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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 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 strom (d) the number of strands of hair on your head (e) the number of air molecules in your classroom. |
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Answer» Solution :In India, average rainfall is 100 cm or 1 m. According to meterologist let AREA of country is A. `A=3.3Mkm^(2)` `=3.3xx10^(6)xx(10^(3))^(2)` `=3.3xx10^(12)m^(2)` Mass M = density `xx` volume = density `xx` Area `xx` height `=10^(3)xx3.3xx10^(12)xx1` `=3.3xx10^(15)kg` (b) Mass of elephant can be FOUND by using law of floatation. Suppose area of bottom of boat is A and its `d_(1)` height is sink in water. Hence, displaced volume of water is, `V_(1)=Ad_(1)` Now, if elephant is BROUGHT in boat, its `d_(2)` height is sink in water. Displaced volume of water is, ` V_(2)=Ad_(2)` `:.` Displaced volume due to elephant, `V=V_(2)-V_(1)=A[d_(2)-d_(1)]` If density of water is `rho`, then Mass of elephant `=Vrho` `=A(d_(2)-d_(1))rho` (c) If balloon is placed at height h from the ground. When storm is not there, then balloon is at position A from point O on ground OA=h. When storm is there, balloon will shift to position B in time .t. by AB=x. Due to displacement of balloon now position B makes an angle of `theta` at O.  `TANTHETA=(AB)/(AO)=(x)/(h)` `:.htantheta=x` `:.(htantheta)/(t)=(x)/(t)` `:.(htantheta)/(t)=v` Thus, speed of storm v can be found. (d) Suppose, person has hairs on head in 8 cm radius and thickness of hair is `10xx10^(-5)cm`. No. of hairs, `=("Area of region of hairs")/("Cross-sectional area of one hair")` `=(pi(0.08)^(2))/(pi((10xx10^(-5))/(2))^(2))` `=(64xx10^(-4))/(25xx10^(-10))` `=2.56xx10^(6)` No. of hairs (e) Volume of 1 mole air molecule at NTP is `22.4xx10^(-3)m^(3)` and no. of molecules in 1 mole are `N_(A)=6.02xx10^(23)`. Size of normal room is `10mxx10mxx5` m, then volume of room `V=500m^(3)` No. of molecules in room, `N=(N_(A))/(22.4xx10^(-3))xxV` `=(6.02xx10^(23)xx500)/(22.4xx10^(-3))` `~~1.34xx10^(28)` |
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