Units and Measurements Class 11 Important Questions

Very Short Answer Questions (1 Mark Questions)

Ques. Define the S.I. Solid Angle Unit.

Ans. The steradian is the SI unit for solid angles. The angle formed by a spherical plane of unit square metre area in the centre of a sphere with a radius of unit length is defined as one steradian.

Ques. Name physical quantities using Electron Volt and Pascal as their units.

Ans. Energy and pressure are physical quantities with electron volt and pascal units, respectively.

Ques. What is the measurement unit for sound amplitude?

Ans. Decibel. A decibel is one-tenth of a bel, and it is a unit of sound pressure level measurement.

Ques. Is a mile the same on land as it is at sea?

Ans. A mile on the water is not the same as a mile on land. A nautical mile is 6,076 feet long and is used on the ocean (1,852 meters). A land mile (also known as a statute mile) is 5,280 feet long (1,609 meters).

Ques. What is the definition of an acre?

Ans. An acre is a measurement unit for land area. The term acre comes from the Old English word acre, which means "a field." The acre was traditionally defined as the area that a yoke of oxen could plough in a day. The area of one acre is 43,560 square feet (4,840 square yards). A hectare is equal to 0.4047 acres (4,047 square meters).

Ques. What is the definition of horsepower?

Ans. Horsepower is a measurement of the force exerted by a horse when it pulls something. Engine and motor output is often referred to as "horsepower," which is a unit of measurement for power or "rate of work."

Ques. What is the definition of the metric system?

Ans. The decimalized metric system is a decimalized measuring system.

The metre, gram, and litre are the three most popular basic units in the metric system.

Read More: Fundamental and Derived Units of Measurement

Short Answer Questions (2 Marks Questions)

Ques. Science requires precise measurements of physical quantities. To determine an aircraft's speed, for example, a precise technique of determining its locations at closely spaced instants of time is required. This was the driving force for the development of radar during World War II. Consider a variety of situations in modern research when exact measurements of length, time, mass, and other variables are required. Also, wherever possible, provide a quantitative estimate of the accuracy required.

Ans. 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 spectroscopy. The mass spectrometer was designed to determine the mass of individual atoms.

Ques. Explain this frequent observation clearly: When looking out the window of a fast-moving train, close trees, homes, and other things appear to move swiftly in the opposite direction of the train's velocity, yet distant objects (hilltops, the Moon, the stars, and so on) appear to be immobile. (In reality, these distant things appear to move with you because you are aware that you are moving.)

Ans. The line of sight is a made-up line that links the observer's eye to the object. When we look at items that are close together, we observe that they move swiftly in the other way when the line-of-sight shifts. The distant objects, on the other hand, appear to be immobile since the line of sight does not shift fast.

Ques. Noah is 48000 grams in weight. How much does he weigh in kilograms?

Ans. Noah's weight is 48000 grams, hence the solution is that. Kilograms and grams are both measuring units, and we divide the amount by 1000 to convert grams to kilograms.

1/1000 kg = 1 g

48000 g = 48000/1000 kg

= 48 kg

As a result, Noah's weight is 48 kg.

Ques. Kate commutes 1.5 miles every day on her bicycle. In a week, how many kilometres does she ride?

Ans. A mile is 1.6 kilometres long. Multiply 1.5 miles by 1.6 kilometres to convert miles to kilometres.

1 mile equals 1.6 kilometres

1.5 miles = 1.5 × 1.6 kilometres

= 2.4 kilometres

Multiply 2.4 kilometres by 7 to account for the fact that there are 7 days in a week.

= 2.4 km × 7 km

= 16.8 kilometres

Kate bikes 16.8 kilometres every week as a result.

Read More: Measurements of Length

Ques. Two caesium clocks, if left to operate for 100 years without being disturbed, are said to differ by only 0.02 seconds. What does this mean for the precision of a conventional caesium clock in measuring a one-second interval?

Ans. Total time = 100 years = 100 × 365 × 24 × 60 × 60 s

In 100 years, the error is 0.02 seconds.

In 1 second, there was an error = 0.02/100 × 365 × 24 × 60 × 60

= 6.34 × 10^-12 s

The precision of a conventional caesium clock in measuring a one-second interval is 10^-12 seconds.

Ques. A grocer's balance measures the mass of a box at 2.30 kg. The box now contains two gold pieces with weights of 20.15 g and 20.17 g. What is?

(a) the box's overall weight,

(b) the mass difference between the parts in order to rectify significant figures?

Ans. The package weighs 2.30 kg.

and the first gold piece weighs 20.15 grams

The second gold piece weighs 20.17 grams.

The total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg

Because 1 has the fewest decimal points, the total mass equals 2.7 kg.

The mass difference is 0.02 g (20.17 – 20.15).

The total mass Equals 0.02 g since 2 is the smallest number of decimal places.

Ques. (a) A thread and a metre scale are handed to you. How will you determine the thread's diameter?

(b) A screw gauge features a circular scale with 200 divisions and a pitch of 1.0 mm. Do you believe that increasing the number of divisions on the circular scale may arbitrarily enhance the screw gauge's accuracy?

(c) Vernier callipers are used to determine the average diameter of a thin brass rod. Why is it assumed that a set of 100 diameter measurements will produce a more trustworthy estimate than a set of only 5 measurements?

Ans. (a) Wrap the thread around a pencil many times to produce a coil with the twists close together. Now, using a metre scale, determine the length of this coil. The diameter of the thread is provided by the relationship Diameter = L/n, where L is the length of the coil and n is the number of turns of the coil.

(b) The screw gauge's least count is equal to the pitch divided by the number of divisions on the circular scale.

Now, as a result, increasing the number of divisions on the circular scale should reduce the screw gauge's least count. As a consequence, the accuracy of the screw gauge will increase. However, this is simply a hypothesis. When the number of turns is raised, there will be many more challenges.

(c) In 100 observations, the likelihood of committing random mistakes is lowered to a greater extent than in 5 observations.

International System of Units

Read More: Unit Conversion: Concept, Types, Volume Measurement

Long Answer Questions (3 Marks Questions)

Ques. Ultrasonic waves are used by a SONAR (sound navigation and ranging) to identify and locate items underwater. The time delay between the creation of a probe wave and the receiving of its echo after reflection from an enemy submarine is determined to be 77.0 seconds in a submarine equipped with a SONAR. What is the hostile submarine's distance? (In water, the speed of sound is 1450 m s^–1).

Ans. v = 1450 m s^–1 is the speed of sound in water.

2t = 77.0 s is the time between creation and receipt of the echo after reflection

t = 77.0/2 = 38. 5 s is the time it takes for sound waves to reach the submersible.

We know, v = d/t 

d = tv, enemy's submarine distance

As a result, d = v × t = (1450 × 38. 5) = 55825 m = 55.8 × 10³ m or 55.8 km is calculated.

Ques. It is well known that during a total solar eclipse, the moon's disc almost fully covers the Sun's disc. Determine the moon's approximate diameter.

Ans. The Moon's distance from Earth is 3.84 × 10⁸ metres.

The Sun's distance from the Earth is 1.496 × 10¹¹ metres.

The diameter of the sun is 1.39 × 10⁹ m.

The angular diameter of the Sun is = 1920′′ = 1920 × 4.85 × 10^-6 rad 

= 9.31 × 10^-3 rad. [1-inch equals 4.85 × 10^-6 rad]

Because the moon's disc fully covers the sun's disc during a total solar eclipse, both the sun and the moon must have the same angular diameter.

As a result, the moon's angular diameter θ is 9.31 × 10^-3 rad.

S = 3.8452 × 10⁸ m is the earth-moon distance.

As a result, D = θ × S= 9.31 × 10^-3 × 3.8452 × 10⁸ m = 35.796 × 10⁵ m is the moon's diameter.

Read More: Angular Momentum

Ques. Calculate the average mass density of a sodium atom with a size of 2.5 as a starting point. (Use the known Avogadro's number and sodium atomic mass values.) When you compare it to the mass density of sodium in its crystalline form, you will notice a significant difference: 970 kg m^–3 Is the order of magnitude of the two densities the same? If so, what is the reasoning behind it?

Ans. 2.5 A = 2.5 × 10^-10 m is the diameter of sodium.

As a result, the radius is 1.25 × 10^-10 metres.

V = (4/3)πr³ is the volume of a sodium atom.

= (4/3) × (22/7) × (1.25 × 10^-10) ³= 8.177 × 10^-30 m³

One mole atom of sodium has a mass of 23 g = 23 × 10^-3 kg.

6.023 × 10²³ atoms make up one mole of sodium.

As a result, M = 23 × 10^-3/6.023 × 10²³ = 3.818 × 10^-26 kg is the mass of one sodium atom.

Therefore,

M/V = 3.818 × 10^-26/8.177 × 10^-30 atomic mass density of sodium

= 4669.2 kg m^-3 = 0.46692 × 10⁴

The density of sodium in its solid state is 4669.2 kg m^-3, whereas it is 970 kg m^-3 in the crystalline phase. As a result, they are in a different order. Atoms are closely packed in the solid state, but in the crystalline state, they organise themselves in a void-filled sequence. Thus, as an outcome, the density of the solid phase is larger than that of the crystalline phase.

Read More: Avogadros Number

Ques. The Sun is a heated plasma (ionized matter) with a temperature of over 107 K in its inner core and over 6000 K on its outer surface. No material can exist in a solid or liquid state at these temperatures. In terms of densities of solids, liquids, and gases, where do you think the Sun's mass density will fall? Check the following data to see whether your assumption is correct: The Sun has a mass of 2.0 × 10³⁰ kilogrammes and a radius of 7.0 × 10⁸ metres.

Ans. Mass = 2 × 10³⁰ kg

Radius = 7 × 10⁸ m
Volume, \(V=\frac{4}{3}\pi r^3\)

\(=\frac{4}{3} \times \frac{22}{7} \times (7 \times 10^8)^3\)

\(= \frac{88}{21} \times 343 \times 10^{24}\ m^3\)

= 1437.33 × 10²⁴ m³

Density = \(\frac{\text{Mass}}{\text{Volume}} = \frac{2 \times 10^{30}}{1437.3 \times 10^{24}} = 1.39 \times 10^3 kg/m^3\)

The density falls in between solids and liquids. Its density is owing to the sun's inner layer's strong gravitational influence on the outer layer.

Read More: Gravitational Force and Law of Gravitation

Very Long Answer Questions (5 Marks Questions)

Ques. In a test to determine the size of an oleic acid molecule In 19 mL of alcohol, 1 mL of oleic acid is dissolved. Then, using alcohol, dilute 1 mL of this solution to 20 mL. Now, 1 drop of this diluted solution is dropped into a shallow trough of water. The solution forms a one-molecule thick film on the water's surface. The film is now uniformly dusted with lycopodium powder, and its diameter is measured. We can compute the thickness of the film using the volume of the drop and the area of the film, which will give us the size of the oleic acid molecule.

Answer the following questions after carefully reading the passage:

1. Why is oleic acid dissolved in alcohol?
2. What role does lycopodium powder play?
3. How much oleic acid would be in each mL of the solution prepared?
4. In what way will you determine the volume of n drops of oleic acid solution?
5. How much oleic acid will be in one drop of this solution?

Ans. (a) Oleic acid is dissolved in alcohol since it does not dissolve in water.

(b) When uniformly dusted, lycopodium powder covers the whole surface of the water.

Oleic acid does not dissolve in water when a drop of the prepared solution is placed on it. Instead, it spreads over the water's surface, pushing the lycopodium powder away from the drop's landing place. As a result, we can calculate the area covered by oleic acid.

(c) The volume of oleic acid in each mL of the solution made = \(\frac{1}{20}\)mL × \(\frac{1}{20}\) = \(\frac{1}{400}\)mL

(d) A burette and measuring cylinder may be used to compute the volume of n droplets of this oleic acid solution, and the number of drops can be counted.

(e) If n drops of the solution equal 1 mL, the amount of oleic acid in one drop is \(\frac{1}{(400)n}\)mL.

Ques. (a) One parsec equals how many astronomical units (A.U.)?

(b) The diameter of Mars is roughly half that of the Earth. It is roughly 1/2 A.U. from the earth when it is closest to it. Calculate the size of it as viewed through the same telescope.

Ans. (a)

1 parsec = \(\left( \frac{1\ \mathrm{A.U.}}{1\ arc\ sec} \right)\)

1 deg = 3600 arc sec

1 arc sec = \(\frac{\pi}{3600 \times 180}\) radians

Therefore, 1 parsec = \(\frac{3600 \times 180}{\pi}\) A.U. = 206265 A.U. ≈ 2 × 10⁵ A.U.

(b ) \(\frac{D_{mars}} {D_{earth}}\) = 1/2, \(\frac {D_{earth}}{D_{sun}}\) = 1/400 

∴\(\frac{D_{mars}} {D_{sun}}\) =1/800. 

At 1 A.U., the sun seems to be 1/2 degree in diameter, whereas mars appear to be 1/1600 degree in diameter.

Mars will seem to be 1/800 degree in diameter at 1/2 A.U. With a magnification of 100, Mars seems to be 1/8-degree 60/8 = 7.5 = arcmin.

Due to atmospheric variations, this exceeds the resolution limit. As a result, it seems to be exaggerated.

Read More: Mirror Formula and Magnification

Ques. (a) Demonstrate that 1 u has an energy equivalent of 931.5 MeV.

(a) A pupil writes 1 u = 931.5 MeV as the relation. The teacher points out that the relationship is erroneous in terms of dimensions. Fill in the missing information.

Ans. Given:

m = 1

u = 1.67 × 10^-27 kg

c = 3 × 10⁸ m/s

According to the formula, E = mc²

E = 1.67 × 10^-27 × 3 × 10⁸ × 3 × 10⁸

= 1.67 × 10^-27+16 × 9 J

\(= \frac{(1.67)(9)(10^{-11})}{(1.6)(10^{-13})}\) MeV

= 939.4 MeV

≅ 931.5 MeV

(b) 931.5 MeV will release 1 u mass converted to total energy.

1 amu = 931.5 MeV, on the other hand, is dimensionally wrong.

E = mc² → 1 uc² ≅ 931.5 MeV, will be dimensionally correct.

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