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(a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during Monosoon, in India. If A is the area of the country, then A = 3.3 million sq. `Km=3.3xx10^(6) (km)2=3.3xx10^(6)xx10^(6) m^(2)=3.3xx10^(12) m^(2)` <br> Mass of rain-bearing clouds <br> = area `xx` height `xx` density `=3.3xx10^(12)xx1xx1000 kg=3.3xx10^(15) kg` <br> (b) Measure the depth of an empty boat in water. Let it be d1. If A be the base area of the boat, then volume of water displaced by boat, `V1=Ad2` <br> Let d2 be the depth of boat in water when the elephant is moved into the boat. Volume of water displaced by (boat + elephant), `V2=Ad2` volume of water displaced by elephant, <br> `V=V2-V1=A(d2-d1)` <br> If p be the density of water, then mass of elephant =mass of water displaced by it <br> `=A(d2-d1)p`. <br> (c) Wind speed can be estimated by floating a gas-filled balloon in air at a known height h. When there is no wind, the balloon is at A. Suppose the wind starts blowing to the right such that the balloon drifts to position B in 1 second. <br> Now, `AB=d=h theta` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XI_C02_E01_022_S01.png" width="80%"> <br> The value of d directly gives the wind speed. <br> (d) Let us assume that the man is not partially bald. Let us further assume that the hair on the head are uniormly distributed. We can estimate the area of the head. The thickness of a hair can be measured by using a screw gauge. The number of hair on the head is clearly the ratio of the area of head to the cross-sectional area of a hair. <br> Assume that the human head is a circle of radius 0.08 m i.e., 8 cm. Let us further assume that the thickness of a human air is `5 xx10^(-5) m`. <br> Number of hair on the head <br> = Area of the head/Area of cross - section of a hair <br> `= pi (0.08)2//pi (5xx10^(-5))=64xx10^(-4)//25xx10^(-10)=2.56xx10^(6)` <br> The number of hair on the human head is of the order of one million. <br> (e) We can determine the volume of the class-room by measuring its length, breadth and height. consider a class room of size `10 m xx 8 m xx 4m`. volume of this room is `320 m^(3)`. We know that 22.41 or `22.4xx10^(-3) m^(3)` of air has `6.02xx10^(23)` molecules (equal to Avogardo's number). <br> Number of molecules of air in the class room <br> `=(6.02xx10^(23)//22.4xx10^(-3))xx320=8.6xx10^(27)`Related Video

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