हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Calculate the change in g value in your district of Tamil nadu. (Hint: Get the latitude of your district of Tamil nadu from Google). What is the difference in g values at Chennai and Kanyakumari? - Physics

Advertisements
Advertisements

प्रश्न

Calculate the change in g value in your district of Tamil nadu. (Hint: Get the latitude of your district of Tamil nadu from Google). What is the difference in g values at Chennai and Kanyakumari?

संख्यात्मक

उत्तर

Variation of ‘g’ value in the latitude to Chennai

`"g’"_"Chennai" = "g" - ω^2"R" cos^2λ`

Here `ω^2"R" = ((2π)/"T")^2 xx "R"`

Period of revolution (T) = 1 day = 86400 sec

Radius of the Earth (R) = 6400 × 103 m

Latitude of Chennai (λ) = 13° = 0.2268 rad

`"g’"_"Chennai" = 9.8 - [((2 xx 3.14)/86400)^2 xx 6400 xx 10^3] xx (cos 0.2268)^2`

= `9.8 - [(3.4 xx 10^-2) xx (0.9744)^2]`

= 9.8 − [0.034 × 0.9494]

= 9.8 − 0.0323

`"g’"_"Chennai"` = 9.7677 ms−2

Variation of ‘g’ value in the latitude of Kanyakumari

`"λ’"_"Kanyakumari"` = 8°35’ = 0.1457 red

`"g’"_"Kanyakumari" = 9.8 - [3.4 xx 10^-2 xx (cos 0.1457)^2]`

= 9.8 − 0.0333

`"g’"_"Kanyakumari"` = 9.7667 ms−2

The difference of ‘g’ value Δg = `"g’"_"Chennai" - "g’"_"Kanyakumari"`

= 9.7677 − 9.7667

Δg = 0.001 ms−2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Gravitation - Evaluation [पृष्ठ ४६]

APPEARS IN

सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 6 Gravitation
Evaluation | Q V. 15. | पृष्ठ ४६

संबंधित प्रश्न

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?


The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?


The acceleration of the moon just before it strikes the earth in the previous question is


Find the height over the Earth's surface at which the weight of a body becomes half of its value at the surface.


A mass of 6 × 1024 kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3 × 108 m s−1. What should be the radius of the sphere?


Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?


If both the mass and the radius of the earth decrease by 1%, then the value of acceleration due to gravity will


One can easily weigh the earth by calculating the mass of the earth by using the formula:


A pebble is thrown vertically upwards from the bridge with an initial velocity of 4.9 m/s. It strikes the water after 2 s. If acceleration due to gravity is 9.8 m/s2. The height of the bridge and velocity with which the pebble strikes the water will respectively be ______.


The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of the earth will, be ______.

(Radius of earth = 6400 km)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×