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NCERT Technical Solutions 11 Physics Chapter 2 Units and Measurements

Question 1.

Fill in the blanks.

(a) the degree of cubic side 1 cm is equal to ……… m3.

(b) A solid cylinder area of ​​2.0 cm and a length of 10.0 cm is equal to mm… (mm) 2.

(c) A vehicle traveling at a speed of 18 km h-1 covers ………… m in 1 s.

(d) Lead congestion 11.3. Its density is …… .. g cm-3 or ……………. kg m-3.

Answer:

 

Question 2.

Fill in the blanks with the appropriate conversion of units

(a) 1 kg m2 s-2 = …………. g cm2 s-2

(b) 1 m = ………………. ly

(c) 3 m s-2 = ………. km h-2

(d) G = 6.67 x 10-11 N m2 (kg) –2 = ……………. (cm) 3 s – 2 g1.

Answer:

 

Question 3.

Calorie can be a unit of heating or energy and is equal to 4.2 J when 1 J = 1 kg m2 s-2. Suppose we use a unit system where the unit weight is equal to α kg, the unit length is equal to β m, the unit of time is γ s. Indicate that the calorie content is 4.2 α-1 β-2 γ2 in terms of new units.

Answer:



Question 4.

Explain this statement clearly:

“Calling an equal value ‘big’ or ‘small’ means nothing but standard specification for comparison”. To this end, rearrange the following statements as needed:

(a) atoms are the smallest particles

(b) the aircraft is traveling at high speed

(c) Jupiter’s mass is incredibly large

(d) the air inside the chamber contains a large number of molecules

(e) A proton is much larger than an electron

(f) the noise speed is much lower than the speed of sunlight.

Answer:

  1. Atoms are very small when compared with a ball.
  2. The plane is traveling at high speed when parked next to a car.
  3. Jupiter’s mass is much larger than Earth.
  4. The air inside the room contains a very large number of molecules when placed close to the number of objects inside the room.
  5. No change is required.
  6. No change is required.

Question 5.

The new length unit is selected as the speed of sunlight during vacuum is uniform. What is the distance between the Sun and the Earth in terms of a new unit if light takes 8 to 20 minutes to cover this distance?

Answer:

New unit length = 3 × 108 ms-1

The distance between the planet and the sun

= (8mins20) × 3 ×10

8

ms

−1

 

= 500 × 3 × 108 ms-1

∴The distance between the earth and the Sun in terms of new units

=

500 × 3 ×10

8

3 ×

10

8

 

= 500 new units.

Question 6.

Which of the following is the most accurate device for measuring length:

(a) vernier caliper with 20 sections in the pay system

(b) 1 mm voice screw gauge and 100 sections on circular scale

(c) a device that will measure the length of the wavelength

Answer:

 

(c) Minimum amount of tool w 6000 A

(the average of the wavelength of actinic ray) = 6 x 10-7 m

∴ (c) is the most accurate tool.

Question 7.

The reader measures the thickness of a person’s hair by looking at it with a magnification microscope 100. He checks 20 times again, and finds that the normal diameter of the hair inside the microscope viewing area is 3.5 mm. What is the measure of hair strength?

Answer:

 

Question 8.

Answer the following:

(a) is provided with a cord and a meter scale. How will you measure the width of the series?

(b) The screw gauge consists of a 1.0 mm pitch and 200 parts on a circular scale. does one think it is possible to extend the accuracy of the screw gauge unnecessarily by increasing the separation value on the circular scale?

(c) The center diameter of a pure copper rod should be measured with vernier caliper. Why might it be a set of 100 scales that are expected to provide a more reliable estimate than just a 5-point scale?

Answer:

(a) Uma; it is done by twisting a known number of turns over the pencil, turning it into close contact. Then the length taken by each turn will know the size of the rope.

(b) Yes, because the minimum value of the screw gauge is provided by

 

i.e. the minimum number of screw gauge is equal to the value of the split on a circular scale. So with the increase in the maximum number of categories in a circular scale, the smallest value will be better. So the accuracy of the screw gauge will increase.

(c) Increasing the number of observations, increases the reliability because the intermediate error is reduced. the most effective value possible

 

Question 9.

The picture of the house occupies a 1.75 cm2 section on a 35 mm slide. The slide is displayed on the screen so the area of ​​the house on the screen is 1.55 m2. What is the expansion of the project screen line layout?

Answer:

Given The location of the house in the photo = 1.75 cm2

House area on screen = 1.55 m2 = 1.55 x 104 cm2

 

Question 10.

State the number of large numbers in the following:

  1. 0.007 m2
  2. 2.64 x 104 kg
  3. 0.2370 gem-3
  4. 6,320 J
  5. 6.032 N nr-2
  6. 0.0006032 m2

Answer:

  1. 1
  2. 3
  3. 4
  4. 4
  5. 4
  6. 4

Question 11.

The length, width, and thickness of the oblong sheet metal are 4.234 m, 1.005 m and a pair of .1 cm respectively. Provide land and sheet capacity to adjust key figures.

Answer:

Length l = 4.234 m,

Diameter b = 1.005 m,

Dimensions t = 2.01 cm = 0.0201 m

Location = 2 × (lb + bt + It)

= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 4.234 × 0.0201)

= 8.72 m2

(Summary of three important values)

Volume = lbt

= 4.234 × 1.005 × 0.0201

= 8.55 × 10-2 m3

Question 12.

The box weight measured by the grocer’s in it’s balance is 2.3 kg. Two pieces of gold for 20.15 g and 20.17 g are added to the box. What

(a) the rest of the box

(b) the difference between a pile of fragments to fix important figures?

Answer:

(a) Total weight of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg

Since the smallest total volume is 2, therefore, the total weight of the box = 2.3 kg.

(b) Weight difference = 2.17 – 2.15 = 0.020 g

As there are two important numbers that the difference in quantity to the necessary values ​​is correct

0.020 g.

Question 13.

Visual value P is defined in four visual a, b, c and d as follows:

 

Percentage error measurements during a, b, c and d are 1%, 3%, 4% and 2% respectively. What’s wrong with sharing P? If the P value calculated using the above correlation appears to be 3.763, what value should you shorten the result for?

Answer:

 

Question 14.

A book with many printing errors contains four different particle removal formulas:

(a) y = sin 2π

t

T

 

(b) y = sin

(c) y =

a

T

sin

t

a

 

(d) y = (a√2) (sin 2π

t

T

+ cos 2π

t

T

)

(a = maximum particle emissions, υ = particle velocity, T = base motion mass). Release incorrect formulas for size reasons.

Answer:

Round of the work of the circle i. e., the angle is unlimited

 

Question 15.

A popular relationship in physics associates ‘moving weight’ m ‘with resting weight’ m0 of particles in terms of its speed v and the speed of sunlight, c. (This relationship began to emerge as a result of Einstein’s theory of relativity because of Albert Einstein). The boy remembers the relationship almost perfectly but forgets where to place the c. He writes:

 

Guessing where you put them lost c.

Answer:

In redesign, we have

 

Since the left hand side has no balance, so the right hand side should be blank. this could be ………………

 

Question 16.

The simplest unit in the atomic scale is considered angstrom and defined by A: 1 A = 10–10 m. The atomic scale is about 0.5 A. What is the atomic volume in m3 of the hydrogen atom molecule?

Answer:

Atomic volume 1 =

Unspecified control sequence \ cfrac

 

Unspecified control sequence \ cfrac

X 3.14 x (0.5 x l (T-10) m3 = 5-23 x 10-31 m3.

According to Avogadro’s hypothesis, one hydrogen molecule contains 6 023 x 1023 atoms.

∴ Atomic molecule of molecules 1 hydrogen atoms = 6.023 x 1023 x 5.23 x 10–31 = 3.15 x 10–7 m3.

Question 17.

A gram is a complete gas molecule at Centigrade level and the pressure takes up 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a hydrogen molecule? (Take the hydrogen molecule scale to 1 A). Why is this rate so great?

Answer:



This high level is due to the gaps in the gas molecules being much larger than the size of molecules.

Question 18.

Explain this common observation: Looking out the window of a speeding train, nearby trees, houses, etc. they appear to be moving rapidly toward the opposite direction of the train, but distant objects (hilly peaks, moon, stars, etc.) appear to be standing still. (In fact, as you know it just goes away, these remote things seem to go with you).

Answer:

Close-up objects reduce the angle of magnitude inside the eye than distant objects. when moving, the transition between these angles is a small number of objects farther away than nearby objects. so the distant objects appear to be standing but the near objects seem to move in the other direction.

 

Question 19.

The ‘parallax’ principle is used during the determination of the distances of the farthest stars. The AB basis is that the line joins the two Earth’s spheres in a six-month orbit along its orbit around the Sun. That is, the base is about the diameter of the earth ≈3 x 1011 m. However, even the nearest stars are so far away that with such an extended base, they show only parallax of the 1st (second) order of the arc about. Parsec can be a simple unit of length on a star scale. is an object object that can represent 1 (second) parallax of the arc from the different edges of the base sufficient for the gap from Earth to the Sun. how much is parsec per meter?

Answer:

Earth’s distance from the Sun = = b = 1/2 × the circumference of the earth ⇒b = 0.5 × 3 × 10 11 mb = 1.5 × 10 11 m 1 parsec = D = b / θ = 1 11 1.5 × 10 11 m = (3600 1) 1.5 × 10 11 = 3600 1 × 180 π 1.5 × 10 11 m = 3 × 10 16 m

 

Question 20.

The nearest star is 4.29 light years. What is the distance in stages? How parallel is this star (called Alpha Centauri) when viewed from two different points of the earth six months apart in its orbit around the Sun?

Answer:

As we know, the unit of astronomy 1 = 9.46 x 1015 m

∴ 4.29 light age = 4.29 x 9.46 x 1015 = 4.058 x 1016 m

 

Question 21.

Accurate estimates of physical value are the latest trend. for example, in order to determine the speed of an enemy aircraft, one must have an accurate way of looking for its locations at very different times of your time. Only then will we hope to wrap it up with a flack. This was one of the motives behind the introduction of radar in World War II. consider different examples in modern science where accurate measurements of length, time, weight etc. And, wherever you like, it provides a glimpse of the amount of precision needed.

Answer:

  • Length: Indoor, sports, satellite navigation
  • Mass: in order to add the correct amount of salt to the solution, the satellite’s size must be accurately measured.
  • Time: To study chemical diversity, research into various activities in the universe.

Question 22.

Just as accurate measurements are needed in science, it is equally important that you be prepared to make difficult price estimates using unconventional ideas and human observations. think of ways to measure the following (when it is difficult to get a measure, try to find the price range):

(a) A mass of rain clouds 

(b) The abundance of elephants

(c) Wind speed during storms

(d) the number of strands of hair on your head.

(e) the number of air molecules in your classroom.

Answer:

(a) For the first time to calculate the total rainfall in India, we are able to calculate its value so by knowing the water level we will measure the load on the clouds.

(b) To measure the weight of an elephant, we take a vessel of the known base A. Measure the depth of the boat in the water. Let x1 Therefore, the water volume is removed by the boat, V1 – AX1

Move the elephant in this boat. The boat plunges headlong into the water. Measure the depth of the boat now in the water. Allow or not x2.

∴ Capacity of water removed by boat and elephant V2 = Ax2

∴ Elephant volume removed V = V2 – V1 = A (x2 – X1)

If p is a water congestion, then the amount of elephant = the amount of water is removed by it.

= V ρ = A (x2 – x1) ρ

(c) Air pressure can give us a measure of speed.

(d) Estimating a number. of hair in each cm2 area of ​​the pinnacle, we can measure the whole number. and the hairs of our head,

(∴ we can measure the location of our head)

(e) we are able to detect air congestion and that is why the ratio is numerical. of 1 cm3 molecules are made to measure no. of molecules in our room.

Question 23.

The Sun can be a hot plasma (ionized matter) with an indirect core at temperatures above 107 K, and its outer surface at temperatures of about 6000 K. At high temperatures, nothing stays in a very solid or liquid state. To what extent does one expect the solar eclipse to occur? within the range of solids and solids or liquids? Check your guess for accuracy on the following data: Sun weight = 2.0 x 1030 kg. Sun radiation = 7.0 x 108 m.

Answer:

 

Solar density is within the range of solid / liquid density and not gases.

Question 24.

When the planet Jupiter is 824.7 million kilometers from Earth, its angular diameter is estimated to be 35-72 arcs. Calculate the width of Jupiter.

Answer:

Given the angular width θ = 35.72 = 35.72 x 4.85 x 10-6 rad

= 173.242 x 10-6 = 1.73 x 10-6 rad

∴Width of Jupiter, D = θ x d = 1.73 x 10-4 x 824.7 x 109 m

= 1426.731 x 105 = 1.43 x 108 m

Question 25.

A man who is walking fast in the rain at a speed v> must tilt his umbrella forward to make a straight angle. The learner finds the following relationship between θ and v: tan θ = υ and checks that the relationship has the correct limit: υ → 0, θ → 0, of course. (We think there is no strong wind and rain that falls directly on a dry person). does one think that this relationship is usually good? If not, guess the appropriate relationship.

Answer:

In accordance with the principle of homogeneity of dimensional calculations,

The size of L.H.S. = Size of R.H.S.

Here υ = tan θ i.e. [L1 T-1] = flawless, faulty.

To correct the death of L.H.S, we find

υ / u = tan θ, when u are at that speed of rain.

Question 26.

It is said that two clocks of cesium if allowed to last for 100 years, without interruption, may differ by only 0.02 seconds. What does this mean for the accuracy of the quality of the cesium enter time measurement- interval of 1 s?

Answer:

Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Question 27.

Estimate the normal density of the sodium atom taking its magnitude of 2.5 Å. (Use known values ​​of Avogadro number and sodium weight). Compare sodium density in its crystalline phase: 970 kg m-3. Is it 2 congestion of the same sequence of magnitude? If so, why?

Answer:

Volume of sodium atom 1,

 

Both congestion is in the order of 103 which is that atoms are fully packed. They belong to the strong class.

 

Question 28.

A unit of simple length on a nuclear scale could be fermi: 1 f = 10-15 m. Nuclear sizes comply with the following empirical correlations: r = r0A1 / 3 where r is the nucleus of the nucleus, A is its main unit, and r0 may last up to 1.2 f. Point out that the law suggests that nuclear density is a constant type for various nuclei. Measure the sodium nucleus density. Compare the average weight of the sodium atom found in 2.20.

Answer:

 

Question 29.

LASER can be a source of very strong, monochromatic, and unidirectional sunlight. These laser light structures will be used to live long distances. The Moon’s surface from the earth is already precisely determined using a laser as a source of sunlight. The radiant laser beam on the Moon takes 2.56 seconds to return after a spot on the Moon. What is the orbit of the moon orbit around the Earth?

Answer:

 

Question 30.

SONAR (noise and ground rotation) uses ultrasonic waves to detect and detect objects under water. on a SONAR-equipped submarine, the time delay between the hunting wave generation and the reception of its echo after exposure from the enemy submarine is found in 77.0 s. What is the enemy submarine gap? (Speed ​​of sound in water = 1450 m s-1).

Answer:

 

Question 31.

The fossil record found by modern astronomers is so far from the light emitted by billions of years. These objects (known as quasars) have many confusing features, which have not yet been fully explained. what is the space in the quasar km where light takes 3’0 billion years to achieve for us?

Answer:

Given t = 3 x 109 years = 3 x 109 x 365.25 x 24 x 60 x 60 s

c = 3 x 105 km s-1 (wave speed e.m.)

∴ d = c x t = 3 x 105 x 3 x 109 x 365.25 x 24 x 60 x 60

= 2840184 x 1016 km = 2.84 x 1022 km.

Question 32.

It is a well-known fact that during a total eclipse, the moon’s disk almost completely closes the Sun’s disk. From this fact and therefore the information, you will be able to collect from examples 1 and a few, determine the approximate width of the moon.

Answer:

Here CD = Sun exposure

AB = month range

and E = Land position

As ΔCDE and ABE are similar,

 

Question 33.

The great physicist of this century (PA.M. Dirac) loved to move the numerical values ​​of basic fixed objects. This led him to an interesting comment. Dirac found that from important constants of natural science (c, e, electron weight, proton weight) so gravity does not change G, it can vary with the size of your time. Moreover, it was a very large number, the size of which is close to the present average of the universe (≈ 15 billion years). In the table of basic elements in the middle of this book, try to see if you can build this number (or another interesting number you can think of). If coincidence with the age of the universe were significant, what would this mean for the consistency of the basics?

Answer:

If we experiment with the basic level of nuclear physics (solar light c, charge in electron e, electron me weight, proton mp weight) and the universal constant of gravitation G, we are able to beat the size of your time.

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