logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

A particle is suspended from a spring and it stretches the spring by 1 cm on the surface of the earth. The same particle will stretch the same spring at a place 800 Km above the earth by

\[(a)\;0.79 \;cm \quad (b)\;0.1\;cm \quad (c)\;1\;cm \quad (d)\;2\;cm \]

1 Answer

For a spring $kx=mg$
$x=\large\frac{mg}{k}$
$\quad= \large\frac{GMm}{k r^2}\qquad$$ \bigg[g= \large\frac{GM}{r^2}\bigg]$
$x\; \alpha \; \large\frac{1}{r^2}$
$\large\frac{x_2}{x_1}=\frac{R^2}{(R+h)^2}$
$x^2 =1 \times \bigg(\large\frac{6400}{7200}\bigg)^2$
$\quad= 0.79 \;cm$
Hence a is the correct answer.

 

answered Aug 24, 2013 by meena.p
edited Mar 14, 2014 by thagee.vedartham
 

Related questions

...