Units and Measurements

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E=mc^{2}, 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 !MeV=1.6\times 10^{-13}J, 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) The energy that is comparable to a given mass can be computed using Einstein's mass-energy relation. $1amu=1u=1.67\times 10^{-27}kg$ On Applying $E=mc^{2}$ we get, E = 931.5 MeV b) As $E=mc^{2}$...

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a) How many astronomical units (AU) make 1 parsec?
b) Consider the sun like a star at a distance of 2 parsec. 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.

a) 1 parsec is the distance at which 1 AU long arc subtends an angle of 1s, according to the definition. Using the definition, we can write, 1 parsec = (3600)(180)/π AU = 206265 AU = 2 × 105 AU b)...

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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:
What will be the volume of oleic acid in one drop of this solution?

The volume of oleic acid in one drop is 1/400mL

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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:
a) What would be the volume of oleic acid in each mL of solution prepared?
b) How will you calculate the volume of n drops of this solution of oleic.

a) 1 mL of oleic acid is found in every 20 mL of oleic acid. This signifies that 1/20 mL of oleic acid is present in each mL of solution. Adding alcohol dilutes 1 mL of this solution to 20 mL. As a...

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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:
a) Why do we dissolve oleic acid in alcohol?
b) What is the role of lycopodium powder?

a) Because oleic acid does not dissolve in water, it is dissolved in alcohol. b) When oleic acid is introduced, lycopodium powder clears the circular area. This makes it possible to measure the area...

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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.

Kepler's third law states that, $T^{2} \propto a^{3}$ i.e., square of time period $\left(T^{2}\right)$ of a satellite revolving around a planet, is proportional to the cube of the radius of the...

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A physical quantity X is related to four measurable quantities a, b, c and d as follows: X=a^{2}b^{3}c^{5/2}d^{-2}. The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

The given physical quantity is $X=a^{2}b^{3}c^{5/2}d^{-2}$ Percentage error in X is given as (∆x/x)(100) Percentage error in a is given as (∆a/a)(100) = 1% Percentage error in b is given as...

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The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as v=\frac{\pi}{8}\times \frac{Pr^{4}}{\eta l} where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula ML^{-1}T^{-1}. Check whether the equation is dimensionally correct.

Dimension of the given physical quantity is as follows, [V] = dimension of volume/dimension of time $=[L^{3}]/[T]$ $=[M^{-1}T^{-2}]$ LHS $=[L^{3}T^{-1}]$ RHS $=[L^{3}T^{-1}]$ LHS = RHS Hence, the...

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(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?
(b) Moon is seen to be of (½)° diameter from the earth. What must be the relative size compared to the earth?

(a) Because the distance between the moon and the earth is greater than the radius of the earth, it is considered as an arc. Let the length of the arc be $R_{e}$ Distance between the moon and the...

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If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which of the following can, also, be taken to express length, mass, and time in terms of the three chosen fundamental quantities?
a) mass of the electron (m_{e})
b) universal gravitational constant (G)
c) charge of the electron (e)
d) mass of proton (m_{p})

Correct answers are a) mass of electron b) universal gravitational constant and d) mass of proton

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It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer From the mentioned examples 2.3 and 2.4,  we get the following data Distance of the Moon from Earth is 3.84 x 108 m Distance of the Sun from Earth is 1.496 x 1011 m And Sun’s diameter is 1.39...

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The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Answer We are given that time taken by light from the quasar to reach the observer is t = 3.0 billion years = 3.0 x 109 years = 3.0 x 109 x 365 x 24 x 60 x 60 s = 94608000 x 109  s...

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A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s–1).

Answer: According to the question, speed of sound in water is v = 1450 m s–1 Time between after reflection, generation and the reception of the echo 2t = 77.0 s Then, the time taken for the sound...

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A LASER is a source of very intense, monochromatic, and the unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?

Answer According to the question, time taken = 2.56 s This is the time taken by the laser beam to return to Earth after getting reflected by the Moon’s surface The speed of laser light = speed of...

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The unit of length convenient on the nuclear scale is a fermi: 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to
about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant
for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer: We know that the expression for Radius of the nucleus is given by - r = r0 A1/3 where ro = 1.2 f = 1.2 x 10-15 m Now, assuming that the nucleus is spherical. Volume of nucleus can be...

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Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg m–3. Are the two densities of the same order of magnitude? If so, why?

Answer According to the question the diameter of sodium is 2.5 Angstrom = 2.5 x 10-10 m Therefore, the radius will be half of the above value, i.e., 1.25 x 10-10 m Expression for the volume of...

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It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?

Answer According to the question, the total time = 100 years = 100 x 365 x 24 x 60 x 60 s And the given error in 100 years is 0.02 sTherefore, error in 1 second = 0.02/100 x 365 x 24 x 60 x 60= 6.34...

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A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit: as v →0, θ →0, as
expected. (We are assuming there is no strong wind and that the rain falls vertically
for a stationary man). Do you think this relation can be correct? If not, guess the
correct relation.

Answer The principle of homogeneity of dimensional equations states that the dimensions of L.H.S are equal to the dimensions of R.H.S. In expression v = tan θ, where tan θ is a trigonometric...

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The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: a mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m.

Answer: According to the question, Mass = 2 x 1030 kg and Radius = 7 x 108 m Expression of volume is \[\]\[V=\frac{4}{3}\pi {{r}^{3}}\] \[V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 7\times...

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The nearest star to our solar system is 4.29 light-years away. How much is this
distance in terms of parsecs? How much parallax would this star (named Alpha
Centauri) show when viewed from two locations of the Earth six months apart in its
orbit around the Sun?

Answer We know that 1 light year is the distance travelled by light in a year's time period. Therefore, 1 light year = 3 x 108 x 365 x 24 x 60 x 60 1 ly = 9.46 x 1015 m As a result, distance...

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The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer: According to the question, diameter of Earth’s orbit is 3 × 1011 m Then, the radius of Earth’s orbit r = 1.5 × 1011 m Then, the distance parallax angle is θ = 1″ (s) θ = 4.847 × 10–6 rad....

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Precise measurements of physical quantities are a need for science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer: For the advancement of science, precise measurement is required. Time intervals are measured using an ultra-short laser pulse. The interatomic separation is determined using X-ray...

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Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hilltops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer: The line of sight is an imaginary line that connects the item and the observer's eye. When we look at adjacent things, we notice that they move quickly in the other way when the line of...

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One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer: According to the question, radius = 0.5 A = 0.5 x 10-10 m Expression for the volume of the sphere is - $\frac{4}{3}\pi {{r}^{3}}$ Upon substituting values, we have - $=\frac{4}{3}\times...

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A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

$m=\frac{{{m}_{0}}}{\sqrt{1-{{v}^{2}}}}$ Guess where to put the missing c. Answer: We are given the following relation - $m=\frac{{{m}_{0}}}{\sqrt{1-{{v}^{2}}}}$ We can reduce the above relation to...

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The photograph of a house occupies an area of 1.75 cmon a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?

Answer Expression for arial Magnification = Area of image / Area of object Upon substituting values => =1.55/1.75 x 104 = 8.857x 103 And we know that linear Magnification = √Arial magnification...

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A student measures the thickness of a human hair by looking at it through a
microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of the hair?

Answer Given, magnification of the microscope is 100Also, average width of the hair is 3.5 mm in the field of view of the microscope. So, the actual thickness of hair will be =3.5 mm/100 = 0.035...

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A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J =1 kg ms–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α–1 β–2 γ2 in terms of the new units.

Answer Given that 1 calorie = 4.2 J = 4.2 kg m2 s–2 Expression for the standard formula for the conversion is as follows - \[\]\[\frac{Unit(given)}{Unit(new)}={{\left( \frac{{{M}_{1}}}{{{M}_{2}}}...

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