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physics ncert solutions class 11th Physics 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

3 months ago

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 $\times 10^{12}$ $m^{2}$

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

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

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

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 $\times 10^{15}$ 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 $\times 10^{12}$ $m^{2}$

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

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

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

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 $\times 10^{15}$ 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 $\times 10^{- 3}$ $m^{3}$ volume

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

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

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2.22 (a) During monsoons, a metrologist records about 215 cm of rainfall in India. SoHeight of water column, h = 215 cm = 2.15 mArea of the country, A = 3.3<math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>12</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Hence, volume of rain water, V = A <math><mo>×</mo><mi>h</mi></math> = 7.09 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>12</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Density of water,p = 1 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> kg/ <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Hence, mass of rain water = p <math><mi></mi><mo>×</mo><mi>V</mi></math> = 7.09 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>15</mn></mrow></mrow></msup></math> kgHence, the total mass of rain-bearing clouds over India is approximately 7.09 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>15</mn></mrow></mrow></msup></math> kg (b) Consider a ship of known base area floating in the sea.Let the depth of the ship be <math><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math>Volume of water displaced by the ship, <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>b</mi></mrow></mrow></msub></math> = A <math><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math>Now, move an elephant on the ship and let the depth of the ship be <math><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math>Volume of water displaced by the ship with elephant on board = A <math><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math>Volume of water displaced by the elephant = A <math><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>-</mo><mi></mi><msub><mrow><mrow><mi>A</mi><mi>d</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math>If the density of the water is p <math><mi></mi><mo>.</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">s</mi></math> of the elephant is = A <math><mi></mi><mi>p</mi><mo>(</mo><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>-</mo><msub><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>)</mo></math> (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 = AWith 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 = <math><mi>?</mi><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Number of strands of hair = <math><mfrac><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mi></mi><mi>s</mi><mi>u</mi><mi>r</mi><mi>f</mi><mi>a</mi><mi>c</mi><mi>e</mi><mi></mi><mi>a</mi><mi>r</mi><mi>e</mi><mi>a</mi></mrow></mrow><mrow><mrow><mi>A</mi><mi>r</mi><mi>e</mi><mi>a</mi><mi></mi><mi>o</mi><mi>f</mi><mi></mi><mi>o</mi><mi>n</mi><mi>e</mi><mi></mi><mi>h</mi><mi>a</mi><mi>i</mi><mi>r</mi></mrow></mrow></mfrac></math> = <math><mrow><mrow><mi>A/</mi></mrow></mrow></math>π<math><msup><mrow><mrow><mn>r</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> (e) Let the volume of the room be VOne mole of air at NTP occupies 22.4 lit = 22.4 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> volumeNumber of molecules in one mole = 6.023 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></msup></math>Number of molecules in room of volume V = <math><mfrac><mrow><mrow><mn>6.023</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>22.4</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></mrow></mrow></mfrac><mi></mi><mo>×</mo><mi>V</mi></math> = 2.69 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>25</mn></mrow></mrow></msup></math> V

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