ½ AU from the earth. Calculate at what size it will disappear when seen through the same telescope.
a) From the definition, 1 parsec is equal to the distance at which 1 AU long arc subtends an angle of 1s.
Using the definition, we can say that
1 parsec = (3600)(180)/π AU
= 206265 AU
= 2 × 105 AU
b) Given that the sun’s angular diameter from the earth is 1/2 degree at 1 AU.
Angular diameter of the sun-like star at a distance of 2 parsec
= [(1/2)/(2)(2)(105)] degree
= 1/8 × 10-5 degree
= 7.5 × 10-5 arcmin
When the sun appears like a start through the telescope that has a magnification of 100, the angular diameter of the star is
= (100)(7.5 × 10-5)
= 7.5 × 10-3 arcmin
But the angular size of the sun appears as 1 arcmin to the eyes as the eyes cannot resolve smaller than 1 arcmin because of atmospheric fluctuations.
c) Given that,
Dmars/Dearth = 1/2
We also know that Dearth/Dsun = 1/100
Therefore, Dmars/Dsun = 1/2 × 1/100
At 1AU, the sun’s diameter = (1/2) degree
Therefore, diameter of mars = (1/400) degree
At 1/2 AU, mars diameter = (1/400)(2) = (1/200) degree
With 100 magnification, mars diameter = (1/2) degree = 30’
Therefore, it can be said that the value is larger than the resolution limit because of atmospheric fluctuations and hence it looks magnified.
Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E = mc2, where c is the speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV where 1 MeV = 1.6 × 10-13J, the masses are measured in unified equivalent of 1u is 931.5 MeV.
a) Show that the energy equivalent of 1 u is 931.5 MeV.
b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
a) Using Einstein’s mass-energy relation, the energy that is equivalent to the given mass can be calculated
1 amu = 1 u = 1.67 × 10-27 kg
Applying E = mc2
E = 931.5 MeV
b) As E = mc2
m = E/c2
Which means that 1u = 931.5 MeV/c2
The dimensionally correct relation of 1 amu = 931.5 MeV
Class 11 Chapter 2 Physics Exemplar Concepts for Units and Measurement| Section Number | Topic |
| 1 | The International System of Units |
| 2. | Measurement of Length |
| 3 | Measurement of Mass |
| 4 | Measurement of Time |
| 5 | Accuracy, Precision of Instruments and Errors in Measurement |
| 6 | Significant Figures |
| 7 | Dimensions of Physical Quantities |
| 8 | Dimensional Formulae and Dimensional Equations |
| 9 | Dimensional Analysis and Its Applications |