- Home
- Standard 11
- Physics
7.Gravitation
hard
પૃથ્વીની સપાટી થી કેટલી ઊંડાઇએ પૃથ્વીની સપાટી થી $1600 \,km$ ઊંચાઈએ ગુરુત્વ પ્રવેગના મૂલ્યથી અડધું હશે ?
A
$4.2 \times {10^6}\,m$
B
$3.19 \times {10^6}\,m$
C
$1.59 \times {10^6}\,m$
D
એક પણ નહીં
Solution
(a) Radius of earth $R = 6400\, km $
$h = \frac{R}{4}$ Acceleration due to gravity at a height h ${g_h} = g{\left( {\frac{R}{{R + h}}} \right)^2}$
$ = g{\left( {\frac{R}{{R + \frac{R}{4}}}} \right)^2}$
$ = \frac{{16}}{{25}}g$
At depth $'d'$ value of acceleration due to gravity
${g_d} = \frac{1}{2}{g_h}$ (According to problem)
$⇒$ ${g_d} = \frac{1}{2}\left( {\frac{{16}}{{25}}} \right)g$ $ \Rightarrow g\left( {1 – \frac{d}{R}} \right)$
$ = \frac{1}{2}\left( {\frac{{16}}{{25}}} \right)\;g$
By solving we get $d = 4.3 \times {10^6}\,m$
Standard 11
Physics
