Stress And Strain

The correct answer is(D): 1.8×10²N.

For $Cu$ wire, $l_%7B1%7D%3D2.2%20%5C%2Cm$ ,
$r_%7B1%7D%3D1.5%20%5C%2Cmm%20%3D1.5%20%5Ctimes%2010%5E%7B-3%7D%20m$
$Y_%7B1%7D%3D1.1%20%5Ctimes%2010%5E%7B11%7D%20%5C%2CN%20%2F%20m%20%5E%7B2%7D$
For steel wire, $l_%7B2%7D%3D1.6%5C%2Cm$ ,
$r_%7B2%7D%3D1.5%20%5C%2C%20mm%20%3D1.5%20%5Ctimes%2010%5E%7B-3%7D%20m$
$Y_%7B2%7D%3D2.0%20%5Ctimes%2010%5E%7B11%7D%20%5C%2C%20N%20%2F%20m%20%5E%7B2%7D$
Let $F$ be the stretching force in both the wires then
For $Cu$ wire, $Y_%7B1%7D%3D%5Cfrac%7BF%7D%7B%5Cpi%20r_%7B1%7D%5E%7B2%7D%7D%20%5Ctimes%20%5Cfrac%7Bl_%7B1%7D%7D%7B%5CDelta%20l_%7B1%7D%7D$
$%5CRightarrow%20F%3D%5Cfrac%7BY_%7B1%7D%20%5Cpi%20r_%7B1%7D%5E%7B2%7D%20%5Ctimes%20%5CDelta%20l_%7B1%7D%7D%7Bl_%7B1%7D%7D$
$%3D%5Cfrac%7B1.1%20%5Ctimes%2010%5E%7B11%7D%7D%7B2.2%7D%20%5Ctimes%20%5Cfrac%7B22%7D%7B7%7D%20%5Ctimes%5Cleft(1.5%20%5Ctimes%2010%5E%7B-3%7D%5Cright)%5E%7B2%7D%20%5Ctimes%200.5%20%5Ctimes%2010%5E%7B-3%7D$
$%5Ccong%201.8%20%5Ctimes%2010%5E%7B2%7D%20N$

The total energy

E

of a satellite in a circular orbit is given by the sum of its kinetic and potential energy. The kinetic energy

K

is:

K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2

where

v

is the orbital velocity of the satellite. The potential energy

U

is given by:

U%20%3D%20-%20%5Cfrac%7BG%20m%20M%7D%7BR%20%2B%20h%7D

where

G

is the gravitational constant,

m

is the mass of the satellite,

M

is the mass of the Earth, and

R%20%2B%20h

is the distance from the center of the Earth to the satellite. Using the fact that the orbital velocity

v

of a satellite is:

v%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7BR%20%2B%20h%7D%7D

we can substitute this into the kinetic energy equation:

K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%5Cleft(%20%5Cfrac%7BGM%7D%7BR%20%2B%20h%7D%20%5Cright)

Thus, the total energy

E

is:

E%20%3D%20K%20%2B%20U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%5Cleft(%20%5Cfrac%7BGM%7D%7BR%20%2B%20h%7D%20%5Cright)%20-%20%5Cfrac%7BG%20m%20M%7D%7BR%20%2B%20h%7D%20%3D%20-%20%5Cfrac%7BG%20m%20M%7D%7B2%20(R%20%2B%20h)%7D

This shows that the total energy

E

is inversely proportional to

(R%20%2B%20h)%5E2

. Therefore, the correct answer is:

%5Cfrac%7B1%7D%7B(R%20%2B%20h)%5E2%7D

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Stress And Strain

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs