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4 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?

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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

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Units and Measurements Class 11th

25. Kindly Consider the following

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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)

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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.

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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5 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.

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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.

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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

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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

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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

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Units and Measurements Class 11th

2.21 Precise measurements of physical quantities are a need of 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.

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5 months ago

Units and Measurements Class 11th

2.20 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?

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Units and Measurements Class 11th

2.19 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) fr

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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).

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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.

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Units and Measurements Class 11th

2.17 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?

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