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NEET Practice Questions (MCQs) with Answers & Solutions

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Reason of weightlessness in a satellite is:

Zero gravity

Centre of mass

Zero reaction force by satellite surface

None

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Explanation

(c)

The reason of weightlessness in a satellite is a zero reaction force by satellite surface.

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Mass M is divided into two parts xM and $(1 - x) M$ . For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is

$\frac{1}{2}$

$\frac{3}{5}$

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Explanation

(a)

$F \propto xm \times (1 - x) m = {xm}^{2} (1 - x) = {xm}^{2} - x^{2} m^{2} For maximum force; \frac{dF}{dx} = 0 \frac{dF}{dx} = m^{2} - 2 {xm}^{2} = 0 \Rightarrow x = 1 / 2$

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If earth is supposed to be a sphere of radius R, if $g_{30 °}$ is value of acceleration due to gravity at latitude of 30^o and g at the pole, the value of $g - g_{30^{ο}}$ is

$\frac{1}{4} ω^{2} R$

$\frac{3}{4} ω^{2} R$

$ω^{2} R$

$\frac{1}{2} ω^{2} R$

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Explanation

(b) Acceleration due to gravity at latitude $λ$ is given by

$g' = g - R ω^{2} \cos^{2} λ a t 30^{ο}, g_{30^{ο}} = g - R ω^{2} \cos^{2} 30^{ο} = g - \frac{3}{4} R ω^{2} s o g - g_{30 °} = \frac{3}{4} ω^{2} R$

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The radii of two planets are respectively $R_{1}$ and $R_{2}$ and their densities are respectively $ρ_{1}$ and $ρ_{2}$ . The ratio of the accelerations due to gravity at their surfaces is

$g_{1} : g_{2} = \frac{ρ_{1}}{{R_{1}}^{2}} : \frac{ρ_{2}}{{R_{2}}^{2}}$

$g_{1} : g_{2} = R_{1} R_{2} : ρ_{1} ρ_{2}$

$g_{1} : g_{2} = R_{1} ρ_{2} : R_{2} ρ_{1}$

$g_{1} : g_{2} = R_{1} ρ_{1} : R_{2} ρ_{2}$

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Explanation

(d) $g = \frac{4}{3} π ρ G R S o \frac{g_{1}}{g_{2}} = \frac{R_{1} ρ_{1}}{R_{2} ρ_{2}}$

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If M the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitational constant is

$\frac{R^{2}}{M}$

$\frac{M}{R^{2}}$

$M R^{2}$

$\frac{M}{R}$

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Explanation

(b) Acceleration due to gravity g= $\frac{G M}{R^{2}}$

$S o, \frac{g}{G} = \frac{M}{R^{2}}$

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Two sphere of mass m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be

$\frac{F}{3}$

$\frac{F}{9}$

3 F

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Explanation

(a) Gravitational force does not depend on the medium.

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Earth binds the atmosphere because of:

Gravity

Oxygen between earth and atmosphere

Both (a) and (b)

None of these

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Explanation

(a)

Various atom of gases present in the earth atmosphere has masses and a gravitational force acts between theses atoms and earth.

also this is the force that holds the various gases near the surface and the atmosphere is present.

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A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is

mgh

$\frac{4}{5} m g h$

$\frac{5}{6} m g h$

$\frac{6}{7} m g h$

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Explanation

Substituting $R = 5 h w e g e t ∆ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} m g h$

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Spot the wrong statement :

The acceleration due to gravity ‘g’ decreases if

We go down from the surface of the earth towards its centre

We go up from the surface of the earth

We go from the equator towards the poles on the surface of the earth

The rotational velocity of the earth is increased

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Explanation

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Which of the following statements is true?

g is less at the earth's surface than at a height above it or a depth below it

g is the same at all places on the surface of the earth

g has its maximum value at the equator

g is greater at the poles than at the equator

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Explanation

(d)

The value of 'g' that is gravity is greater at the poles because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

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