JEE Physics Gravitation MCQs Set 11

Answer: (b) 32 N

Explanation:
1. At height \(\frac R2\) the distance from the centre is \(\frac{3R}{2}\), so \(F' = 72\left(\frac{R}{1.5R}\right)^2 = 72\times\frac49\).
2. So \(F' = 32\) N.

Teacher's Note:
1. Weight scales as the inverse square of distance from the centre.
2. Measure the distance from the centre, not the surface.

Answer: (b) d = 2h

Explanation:
1. The decrease at height h is \(\frac{2h}{R}g\).
2. The decrease at depth d is \(\frac dRg\).
3. Equating, \(d = 2h\).

Teacher's Note:
1. g falls twice as fast with height as with depth.
2. Remember the factor 2 for height.

Answer: (d) remains the same

Explanation:
1. Mass is the quantity of matter in the body and does not depend on location.
2. Only the weight changes with g.

Teacher's Note:
1. Mass is constant, weight is not.
2. Do not confuse the two.

Answer: (a) decrease by 1 percent

Explanation:
1. At height h the decrease is \(\frac{2h}{R} = 2\%\), so \(\frac hR = 1\%\).
2. At depth h the decrease is \(\frac hR = 1\%\).
3. The weight decreases by 1 percent.

Teacher's Note:
1. Height gives twice the change of depth.
2. Use the small-change formulas.

Answer: (b) remain constant

Explanation:
1. The poles lie on the axis of rotation, so the centrifugal term \(\omega^2R\cos^2\lambda\) is already zero there.
2. Stopping the rotation therefore leaves g at the poles unchanged.

Teacher's Note:
1. At the poles latitude is 90 degrees and the effect of rotation is zero.
2. Only g at other latitudes would change.

Answer: (b) decreases

Explanation:
1. The effective g at latitude \(\lambda\) is \(g - \omega^2R\cos^2\lambda\).
2. A larger \(\omega\) reduces it everywhere except at the poles, where \(\cos\lambda = 0\).
3. So the weight decreases.

Teacher's Note:
1. Spin lowers effective gravity.
2. The poles are not affected.

Answer: (a) \(\sqrt{\frac{g}{2R}}\)

Explanation:
1. At the equator \(g_{eff} = g - \omega^2R\).
2. Setting \(g - \omega^2R = \frac g2\) gives \(\omega^2 = \frac{g}{2R}\).
3. So \(\omega = \sqrt{\frac{g}{2R}}\).

Teacher's Note:
1. At the equator the whole centrifugal term acts against g.
2. Solve for omega.

Answer: (c) \(g_A < g_B\)

Explanation:
1. \(g_{eff} = g - \omega^2R\cos^2\lambda\).
2. At 30 degrees \(\cos^2\lambda = 0.75\), at 60 degrees it is 0.25.
3. The reduction is larger at A, so \(g_A < g_B\).

Teacher's Note:
1. Latitude closer to the equator means a smaller g.
2. The reduction varies as the square of the cosine of latitude.

Answer: (c) \(\pi\sqrt{\frac Rg}\)

Explanation:
1. Weightless means \(g - \omega^2R\cos^2\lambda = 0\).
2. At \(\lambda = 60^\circ\), \(\cos^2\lambda = \frac14\), so \(\omega^2 = \frac{4g}{R}\) and \(\omega = 2\sqrt{\frac gR}\).
3. Then \(T = \frac{2\pi}{\omega} = \pi\sqrt{\frac Rg}\).

Teacher's Note:
1. Use the latitude formula for effective g.
2. T is two pi over omega.

Answer: (c) \(-10G\) J/kg

Explanation:
1. \(V = -\frac{GM}{r} = -\frac{G(1)}{0.1}\).
2. So \(V = -10G\) J/kg.

Teacher's Note:
1. Potential is negative for an attractive force.
2. Use V equals minus GM over r.

Answer: (d) \(-\frac{4GM}{r}\)

Explanation:
1. The centre of mass is at the midpoint, at distance \(\frac r2\) from each sphere.
2. Potential is a scalar: \(V = -2\cdot\frac{GM}{r/2} = -\frac{4GM}{r}\).

Teacher's Note:
1. Potentials add as scalars.
2. Use the distance from each mass to the point.

Answer: (b) the gravitational field is zero but the potential is \(-\frac{GM}{R}\)

Explanation:
1. Inside a uniform shell the field is zero.
2. The potential is constant inside and equals its value at the surface, \(-\frac{GM}{R}\).

Teacher's Note:
1. Field zero does not mean potential zero.
2. Potential inside a shell is constant.

Answer: (d) acceleration

Explanation:
1. In free fall the acceleration is g for every mass.
2. Kinetic energy \(mgh\) and momentum \(m\sqrt{2gh}\) depend on the mass, so they differ.
3. Only the acceleration (and speed) is the same.

Teacher's Note:
1. Free fall acceleration is independent of mass.
2. Energy and momentum depend on mass.

Answer: (c) \(-GMm_m\left(\frac{80}{r_1} + \frac{1}{r_2}\right)\)

Explanation:
1. The potential energy is the sum of the two terms: \(U = -\frac{GM_em}{r_1} - \frac{GM_mm}{r_2}\).
2. Substituting \(M_e = 80M_m\) gives \(U = -GM_mm\left(\frac{80}{r_1} + \frac1{r_2}\right)\).

Teacher's Note:
1. Potential energy is a scalar sum.
2. Both attractions give negative terms.

Answer: (a) infinity

Explanation:
1. Potential energy is negative at every finite distance and rises toward zero as the distance increases.
2. Zero is the highest value, reached at infinity.

Teacher's Note:
1. Gravitational potential energy is always negative with the zero at infinity.
2. The maximum is at infinity.

Answer: (b) 1 J

Explanation:
1. A uniform field gives a constant potential gradient of \(\frac{4}{40} = 0.1\) J/kg per metre.
2. For 5 m the potential difference is 0.5 J/kg.
3. Work \(= m\,\Delta V = 2\times0.5 = 1\) J.

Teacher's Note:
1. Work equals mass times the potential difference.
2. Potential varies linearly in a uniform field.

Answer: (c) 40 m

Explanation:
1. The landing speed is \(v = \sqrt{2gh}\), and the safe landing speed is the same.
2. So \(g_1h_1 = g_2h_2\), giving \(h_2 = 2\times\frac{9.8}{0.49} = 40\) m.

Teacher's Note:
1. Same safe speed means g times h is constant.
2. Lower g allows a greater height.

Answer: (c) \(\frac{U}{2r}\)

Explanation:
1. \(U = -\frac{GMm}{2r}\), so \(GMm = 2r|U|\).
2. The force is \(\frac{GMm}{(2r)^2} = \frac{2r|U|}{4r^2} = \frac{|U|}{2r}\).

Teacher's Note:
1. Force is the negative gradient of potential energy.
2. Eliminate GMm using U.

Answer: (a) \(\frac{GMmx}{R(R+x)}\)

Explanation:
1. \(\Delta U = -\frac{GMm}{R+x} - \left(-\frac{GMm}{R}\right) = GMm\left(\frac1R - \frac1{R+x}\right)\).
2. This simplifies to \(\frac{GMmx}{R(R+x)}\).

Teacher's Note:
1. Change in potential energy is final minus initial.
2. Common denominator gives the result.

Answer: (c) \(mgR_E\frac{n}{n+1}\)

Explanation:
1. \(\Delta U = GMm\left(\frac1{R_E} - \frac{1}{R_E + nR_E}\right) = \frac{GMm}{R_E}\cdot\frac{n}{n+1}\).
2. With \(GM = gR_E^2\) this is \(mgR_E\frac{n}{n+1}\).

Teacher's Note:
1. Use GM equals gR squared.
2. For small n the change is about n times mgR.

Answer: (d) \(-\frac{11GMm}{8R}\)

Explanation:
1. Inside a uniform sphere \(U = -\frac{GMm}{2R^3}(3R^2 - r^2)\).
2. At \(r = \frac R2\): \(U = -\frac{GMm}{2R^3}\left(3R^2 - \frac{R^2}{4}\right) = -\frac{GMm}{2R}\cdot\frac{11}{4}\).
3. So \(U = -\frac{11GMm}{8R}\).

Teacher's Note:
1. Use the interior potential formula.
2. At the surface it gives -GMm/R and at the centre -3GMm/2R.

Answer: (b) \(-\frac{3GM^2}{a}\), \(\frac{3GM^2}{2a}\)

Explanation:
1. The system has three pairs, each with energy \(-\frac{GM^2}{a}\), so \(U = -\frac{3GM^2}{a}\).
2. With side 2a, \(U' = -\frac{3GM^2}{2a}\).
3. Work done equals \(U' - U = \frac{3GM^2}{a} - \frac{3GM^2}{2a} = \frac{3GM^2}{2a}\).

Teacher's Note:
1. Count the pairs: three bodies make three pairs.
2. External work equals the increase in potential energy.

Answer: (b) \(\sqrt{\frac{2GM}{3R}}\)

Explanation:
1. Energy conservation gives \(\frac12v^2 = GM\left(\frac1R - \frac{1}{1.5R}\right) = \frac{GM}{3R}\).
2. So \(v = \sqrt{\frac{2GM}{3R}}\).

Teacher's Note:
1. Use energy conservation with the exact potential.
2. The height is measured from the surface.

Answer: (d) \(\sqrt{2Rg}\)

Explanation:
1. Energy conservation gives \(\frac12v^2 = \frac{GM}{R}\).
2. With \(GM = gR^2\), \(v = \sqrt{2gR}\), which equals the escape velocity.

Teacher's Note:
1. Falling from infinity gives the escape velocity at the surface.
2. Use GM equals gR squared.

Answer: (b) \(\sqrt{gR}\)

Explanation:
1. Energy conservation: \(\frac12v^2 = GM\left(\frac1R - \frac{1}{2R}\right) = \frac{GM}{2R}\).
2. So \(v^2 = \frac{GM}{R} = gR\) and \(v = \sqrt{gR}\).

Teacher's Note:
1. Do not use \(v = \sqrt{2gh}\) for large heights.
2. g is not constant over a height equal to R.

Answer: (b) (ii) and (iv)

Explanation:
1. \(v_e = \sqrt{\frac{2GM}{r}}\) depends on the mass of the earth and on the distance r from the centre, i.e., the launch height.
2. It is independent of the mass of the body and of the direction of projection.

Teacher's Note:
1. Escape speed depends on the planet and on the place.
2. It does not depend on the projectile.

Answer: (a) 11.2 km/s

Explanation:
1. Escape velocity is a scalar condition based on energy.
2. It does not depend on the direction of projection, so it remains 11.2 km/s.

Teacher's Note:
1. The escape speed is the same in every direction.
2. Neglect air resistance and rotation.