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8 months ago

Class 11th Units and Measurements

2.28 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 A^1/3

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 sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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Class 11th Units and Measurements

2.27 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/m³. Are the two densities of the same order of magnitude? If so, why?

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Vishal Baghel

Contributor-Level 10

2.27

The diameter of the sodium atom = 2.5 Å = 2.5 * 10^-10 m

Radius of the sodium atom = 1.25 * 10^-10 m

Volume of the sodium atom = (4/3)(π) r³= (4/3) * (22/7) * (1.25 * 10^-10) m³= 8.18 * 10^-30

Mass of 1 mole atom of sodium = 23 g = 23 * 10^-3 kg

1 mole of sodium contains 6.023 * 10²³ atoms.

Hence mass of 1 sodium atom = (23 * 10^-3)/ (6.023 * 10²³) = 3.818 * 10^-26 kg

Atomic mass density of sodium = M/V = (3.818 * 10^-26)/ (8.18 * 10^-30) kg/m³= 4667.48kg/m³

The density of sodium in its solid state is 4667.48 kg/m³but in the crystalline phase is 970 kg/m^3.

In crystalline phase there are voids in between atoms, hence the density

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8 months ago

Class 11th Units and Measurements

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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Class 11th Units and Measurements

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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8 months ago

Class 11th Units and Measurements

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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8 months ago

Class 11th Units and Measurements

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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8 months ago

Class 11th Units and Measurements

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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8 months ago

Class 11th Units and Measurements

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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8 months ago

Class 11th Units and Measurements

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 ?

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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)

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8 months ago

Class 11th Units and Measurements

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

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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

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