JEE Physics Gravitation MCQs Set 04

Answer: (c) controls the motion of satellites and planets

Explanation:
1. The law of gravitation is universal
2. It holds for planets, satellites and falling bodies alike

Teacher's Note:
1. Gravitation is a universal attraction between any two masses.
2. Atomic electrons are held by electrostatic, not gravitational, forces.

Answer: (a) \( 4G \)

Explanation:
1. All forces point along +x and add up
2. \( F = 3G\left(1 + \frac14 + \frac1{16} + \cdots\right) \)
3. The GP sum is \( \frac{1}{1-1/4} = \frac43 \), so \( F = 3G \times \frac43 = 4G \)

Teacher's Note:
1. Forces along one line simply add.
2. Use the sum of an infinite GP with ratio 1/4.

Answer: (d) 6 N

Explanation:
1. The force depends only on m1, m2, d and G
2. None of these changes on the moon
3. So the force remains 6 N

Teacher's Note:
1. G is a universal constant.
2. Do not confuse the mutual force with the weight of a body.

Answer: (d) none of these

Explanation:
1. G is a universal constant, the same everywhere
2. It does not depend on the masses, material, medium or acceleration

Teacher's Note:
1. G is a constant of nature.
2. It is different from g, which varies from place to place.

Answer: (a) \( \sqrt{\frac{3GM}{L^3}} \)

Explanation:
1. Net force on one body: \( 2\frac{GM^2}{L^2}\cos 30^\circ = \sqrt3\,\frac{GM^2}{L^2} \) towards the centroid
2. Orbit radius \( r = L/\sqrt3 \)
3. \( M\omega^2 r = \sqrt3\frac{GM^2}{L^2} \Rightarrow \omega^2 = \frac{3GM}{L^3} \)

Teacher's Note:
1. Resolve the two equal pulls along the line to the centroid.
2. The orbital radius of each body is the circumradius \( L/\sqrt3 \).

Answer: (c) \( \frac{4\sqrt2\,Gm^2}{a^2} \)

Explanation:
1. Distance to each corner: \( r^2 = a^2/2 \)
2. Opposite pair \( (m,3m) \): net \( \frac{Gm(2m)}{a^2/2} = \frac{4Gm^2}{a^2} \)
3. Opposite pair \( (2m,4m) \): net \( \frac{4Gm^2}{a^2} \), perpendicular to the first
4. Resultant \( = \sqrt2 \times \frac{4Gm^2}{a^2} \)

Teacher's Note:
1. Pair up opposite corners first.
2. The two net forces are at right angles.

Answer: (b) the gravitational effect of the moon on the earth

Explanation:
1. Tides come from the difference in the moon's pull on different parts of the ocean
2. The moon is much closer than the sun, so its effect dominates

Teacher's Note:
1. The nearer body gives the larger tidal effect.
2. The sun contributes a smaller secondary tide.

Answer: (b) against the gravitation of the earth in the onward journey

Explanation:
1. The earth is far more massive than the moon
2. Escaping the earth's field needs the most energy, at lift-off on the onward journey

Teacher's Note:
1. Escape energy is greater for a larger, more massive body.
2. The return journey starts from the weak lunar field.

Answer: (d) \( \frac{9D}{10} \)

Explanation:
1. At distance \( x \) from the earth: \( \frac{81M}{x^2} = \frac{M}{(D-x)^2} \)
2. \( 9(D-x) = x \)
3. \( x = \frac{9D}{10} \)

Teacher's Note:
1. Equate the magnitudes of the two fields on the line joining them.
2. Take square roots to avoid solving a quadratic.

Answer: (d) fly off tangentially from the orbit with velocity v

Explanation:
1. With no force, the body obeys Newton's first law
2. It continues along the tangent at the instant of release with speed v

Teacher's Note:
1. Velocity is tangential in circular motion.
2. Without a centripetal force, the path becomes a straight line.

Answer: (b) \( \left(\frac{4\pi}{3} mG\rho\right) r \)

Explanation:
1. Only the mass inside radius \( r \) attracts: \( M' = \frac{4}{3}\pi r^3\rho \)
2. \( F = \frac{GM'm}{r^2} = \frac{4\pi}{3}G\rho m\, r \)

Teacher's Note:
1. Inside a uniform sphere, the force grows linearly with r.
2. The outer shell exerts no net force.

Answer: (d) 0.4

Explanation:
1. \( g \propto \frac{M}{R^2} \)
2. \( \frac{g_M}{g_E} = \frac{0.1}{(0.5)^2} = 0.4 \)

Teacher's Note:
1. Diameter ratio equals radius ratio.
2. Squaring 0.5 gives 0.25 in the denominator.

Answer: (d) zero

Explanation:
1. The three pulls have equal magnitude and are 120 degrees apart
2. Their vector sum is zero

Teacher's Note:
1. Symmetry gives zero net force at the centroid.
2. No numerical calculation is needed.

Answer: (b) \( \frac{g}{G}R_E^2 \)

Explanation:
1. \( g = \frac{GM}{R_E^2} \)
2. \( M = \frac{gR_E^2}{G} \)

Teacher's Note:
1. Surface gravity links mass and radius.
2. Measure g and R to find M.

Answer: (c) zero

Explanation:
1. Inside the earth \( g' = g\left(1-\frac dR\right) \)
2. At the centre \( d = R \), so \( g' = 0 \)

Teacher's Note:
1. g falls linearly to zero at the centre.
2. The pulls from all sides cancel.

Answer: (c) both of the above

Explanation:
1. Shape and mass distribution of the earth affect g
2. Rotation reduces the effective g away from the poles

Teacher's Note:
1. g varies with latitude and altitude.
2. Both structure and rotation matter.

Answer: (d) \( h = (\sqrt2 - 1)R \)

Explanation:
1. \( \frac{g'}{g} = \frac{R^2}{(R+h)^2} = \frac12 \)
2. \( R + h = \sqrt2 R \)
3. \( h = (\sqrt2-1)R \)

Teacher's Note:
1. Use the inverse-square form.
2. Do not use the small-height approximation here.

Answer: (c) 10 m

Explanation:
1. Safe landing speed is the same: \( v^2 = 2gh \)
2. \( h' = h\frac{g}{g'} = 2\times\frac{9.8}{1.96} = 10 \) m

Teacher's Note:
1. Equal landing speed means gh is constant.
2. Lower g allows greater height.

Answer: (c) increase

Explanation:
1. \( g_{eff} = g - \omega^2 R\cos^2\lambda \)
2. With \( \omega = 0 \) the reduction disappears, so g increases

Teacher's Note:
1. Rotation lowers effective g except at the poles.
2. Stopping rotation removes that reduction.

Answer: (b) \( \sqrt2 \) times

Explanation:
1. \( g \propto M/R^2 \) so \( g_p = \frac{2}{4}g = \frac g2 \)
2. \( T \propto \frac{1}{\sqrt g} \)
3. \( T_p = \sqrt2\,T \)

Teacher's Note:
1. Diameter doubling means radius doubling.
2. Smaller g gives a longer period.

Answer: (d) will fall with acceleration \( g/6 \)

Explanation:
1. The moon has no atmosphere, so there is no buoyant force
2. The balloon falls freely with the lunar g = g/6

Teacher's Note:
1. Buoyancy needs a surrounding fluid.
2. Free fall acceleration is the local g.

Answer: (c) \( \rho = \frac{3g}{4\pi GR} \)

Explanation:
1. \( g = \frac{GM}{R^2} = \frac{G}{R^2}\cdot\frac{4}{3}\pi R^3\rho = \frac{4}{3}\pi G\rho R \)
2. \( \rho = \frac{3g}{4\pi GR} \)

Teacher's Note:
1. Write M in terms of density and volume.
2. Check the dimensions of your answer.