JEE Physics Gravitation MCQs Set 10

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

Explanation:
1. Newton's law of gravitation is universal and applies to all masses everywhere.
2. It provides the centripetal force that keeps planets around the Sun and satellites around planets.
3. Electrons are held in atoms by electric forces, not gravity.

Teacher's Note:
1. Gravitation is a universal force between all masses.
2. It is far too weak to matter inside atoms.

Answer: (c) 10 N

Explanation:
1. The force \(F = \frac{Gm_1m_2}{d^2}\) depends only on the masses and the separation.
2. G is the same on the moon, so the force remains 10 N.

Teacher's Note:
1. Mutual gravitational force does not depend on location.
2. Weight changes with g, but this mutual force does not.

Answer: (a) 1 : 1

Explanation:
1. The two forces form an action-reaction pair under Newton's third law.
2. They are equal in magnitude and opposite in direction, so the ratio is 1 : 1 whatever the masses.

Teacher's Note:
1. Gravitational forces always come in equal and opposite pairs.
2. A larger mass gets a smaller acceleration, not a larger force.

Answer: (b) \(\frac{Gm^2}{a^2}\left(4\hat i + \frac14\hat j\right)\)

Explanation:
1. The mass at \(\left(\frac a2, 0\right)\) is at distance \(\frac a2\) along +x, so it pulls with \(\frac{Gm^2}{(a/2)^2} = \frac{4Gm^2}{a^2}\) along \(\hat i\).
2. The mass at (0, 2a) is at distance 2a along +y and pulls with \(\frac{Gm^2}{4a^2}\) along \(\hat j\).
3. Adding the vectors gives \(\frac{Gm^2}{a^2}\left(4\hat i + \frac14\hat j\right)\).

Teacher's Note:
1. Resolve the forces along the axes and add as vectors.
2. Force falls as the square of distance.

Answer: (d) zero

Explanation:
1. The three equal masses are placed symmetrically at equal distances from O.
2. Their three equal forces of attraction on the 4 kg mass are 120 degrees apart, so their vector sum is zero.

Teacher's Note:
1. Symmetry gives zero net force at the centre.
2. Three equal vectors at 120 degrees cancel.

Answer: (c) \(\frac{4Gm^2}{3l^2}\)

Explanation:
1. The two masses at the ends of that side are each at distance \(\frac l2\) from the midpoint and pull in opposite directions, so they cancel.
2. The third mass is at the apex, at distance \(\frac{\sqrt3}{2}l\), so its pull is \(\frac{Gm^2}{(3l^2/4)} = \frac{4Gm^2}{3l^2}\).

Teacher's Note:
1. Equal opposite forces cancel.
2. Only the third mass contributes.

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

Explanation:
1. The distance from the centre to each corner is \(r = \frac{a}{\sqrt2}\), so \(r^2 = \frac{a^2}{2}\).
2. Opposite corners m and 3m give a net \(\frac{Gm(3m-m)}{r^2} = \frac{4Gm^2}{a^2}\), and 2m and 4m give \(\frac{Gm(4m-2m)}{r^2} = \frac{4Gm^2}{a^2}\), at right angles to the first.
3. The resultant is \(\sqrt2\cdot\frac{4Gm^2}{a^2} = \frac{4\sqrt2\,Gm^2}{a^2}\).

Teacher's Note:
1. Pair up opposite corners first.
2. Perpendicular forces combine as the square root of the sum of squares.

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

Explanation:
1. Tides are caused by the difference in the gravitational pull of the moon on different parts of the earth.
2. The moon is much nearer than the sun, so its effect is larger (about twice as large as the sun's).

Teacher's Note:
1. Tidal force depends on the variation of the pull across the earth.
2. The nearer moon dominates the sun.

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

Explanation:
1. With no net force, Newton's first law says the satellite keeps its instantaneous velocity.
2. That velocity is along the tangent to the orbit, so it moves off in a straight line with speed v.

Teacher's Note:
1. Without force, motion is uniform and in a straight line.
2. The velocity of a circular motion is tangential.

Answer: (c) \(\frac{GMm}{(R+r)^2} = M\omega^2 R\)

Explanation:
1. The two bodies stay on opposite sides of the centre of mass, so they are at separation \(R + r\).
2. The gravitational force is \(\frac{GMm}{(R+r)^2}\).
3. For the body of mass M moving in a circle of radius R this force is the centripetal force \(M\omega^2R\).

Teacher's Note:
1. Use the separation between the bodies in the force and the orbital radius in the centripetal force.
2. Both bodies have the same angular velocity.

Answer: (a) \(\frac{6D}{7}\)

Explanation:
1. Let the point be at distance x from the earth. The fields cancel when \(\frac{36M}{x^2} = \frac{M}{(D-x)^2}\).
2. Taking square roots, \(\frac6x = \frac{1}{D-x}\), so \(6D - 6x = x\).
3. Hence \(x = \frac{6D}{7}\).

Teacher's Note:
1. At a neutral point the two fields are equal and opposite.
2. Take square roots to avoid solving a quadratic.

Answer: (a) \(\frac{4\rho G\pi r}{3}\)

Explanation:
1. Only the mass inside radius r attracts the point: \(M' = \frac43\pi r^3\rho\).
2. The field is \(\frac{GM'}{r^2} = \frac{4}{3}\pi G\rho r\).

Teacher's Note:
1. Inside a uniform sphere the field grows linearly with r.
2. Only the inner sphere attracts the point.

Answer: (c) \(\frac{GMmr}{R^3}\)

Explanation:
1. Inside a uniform sphere the field is \(g' = \frac{GMr}{R^3}\).
2. The force on the particle is \(mg' = \frac{GMmr}{R^3}\).

Teacher's Note:
1. Force inside the earth is proportional to distance from the centre.
2. The motion in a tunnel is simple harmonic.

Answer: (c) on both of the above

Explanation:
1. g depends on the mass and radius of the earth (its size and structure).
2. It also depends on the rotation, since the effective value is reduced by the centrifugal term \(\omega^2R\cos^2\lambda\).
3. So it depends on both.

Teacher's Note:
1. Effective g varies with latitude.
2. Both shape and spin matter.

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

Explanation:
1. \(g = \frac{GM}{R^2}\) with \(M = \frac43\pi R^3\rho\).
2. So \(g = \frac43\pi G\rho R\).
3. Hence \(\rho = \frac{3g}{4\pi GR}\).

Teacher's Note:
1. Express M in terms of density.
2. g is proportional to \(\rho R\).

Answer: (b) \(\frac R2\)

Explanation:
1. \(g = \frac43\pi G\rho R\), so g is proportional to \(\rho R\).
2. For equal g with density doubled, the radius must be halved: \(R' = \frac R2\).

Teacher's Note:
1. g is proportional to density times radius.
2. Compare ratios.

Answer: (c) g' = 9g

Explanation:
1. g is proportional to \(\rho R\).
2. With the same density and nine times the radius, \(g' = 9g\).

Teacher's Note:
1. At equal density, g is proportional to the radius.
2. Not the mass.

Answer: (d) will fall down with an acceleration \(\frac g6\)

Explanation:
1. The moon has no atmosphere, so there is no buoyant force.
2. The balloon is acted on only by gravity and falls with the moon's acceleration of free fall, \(\frac g6\).

Teacher's Note:
1. Buoyancy needs a surrounding fluid.
2. In vacuum every object falls with the same acceleration.

Answer: (b) 400 gm wt

Explanation:
1. \(g' = g\cdot\frac{M'}{M}\cdot\frac{R^2}{R'^2} = g\cdot\frac17\cdot4 = \frac{4g}{7}\).
2. The weight is \(700\times\frac47 = 400\) gm wt.

Teacher's Note:
1. Mass stays the same, weight changes with g.
2. Use the ratio of g values.

Answer: (c) equal

Explanation:
1. \(g' = g\cdot\frac{4}{(2)^2} = g\).
2. The time period \(T = 2\pi\sqrt{l/g}\) depends only on g and l, so it is equal.

Teacher's Note:
1. Same g gives the same pendulum period.
2. Compare M over R squared.

Answer: (c) increase by 3 percent

Explanation:
1. \(g\propto\frac{1}{R^2}\), so \(\frac{\Delta g}{g} = -2\frac{\Delta R}{R} = -2(-1.5\%) = +3\%\).
2. So g increases by 3 percent.

Teacher's Note:
1. Fractional change in g is minus twice the fractional change in R.
2. Use the small change approximation.

Answer: (d) \(g_p > g_e\)

Explanation:
1. The earth is slightly flattened, so the radius at the poles is smaller than at the equator.
2. Also, at the equator the rotation reduces effective g by \(\omega^2R\).
3. Both effects make \(g_p > g_e\).

Teacher's Note:
1. Two effects: shape and rotation.
2. Rotation has no effect at the poles.

Answer: (b) at the equator

Explanation:
1. A spring balance reads weight, \(mg\). For a reading of 2 kg wt, \(m = \frac{2\times 9.8}{g}\).
2. The mass is largest where g is smallest, which is the equator.

Teacher's Note:
1. A spring balance measures weight, not mass.
2. Lower g means more mass for the same reading.

Answer: (b) decreases

Explanation:
1. The earth bulges at the equator (larger R) and the rotation lowers the effective g by \(\omega^2R\cos^2\lambda\).
2. So g decreases steadily from pole to equator.

Teacher's Note:
1. g is highest at the poles.
2. It falls as you move toward the equator.

Answer: (a) 0.98g m/s squared

Explanation:
1. For \(h \ll R\), \(g' = g\left(1 - \frac{2h}{R}\right)\).
2. Here \(\frac{2h}{R} = \frac{2\times64}{6400} = 0.02\), so \(g' = 0.98g\).

Teacher's Note:
1. Use the first-order formula for small heights.
2. 64 km is 1 percent of R.

Answer: (a) \(45\times10^6\) m

Explanation:
1. \(g' = g\left(\frac{R}{R+h}\right)^2 = \frac{g}{64}\), so \(\frac{R}{R+h} = \frac18\).
2. Hence \(R + h = 8R\) and \(h = 7R = 7\times6400\) km \(= 44800\) km \(\approx 45\times10^6\) m.

Teacher's Note:
1. Use the exact inverse-square formula for large heights.
2. Take the square root first.

Answer: (d) \(\frac W4\)

Explanation:
1. At height R the distance from the centre is 2R, so g becomes \(\frac g4\).
2. So the weight is \(\frac W4\).

Teacher's Note:
1. Doubling the distance cuts g to a quarter.
2. Weight is proportional to g.