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

0 5 Views | Posted 5 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 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 = 3.3×1012 m2

    Hence, volume of rain water, V = A ×h = 7.09 ×1012 m3

    Density of water,p = 1 ×103 kg/ m3

    Hence, mass of rain water = p
    ×V
     = 7.09 ×1015 kg

    Hence, the total mass of rain-bearing clouds over India is approximately 7.09 ×1015 kg

     

    (b) Consider a ship of known base area floating in the sea.

    Let the depth of the ship be d1

    Volume of water displaced by the ship, Vb = A d1

    Now, move an elephant on the ship and let the depth of the ship be&nbs

    ...more

Similar Questions for you

V
Vishal Baghel

According to question, we can write

Total moles of gas = n = nOxygen + nOxygen = 1 6 2 + 1 2 8 3 2 = 1 2 m o l e s  

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3 2 7 × 1 0 4 c m 3  

V
Vishal Baghel

K = Q r Δ x Δ A T

M L 2 T - 2 ( L ) L 2 ( θ ) ( T ) M 1 L 1 - T - 3 θ - 1

V
Vishal Baghel

Viscosity = Pascal . Second

P X A Y T Z = [ M 1 L 1 T 1 ]        

[ M 1 L + 1 T 1 ] x [ L 2 ] y [ T 1 ] z = M 1 L 1 T 1

= x = 1,     x + 2y = -1,      -x + z = 1

y = -1,              Z = 0

Viscosity = [P1A-1T0]

P
Payal Gupta

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10-2 cm

V
Vishal Baghel

Δ Q = m s Δ T s = Δ Q m Δ T [ s ] = M L 2 T 2 M Q = L 2 T 2 Q 1

Δ Q = m L [ L ] = M L 2 T 2 M = L 2 T 2

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post