JEE Physics Gravitation MCQs Set 14

Answer: (c) \(10^{-2}\) m

Explanation:
1. The escape speed must equal the speed of light: \(c = \sqrt{\frac{2GM}{R}}\).
2. \(R = \frac{2GM}{c^2} = \frac{2\times6.67\times10^{-11}\times5.98\times10^{24}}{9\times10^{16}}\approx8.9\times10^{-3}\) m.
3. This is about \(10^{-2}\) m.

Teacher's Note:
1. Set the escape velocity equal to c.
2. The earth would be about 1 cm in size.

Answer: (a) \(GMK = 4\pi^2\)

Explanation:
1. For a circular orbit, \(\frac{GMm}{r^2} = m\left(\frac{2\pi}{T}\right)^2r\), so \(T^2 = \frac{4\pi^2}{GM}r^3\).
2. Comparing with \(T^2 = Kr^3\), \(K = \frac{4\pi^2}{GM}\), i.e. \(GMK = 4\pi^2\).

Teacher's Note:
1. Equate gravity to the centripetal force.
2. K contains the mass of the sun.

Answer: (a) the acceleration of S is always directed towards the centre of the earth

Explanation:
1. The only force is the earth's gravity, always directed toward the centre, so the acceleration is always toward the centre.
2. The angular momentum is constant in both magnitude and direction, the total energy is constant, and the speed (so the momentum) changes along the orbit.

Teacher's Note:
1. A central force gives acceleration toward the centre.
2. Energy and angular momentum are conserved.

Answer: (a) 2600 km

Explanation:
1. With r the distance from the centre, \(V = -\frac{GM}{r}\) and \(g = \frac{GM}{r^2}\), so \(\frac{|V|}{g} = r\).
2. \(r = \frac{5.4\times10^7}{6.0} = 9\times10^6\) m = 9000 km.
3. The height is \(9000 - 6400 = 2600\) km.

Teacher's Note:
1. The ratio of potential to field gives the distance from the centre.
2. Subtract the radius of the earth.

Answer: (b) \(1 : 2\sqrt2\)

Explanation:
1. \(v_e = \sqrt{\frac{2GM}{R}}\) with \(M = \frac43\pi R^3\rho\), so \(v_e\propto R\sqrt\rho\).
2. For the planet, \(v_p = v_e\times2\times\sqrt2\).
3. So \(v_e : v_p = 1 : 2\sqrt2\).

Teacher's Note:
1. Escape speed is proportional to R times root of density.
2. Substitute the ratios.

Answer: (d) \(-\frac{mg_0R^2}{2(R+h)}\)

Explanation:
1. The total energy of a satellite in a circular orbit is \(-\frac{GMm}{2(R+h)}\).
2. With \(GM = g_0R^2\), this is \(-\frac{mg_0R^2}{2(R+h)}\).

Teacher's Note:
1. Total energy is negative for a bound satellite.
2. Replace GM by g0 R squared.

Answer: (c) d = 2 km

Explanation:
1. The fall at height h is \(\frac{2h}{R}g\) and at depth d is \(\frac dRg\).
2. Equal values give \(d = 2h = 2\) km.

Teacher's Note:
1. The fall with height is twice the fall with depth.
2. Use the small-height approximation.

Answer: (d) g on the earth will not change.

Explanation:
1. The sun's mass has almost no effect on g at the earth's surface, which is \(\frac{GM_e}{R^2}\).
2. With G ten times larger, g becomes ten times larger.
3. That makes raindrops fall faster, walking harder and the pendulum period shorter, so only (d) is incorrect.

Teacher's Note:
1. g depends on G times the earth's mass.
2. A bigger G means a bigger g.

Answer: (b) \(K_A > K_B > K_C\)

Explanation:
1. Kinetic energy is largest at the point nearest the Sun and smallest at the farthest point.
2. B is at a distance from the Sun that is between those at A and C.
3. So \(K_A > K_B > K_C\).

Teacher's Note:
1. Speed and kinetic energy fall as the planet moves away from the Sun.
2. Compare the distances from the Sun.

Answer: (d) 100 N

Explanation:
1. Inside the earth \(g' = g\frac rR\).
2. Half way down, \(r = \frac R2\), so the weight is \(200\times\frac12 = 100\) N.

Teacher's Note:
1. Inside a uniform earth, weight is proportional to the distance from the centre.
2. Half the radius means half the weight.

Answer: (c) \(\frac12mgR\)

Explanation:
1. \(W = GMm\left(\frac1R - \frac1{2R}\right) = \frac{GMm}{2R}\).
2. With \(GM = gR^2\), \(W = \frac12mgR\).

Teacher's Note:
1. Use the exact potential, not mgh.
2. GM equals gR squared.

Answer: (a) \(6\sqrt2\) h

Explanation:
1. The orbit radii are \(7R_E\) and \(3.5R_E\), a ratio of 2.
2. \(T\propto r^{3/2}\), so \(T = 24\left(\frac12\right)^{3/2} = \frac{24}{2\sqrt2} = 6\sqrt2\) h.

Teacher's Note:
1. Add R_E to each height.
2. Use Kepler's third law.

Answer: (b) \(\frac{GMmh}{R(R+h)}\)

Explanation:
1. \(\Delta U = -\frac{GMm}{R+h} + \frac{GMm}{R} = GMm\frac{h}{R(R+h)}\).
2. This reduces to mgh only when h is much smaller than R.

Teacher's Note:
1. Take final minus initial.
2. The result is positive when raising a body.

Answer: (d) S2 and S4

Explanation:
1. A physical balance compares masses, so the reading is the same everywhere (S2 is true, S1 false).
2. A spring balance reads weight, which is larger at the pole where g is larger (S4 true, S3 false).

Teacher's Note:
1. Beam balance compares masses, spring balance measures weight.
2. g is largest at the poles.

Answer: (c) \(\frac{4GM_1m}{(R_1 + R_2)^2}\)

Explanation:
1. The particle lies between the shells, so it is inside the outer shell (no force from \(M_2\)).
2. It lies outside the inner shell, which acts as a point mass \(M_1\) at the centre, so \(F = \frac{GM_1m}{\left(\frac{R_1+R_2}{2}\right)^2}\).
3. This equals \(\frac{4GM_1m}{(R_1+R_2)^2}\).

Teacher's Note:
1. A uniform shell exerts no force inside it.
2. Outside, it acts as a point mass at its centre.

Answer: (b) -180 J

Explanation:
1. The gravitational force on the particle is \(m\vec I = (20\hat i + 20\hat j)\) N and the displacement is \(5\hat i + 4\hat j\) m.
2. Work by the field \(= 20\times5 + 20\times4 = 180\) J.
3. The agent moves it slowly (no change in kinetic energy), so its work is the negative of this: -180 J.

Teacher's Note:
1. External work is minus the work done by the field.
2. Slow motion means no kinetic energy change.

Answer: (a) \(\sqrt{\frac{4G(M_1 + M_2)}{d}}\)

Explanation:
1. At the midpoint the potential is \(V = -\frac{G(M_1 + M_2)}{d/2} = -\frac{2G(M_1 + M_2)}{d}\).
2. Escape needs \(\frac12v^2 = \frac{2G(M_1 + M_2)}{d}\).
3. So \(v = \sqrt{\frac{4G(M_1 + M_2)}{d}}\).

Teacher's Note:
1. Potential adds as a scalar.
2. Each body is at distance d over 2.

Answer: (d) \(\frac{R^2g}{\omega^2}\)

Explanation:
1. For a geostationary satellite, gravity gives the centripetal force: \(\frac{GM}{r^2} = \omega^2r\).
2. So \(r^3 = \frac{GM}{\omega^2}\), and with \(GM = gR^2\) we get \(r^3 = \frac{R^2g}{\omega^2}\).

Teacher's Note:
1. The satellite has the same omega as the earth.
2. GM equals gR squared.

Answer: (b) \(R^{-5/4}\)

Explanation:
1. The force supplies the centripetal force: \(m\omega^2R\propto R^{-3/2}\).
2. So \(\omega^2\propto R^{-5/2}\) and \(\omega\propto R^{-5/4}\).

Teacher's Note:
1. Equate the force law to m omega squared R.
2. Take the square root of the exponent.

Answer: (a) \(m\sqrt{GMR_0}\)

Explanation:
1. The orbital speed is \(v = \sqrt{\frac{GM}{R_0}}\).
2. \(L = mvR_0 = m\sqrt{\frac{GM}{R_0}}R_0 = m\sqrt{GMR_0}\).

Teacher's Note:
1. Angular momentum is mvr for a circular orbit.
2. The mass of the satellite appears, not the mass of the earth.

Answer: (d) \(\frac{v_p - v_a}{v_p + v_a}\)

Explanation:
1. The eccentricity is \(e = \frac{r_a - r_p}{r_a + r_p}\).
2. Angular momentum gives \(\frac{r_a}{r_p} = \frac{v_p}{v_a}\), so \(e = \frac{v_p - v_a}{v_p + v_a}\).

Teacher's Note:
1. e is defined from the extreme distances.
2. Replace distances by inverse speeds.

Answer: (d)

Explanation:
1. Inside a uniform solid sphere the field grows linearly with r, \(I = \frac{GMr}{R^3}\), reaching \(\frac{GM}{R^2}\) at the surface.
2. Outside it falls as \(\frac{GM}{r^2}\).
3. So the magnitude of I rises in a straight line up to R and then decays as an inverse square curve.

Teacher's Note:
1. Linear inside, inverse square outside.
2. The peak is at the surface.

Answer: (c)

Explanation:
1. Inside the shell the potential is constant and equal to \(-\frac{GM}{R}\).
2. Outside it is \(-\frac{GM}{r}\), which rises from \(-\frac{GM}{R}\) toward zero at infinity.
3. Only curve (c) shows a constant negative value inside and a rising curve outside.

Teacher's Note:
1. The potential is negative everywhere and flat inside a shell.
2. It approaches zero at infinity.

Answer: (d)

Explanation:
1. Inside the earth \(g = \frac{GMr}{R^3}\), so g increases linearly with r up to the surface.
2. Outside, \(g = \frac{GM}{r^2}\) decreases with r.
3. Figure (d) shows exactly this: a straight line up to R followed by an inverse square fall.

Teacher's Note:
1. g is zero at the centre and maximum at the surface.
2. Linear inside and inverse square outside.

Answer: (a)

Explanation:
1. The intensity of the gravitational field is directed toward the centre, so with outward r positive its value is negative.
2. Inside, its magnitude grows linearly with r, and outside it falls as \(\frac{1}{r^2}\).
3. Only (a) shows a negative value with a linear increase in magnitude up to R and an inverse square decay afterwards.

Teacher's Note:
1. Field vector points inward, so it is negative.
2. Its magnitude is largest at the surface.

Answer: (d)

Explanation:
1. From the centre to the surface g increases linearly, \(g\propto r\).
2. Beyond the surface g falls as \(\frac{1}{r^2}\).
3. Graph (d) shows a straight line to R and then an inverse square decay.

Teacher's Note:
1. g is zero at the centre.
2. Maximum at the surface.