JEE Physics Gravitation MCQs Set 18

Answer: (b) 1

Explanation:
1. Electric charge is a fundamental constant of nature and does not depend on location or gravity.
2. So the ratio is 1.

Teacher's Note:
1. Gravity changes the drop's weight, not its charge.
2. Charge is quantised in units of e everywhere.

Answer: (b) 110 km/s

Explanation:
1. \( v_e=\sqrt{2GM/R}\propto\sqrt{M/R} \).
2. Ratio = √(10/(1/10)) = √100 = 10, so 110 km/s.

Teacher's Note:
1. Escape speed varies as √(M/R).
2. Mass up by 10 and radius down by 10 multiplies M/R by 100.

Answer: (c) 2R

Explanation:
1. \( g'=g\frac{R^2}{(R+h)^2}=\frac g9 \) gives R + h = 3R.
2. So h = 2R.

Teacher's Note:
1. g falls as 1/r² measured from the centre.
2. Subtract R from the distance to get the height.

Answer: (a) \( -\frac{9Gm}{r} \)

Explanation:
1. Field zero at distance x from m: \( \frac m{x^2}=\frac{4m}{(r-x)^2} \) gives x = r/3.
2. Potential = \( -\frac{Gm}{r/3}-\frac{4Gm}{2r/3}=-\frac{3Gm}r-\frac{6Gm}r=-\frac{9Gm}r \).

Teacher's Note:
1. The neutral point lies nearer the smaller mass.
2. Add potentials as scalars.

Answer: (c) \( \sqrt{\frac{Gm}{4R}} \)

Explanation:
1. The particles are diametrically opposite, so separation = 2R.
2. \( \frac{Gm^2}{(2R)^2}=\frac{mv^2}R \) gives \( v=\sqrt{Gm/4R} \).

Teacher's Note:
1. The gravitational force uses the separation 2R, but the centripetal force uses the radius R.
2. Do not mix the two distances.

Answer: (a) 2 hr

Explanation:
1. \( T\propto r^{3/2} \): \( T=24\left(\frac{6400}{36000}\right)^{3/2}\approx1.8 \) hr.
2. This is about 2 hr.

Teacher's Note:
1. A near-surface satellite has an orbital radius of nearly R.
2. Geostationary period is 24 hr.

Answer: (d) 100R

Explanation:
1. Potential inside a uniform sphere: \( V=-\frac{GM(3R^2-r^2)}{2R^3} \); at r = 0.99R, V = −1.00995 GM/R.
2. Initial KE = ½v_e² = GM/R, so E = −0.00995 GM/R.
3. At the top: −GM/r = E, r ≈ 100.5R, i.e. height ≈ 100R.

Teacher's Note:
1. Inside the Moon the potential formula changes.
2. Total energy is only slightly negative, so the projectile goes very far.

Answer: (a) T_A = T_B

Explanation:
1. The two stars always stay on opposite sides of the centre of mass.
2. So they complete one revolution in the same time: T_A = T_B.

Teacher's Note:
1. Both stars have the same angular speed, so the same period.
2. Kepler's relation T² ∝ R³ applies to bodies orbiting one centre of force, not to the two stars here.

Answer: (a) 31.2 G, directed from P towards the mid-point of AB

Explanation:
1. PA = PB = \( \sqrt{8^2+6^2}=10 \) cm = 0.1 m.
2. Acceleration due to each sphere: \( a_1=\frac{G(0.260)}{(0.1)^2}=26G \), along PA and PB.
3. The components parallel to AB cancel. Each makes an angle with the vertical whose cosine is 6/10, so the vertical components add: \( 2\times26G\times\frac{6}{10}=31.2G \).
4. Direction: perpendicular to AB, from P towards the mid-point of AB.

Teacher's Note:
1. The acceleration does not depend on the mass of the small sphere.
2. Resolve along and perpendicular to AB, using the symmetry of the arrangement.

Answer: (c) \( \frac{mg}{R}(H_1-H_2)\ell \)

Explanation:
1. At a small height H, \( g_H\approx g\left(1-\frac{2H}{R}\right) \).
2. Torque of the right block minus that of the left block: \( \frac\ell2 mg\left[\left(1-\frac{2H_2}R\right)-\left(1-\frac{2H_1}R\right)\right] \).
3. This equals \( \frac{mg}{R}(H_1-H_2)\ell \), and it tends to turn the beam so that the lower pan goes down.

Teacher's Note:
1. Both weights act at equal arms ℓ/2, so only the difference in g matters.
2. Use the binomial form g(1 − 2h/R) for small heights.

Answer: (c) \( \frac{\sqrt3Gm^2}{2d^2} \) in the direction SQ only

Explanation:
1. SQ = √3 d, SP = 2√3 d and SR = 2d.
2. Forces: \( F_{SQ}=\frac{\sqrt3Gm^2}{3d^2} \), \( F_{SP}=\frac{\sqrt3Gm^2}{12d^2} \), \( F_{SR}=\frac{Gm^2}{4d^2} \).
3. Horizontal components: from SP, \( F_{SP}\cdot\frac{\sqrt3}{2}=\frac{Gm^2}{8d^2} \) towards P; from SR, \( F_{SR}\cdot\frac12=\frac{Gm^2}{8d^2} \) towards R. They cancel.
4. Vertical components: \( \sqrt3\frac{Gm^2}{d^2}\left(\frac13+\frac1{24}+\frac18\right)=\frac{\sqrt3Gm^2}{2d^2} \) along SQ.

Teacher's Note:
1. Resolve each force into components parallel and perpendicular to the line TU.
2. The symmetric choice of masses and distances makes the horizontal parts cancel.

Answer: (d) \( \frac{2\pi Gm}{\ell^2} \) along +y axis

Explanation:
1. Radius of the semicircle: \( R=\ell/\pi \); mass per unit length \( \lambda=m/\ell \).
2. The x-components cancel by symmetry. The y-component of the field of an element is \( \frac{G\lambda R\,d\theta}{R^2}\sin\theta \).
3. Total: \( E=\frac{G\lambda}{R}\int_0^\pi\sin\theta\,d\theta=\frac{2G\lambda}{R}=\frac{2\pi Gm}{\ell^2} \), directed towards the wire, i.e. along +y.

Teacher's Note:
1. Use symmetry to drop the component along x.
2. The field always points towards the mass.

Answer: (a) \( \frac{G(M_1+M_2)m}{p^2} \), \( \frac{GM_1m}{q^2} \) and zero

Explanation:
1. At A, outside both shells, both act as point masses at O: \( \frac{G(M_1+M_2)m}{p^2} \).
2. At B, only the inner shell M1 lies inside radius q; the outer shell exerts no force on a point inside it: \( \frac{GM_1m}{q^2} \).
3. At C, inside both shells, the field of a shell is zero: force = 0.

Teacher's Note:
1. Shell theorem: a uniform shell attracts an outside point as if its mass were at the centre, and exerts no force inside.
2. Include only the shells that lie inside the point's distance from the centre.

Answer: (d) c

Explanation:
1. Combine this hemisphere with an identical one placed to complete the sphere. Inside a complete shell the field is zero.
2. By reflection symmetry the components of the two halves' fields parallel to the open face are equal, so each must vanish when the sum is zero; only the component perpendicular to the face remains.
3. This component is directed towards the mass of the shell, i.e. into the bowl, along arrow c.

Teacher's Note:
1. The field at any point of the plane of a hemispherical shell's open face is perpendicular to that plane.
2. Gravitational field points towards the attracting mass.

Answer: (c) \( V_A=V_B=V_C \)

Explanation:
1. Complete the sphere with a second, mirror-image hemisphere of mass M. The full shell (mass 2M) has constant potential \( -\frac{2GM}{R} \) everywhere inside it.
2. For any point in the base plane the two hemispheres contribute equally, since they are mirror images.
3. So each contributes \( -\frac{GM}{R} \), and the potential is the same at A, B and C.

Teacher's Note:
1. Potential is a scalar, so equal contributions from mirror halves simply halve the total.
2. Inside a shell the potential is constant.

Answer: (c) A and B are the kinetic and potential energies and C is the total energy of the system

Explanation:
1. For a circular orbit \( KE=\frac{GMm}{2r} \), \( PE=-\frac{GMm}{r} \), \( E=-\frac{GMm}{2r} \).
2. KE is positive, so it is curve A. PE is twice as negative as the total energy, so the lowest curve B is PE and the middle curve C is the total energy.

Teacher's Note:
1. Total energy = −KE = PE/2 for circular orbits.
2. The curve nearest the axis on the negative side is the total energy.