1
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
The value of the acceleration due to gravity is g1 at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio $$\left( {{d \over R}} \right)$$ equals :
A
$${5 \over 9}$$
B
$${1 \over 9}$$
C
$${7 \over 9}$$
D
$${4 \over 9}$$
2
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:
A
2
B
1
C
$$\sqrt 2$$
D
$${1 \over {\sqrt 2 }}$$
3
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:
A
$${A{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}$$
B
$${A{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}$$
C
$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{1/2}}}}$$
D
$${A \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$
4
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
The mass density of a planet of radius R varies with the distance r from its centre as
$$\rho$$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
Then the gravitational field is maximum at :
A
$$r = {1 \over {\sqrt 3 }}R$$
B
r = R
C
$$r = \sqrt {{3 \over 4}} R$$
D
$$r = \sqrt {{5 \over 9}} R$$
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Optics
Modern Physics
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