Tags Units and Measurements

Units and Measurements

Get insights from 91 questions on Units and Measurements, answered by students, alumni, and experts. You may also ask and answer any question you like about Units and Measurements

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

10 months ago

Units and Measurements Class 11th

2.26 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?

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

2.26

Total time in 100 years = 100 * 365 * 24 * 60 * 60 s

Error in 100 years = 0.02 sec

Error in 1 sec= 0.02 / (100 * 365 * 24 * 60 * 60) s = 6.34 * 10^-12

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

25. Kindly Consider the following

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

2.25 In the equation v = tan(θ) , the dimension v = L¹T^-1.Since tan? is dimensionless, the equation v = tan(θ) is incorrect.

To make it correct v must be divided by the velocity of rain u and the correct expression should be v/u = tan(θ)

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

2.24 When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

0 Follower 17 Views

Vishal Baghel

Contributor-Level 10

Kindly go through the solution

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

2.23 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: mass of the Sun = 2.0 * 10³⁰ kg, radius of the Sun = 7.0 * 10⁸ m.

0 Follower 58 Views

Vishal Baghel

Contributor-Level 10

2.23

Mass of the Sun = 2.0 * 10³⁰ kg

Radius of the Sun = 7.0 * 10⁸ m

Volume of the Sun = (4/3)(π) r³= 4/3 x(π) x (7.0 * 10⁸)³= 1.436 * 10²⁷

Density of the Sun = Mass / volume = 2.0 * 10³⁰ kg / 1.436 * 10²⁷kg/ m³= 1.39 * 10³ kg/ m³

The density is in the range of solids and liquids.

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

2.22 Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) The total mass of rain-bearing clouds over India during the Monsoon

(b) The mass of an elephant

(c) The wind speed during a storm

(d) The number of strands of hair on your head

(e) The number of air molecules in your classroom

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

2.22

(a) During monsoons, a metrologist records about 215 cm of rainfall in India. So

Height of water column, h = 215 cm = 2.15 m

Area of the country, A = 3.3 $* 10^{12}$ $m^{2}$

Hence, volume of rain water, V = A $* h$ = 7.09 $* 10^{12}$ $m^{3}$

Density of water,p = 1 $* 10^{3}$ kg/ $m^{3}$

Hence, mass of rain water = p
$* V$ = 7.09 $* 10^{15}$ kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 $* 10^{15}$ kg

(b) Consider a ship of known base area floating in the sea.

Let the depth of the ship be $d_{1}$

Volume of water displaced by the ship, $V_{b}$ = A $d_{1}$

Now, move an elephant on the ship and let the depth of the ship be&nbs

...more

2.22

(a) During monsoons, a metrologist records about 215 cm of rainfall in India. So

Height of water column, h = 215 cm = 2.15 m

Area of the country, A = 3.3 $* 10^{12}$ $m^{2}$

Hence, volume of rain water, V = A $* h$ = 7.09 $* 10^{12}$ $m^{3}$

Density of water,p = 1 $* 10^{3}$ kg/ $m^{3}$

Hence, mass of rain water = p
$* V$ = 7.09 $* 10^{15}$ kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 $* 10^{15}$ kg

(b) Consider a ship of known base area floating in the sea.

Let the depth of the ship be $d_{1}$

Volume of water displaced by the ship, $V_{b}$ = A $d_{1}$

Now, move an elephant on the ship and let the depth of the ship be $d_{2}$

Volume of water displaced by the ship with elephant on board = A $d_{2}$

Volume of water displaced by the elephant = A $d_{2} - {A d}_{1}$

If the density of the water is p
$. T h e n t h e m a s s$ of the elephant is = A $p (d_{2} - d_{1})$

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of the wind speed.

(d) Area of the head surface carrying hair = A

With the help of a screw gauge, the diameter and hence the radius of a hair can be determined. Let it be r, so that area of one hair = $? r^{2}$

Number of strands of hair = $\frac{T o t a l s u r f a c e a r e a}{A r e a o f o n e h a i r}$ = $A/$ π $r^{2}$

(e) Let the volume of the room be V

One mole of air at NTP occupies 22.4 lit = 22.4 $* 10^{- 3}$ $m^{3}$ volume

Number of molecules in one mole = 6.023 $* 10^{23}$

Number of molecules in room of volume V = $\frac{6.023 * 10^{23}}{22.4 * 10^{- 3}} * V$ = 2.69 $* 10^{25}$ V

less

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

2.18 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 (hill tops, 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).

0 Follower 79 Views

Vishal Baghel

Contributor-Level 10

When we look out of the window of a fast-moving train, a line of sight is created (imaginary) between our eyes and the object outside. The line of sight of the distant object does not change, whereas the line of sight of the objects closer to the train changes rapidly as the train moves. So the nearby objects appear to move rapidly in the opposite direction, but the distant object remained stationary.

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

2.6 Which of the following is the most precise device for measuring length?

(a) A vernier calipers with 20 divisions on the sliding scale

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) An optical instrument that can measure length to within a wavelength of light ?

0 Follower 45 Views

Vishal Baghel

Contributor-Level 10

2.6

(a) The least count of a vernier calliper = 0.1 mm

(b) The least count of the screw gauge = 1/100 mm = 0.01 mm

(c) The least count of the optical instrument = wave length of the light = 10^-5cm = 10^-3 mm = 0.001 mm

Hence the answer is (c)

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

Q.2.4 Explain this statement clearly:

“To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:

(a) Atoms are very small objects

(b) A jet plane moves with great speed

(c) The mass of Jupiter is very large

(d) The air inside this room contains a large number of molecules

(e) A proton is much more massive than an electron

(f) The speed of sound is much smaller than the speed of light

0 Follower 339 Views

Vishal Baghel

Contributor-Level 10

2.4

(a) Atoms are very small compared to a cricket ball

(b) Compared to a car on a busy city road, a jet plane moves with great speed

(c) When compared to the average mass of a human being, mass of Jupiter is very large.

(d) When compared to the air inside a football, the air inside this room contains a large number of molecules

(e) A proton has higher mass compared to that of an electron

(f) The speed of sound is much smaller when calling your friend from your balcony than signaling him with a torch light

0 0

New answer posted

10 months ago

Units and Measurements Class 11th

Q.2.1 Fill in the blanks

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

2.1

(a) The volume of a cube of side 1 ${c m}^{3}$ is equal to $10^{- 6} m^{3}$

The volume of the cube = 1 cm * 1 cm * 1 cm = 1 cm³

1 cm3 = 1 * 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 * 10⁴ (mm)²

The surface area of a solid cylinder = 2? r (r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 * 22/7 * 20 * (100 +20) ${m m}^{2}$ = 1.51 * $10^{4}$ ${m m}^{2}$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 * 10³.kg m^–3.

Relative density o

...more

2.1

(a) The volume of a cube of side 1 ${c m}^{3}$ is equal to $10^{- 6} m^{3}$

The volume of the cube = 1 cm * 1 cm * 1 cm = 1 cm³

1 cm3 = 1 * 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 * 10⁴ (mm)²

The surface area of a solid cylinder = 2? r (r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 * 22/7 * 20 * (100 +20) ${m m}^{2}$ = 1.51 * $10^{4}$ ${m m}^{2}$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 * 10³.kg m^–3.

Relative density or Specific gravity = Density of the subject / density of water

Density of water = 1 gm/cc= 10^-3 / 10^-6

less

0 0

New question posted

10 months ago

Units and Measurements Class 11th

A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m² s^–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 4.2 α ^–1 β ^–2 γ ² in terms of the new units.

0 Follower 123 Views

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Units and Measurements

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience