JEE PYQ: Gravitation Question 38
Question 38 - 2019 (10 Apr Shift 1)
The value of acceleration due to gravity at Earth’s surface is $9.8$ ms$^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9$ ms$^{-2}$, is close to: (Radius of earth $= 6.4 \times 10^6$ m)
(1) $2.6 \times 10^6$ m
(2) $6.4 \times 10^6$ m
(3) $9.0 \times 10^6$ m
(4) $1.6 \times 10^6$ m
Show Answer
Answer: (1)
Solution
$\frac{g}{2} = \frac{g}{(1+h/R)^2}$. $(1+h/R)^2 = 2$. $h = R(\sqrt{2}-1) = 6.4 \times 10^6 \times 0.414 = 2.6 \times 10^6$ m.
BETA
