JEE Physics Gravitation MCQs Set 21

Answer: (b) 2

Explanation:
1. The centre of mass condition is \( m_1r_1=m_2r_2 \). With \( r_{light}=2r_{heavy} \), \( \dfrac{m_{heavy}}{m_{light}}=\dfrac{r_{light}}{r_{heavy}}=2 \).

Teacher's Note:
1. The lighter star sits farther from the centre of mass.
2. Both stars have the same angular velocity and the same period.

Answer: (a) 11

Explanation:
1. The galaxy's mass \( Nm_s \) acts at its centre: \( \dfrac{GNm_s}{R^2}=\dfrac{4\pi^2R}{T^2} \).
2. \( N=\dfrac{4\pi^2R^3}{Gm_sT^2}=\dfrac{40\times10^{60}}{\frac{20}{3}\times10^{-11}\times2\times10^{30}\times10^{30}}=3\times10^{11} \), so \( a=11 \).

Teacher's Note:
1. For a body orbiting at the edge of a uniform sphere, the sphere acts as if all its mass is at the centre.
2. Substitute carefully; the powers of ten must add correctly.

Answer: (c) 9.5 cm/s

Explanation:
1. Distance from the starting point to the ring: \( \sqrt{8^2+6^2}=10\text{ m} \).
2. Energy conservation: \( \tfrac12v^2=GM_R\left(\dfrac18-\dfrac1{10}\right)=\dfrac{GM_R}{40} \).
3. \( v^2=\dfrac{GM_R}{20}=\dfrac{6.67\times10^{-11}\times2.7\times10^9}{20}\approx9.0\times10^{-3}\text{ m}^2/\text{s}^2 \), so \( v\approx0.095\text{ m/s}=9.5\text{ cm/s} \).

Teacher's Note:
1. The mass of the particle cancels out of the speed.
2. Convert the final answer from m/s to cm/s.

Answer: (b) 15

Explanation:
1. Gravitational self-energy of a uniform sphere \( U=\dfrac35\dfrac{GM^2}{R} \).
2. Using \( GM=gR^2 \): \( |U|=\dfrac35gMR=0.6\times10\times2.5\times10^{31}=1.5\times10^{32}\text{ J}=15\times10^{31}\text{ J} \).

Teacher's Note:
1. The binding energy is the energy to separate all the pieces to infinity.
2. The factor 3/5 comes from adding the energy of successive shells.

Answer: (d) \( 30^\circ \)

Explanation:
1. The orbit radius is \( 2R \). The signal reaches places from where the satellite is at or above the horizon, i.e. the line of sight is tangent to the earth.
2. If the angle at the centre between the satellite and the farthest place is \( \theta \), then \( \cos\theta=\dfrac R{2R} \), so \( \theta=60^\circ \) from the equator.
3. Co-latitude is measured from the pole: \( 90^\circ-60^\circ=30^\circ \).

Teacher's Note:
1. Tangent lines are perpendicular to the radius at the point of contact.
2. Latitude plus co-latitude is 90 degrees.

Answer: (c) (1) and (4) only

Explanation:
1. For two bodies about their centre of mass, \( T=2\pi\sqrt{\dfrac{r^3}{G(m_1+m_2)}}=2\pi\sqrt{\dfrac{r^3}{3Gm}} \).
2. So \( T\propto r^{3/2} \) and \( T\propto m^{-1/2} \).

Teacher's Note:
1. The total mass of the system appears in the formula, so a larger mass shortens the period.
2. This is Kepler's third law for a two-body system.

Answer: (a) (2) and (3) only

Explanation:
1. The earth rotates west to east once in 24 h. If the satellite also goes west to east, the relative angular speed is \( 2\pi\left(\dfrac1{1.5}-\dfrac1{24}\right)=2\pi\cdot\dfrac{15}{24} \), so the time is \( \dfrac{24}{15}=1.6\text{ h} \).
2. If it goes east to west, relative angular speed is \( 2\pi\left(\dfrac{16}{24}+\dfrac1{24}\right)=2\pi\cdot\dfrac{17}{24} \), so the time is \( \dfrac{24}{17}\text{ h} \).

Teacher's Note:
1. Relative angular velocity is the difference when the two turn in the same sense and the sum when they turn in opposite senses.
2. The point P is carried along by the earth's rotation.

Answer: (d) (2) and (3) only

Explanation:
1. For a circular orbit \( v=\sqrt{GM/r} \), so the speed and kinetic energy \( \tfrac{GMm}{2r} \) increase as \( r \) decreases.
2. Potential energy \( -\dfrac{GMm}{r} \) and total energy \( -\dfrac{GMm}{2r} \) both become more negative, i.e. decrease.

Teacher's Note:
1. The satellite speeds up as it moves to a lower orbit.
2. The total energy is the negative of the kinetic energy.

Answer: (b) 3 km/s

Explanation:
1. \( g=\\tfrac43\\pi G\\rho R \), so \( \\dfrac{R_p}{R_e}=\\dfrac{g_p/g_e}{\\rho_p/\\rho_e}=\\dfrac{\\sqrt6/11}{2/3}=\\dfrac{3\\sqrt6}{22} \).
2. Escape speed \( v=\\sqrt{2gR} \): \( \\dfrac{v_p}{v_e}=\\sqrt{\\dfrac{\\sqrt6}{11}\\cdot\\dfrac{3\\sqrt6}{22}}=\\sqrt{\\dfrac{18}{242}}=\\dfrac3{11} \).
3. So \( v_p=3\\text{ km/s} \).

Teacher's Note:
1. Find the radius ratio first from g and the density.
2. Then take the product g times R under the square root.

Answer: (a) \( mV^2 \)

Explanation:
1. Escape speed from the orbit is \( \\sqrt2V \).
2. Kinetic energy \( =\\tfrac12m(\\sqrt2V)^2=mV^2 \).

Teacher's Note:
1. The escape speed at a given distance is \( \\sqrt2 \) times the circular orbital speed there.
2. Total energy of the object must be zero.

Answer: (c) (1) and (4) only

Explanation:
1. Potential energy at the midpoint \( =-\\dfrac{2GMm}{L} \), so escape needs \( \\tfrac12mv^2=\\dfrac{2GMm}{L}\\Rightarrow v=2\\sqrt{\\dfrac{GM}{L}} \).
2. The two fixed bodies exert only conservative forces, so the energy of the particle stays constant.

Teacher's Note:
1. The bodies are fixed, so no external work is done on the particle.
2. The direction of projection does not change the escape speed.

Answer: (b) 108 N

Explanation:
1. Planet radius \( R=6\\times10^5\\text{ m} \) and surface gravity \( g_p=\\dfrac{g}{10}=1\\text{ m/s}^2 \). Inside, \( g(r)=g_p\\dfrac rR \).
2. Force at the top equals the weight of the wire: \( \\int_{0.8R}^{R}\\lambda g_p\\dfrac rR\\,dr=\\dfrac{\\lambda g_p}{2R}(R^2-0.64R^2)=0.18\\lambda g_pR \).
3. \( =0.18\\times10^{-3}\\times1\\times6\\times10^5=108\\text{ N} \).

Teacher's Note:
1. Gravity inside a uniform planet increases linearly with distance from the centre.
2. Integrate the weight of each element of the wire.

Answer: (d) 42 km/s

Explanation:
1. Required: \( \\tfrac12v^2=\\dfrac{GM_e}{R_e}+\\dfrac{GM_s}{r} \).
2. The sun term is \( \\dfrac{3\\times10^5}{2.5\\times10^4}\\dfrac{GM_e}{R_e}=12\\dfrac{GM_e}{R_e} \), so \( v^2=13\\,v_e^2 \).
3. \( v=\\sqrt{13}\\times11.2=40.4\\text{ km/s} \), closest to 42 km/s.

Teacher's Note:
1. Potentials from the earth and the sun add at the launch point.
2. Ignore the motion of the earth around the sun, as stated.

Answer: (c) P-3, Q-2, R-4, S-1

Explanation:
1. \( v\\propto R^{-1/2}\\Rightarrow\\dfrac{v_1}{v_2}=\\sqrt{4}=2 \) (value 3).
2. \( L=mvR\\Rightarrow\\dfrac{L_1}{L_2}=2\\times2\\times\\dfrac14=1 \) (value 2).
3. \( K=\\tfrac12mv^2\\Rightarrow\\dfrac{K_1}{K_2}=2\\times4=8 \) (value 4).
4. \( T\\propto R^{3/2}\\Rightarrow\\dfrac{T_1}{T_2}=\\left(\\dfrac14\\right)^{3/2}=\\dfrac18 \) (value 1).

Teacher's Note:
1. The satellite closer to the planet moves faster and has a shorter period.
2. Angular momentum is mass times speed times radius.

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

Explanation:
1. Let the point be at distance \( x \) from \( m \): \( \\dfrac{Gm}{x^2}=\\dfrac{4Gm}{(r-x)^2}\\Rightarrow r-x=2x\\Rightarrow x=\\dfrac r3 \).
2. Potential \( =-\\dfrac{Gm}{r/3}-\\dfrac{4Gm}{2r/3}=-\\dfrac{3Gm}{r}-\\dfrac{6Gm}{r}=-\\dfrac{9Gm}{r} \).

Teacher's Note:
1. Zero field means the two pulls are equal and opposite, but the potentials still add.
2. The point is nearer the smaller mass.

Answer: (d) \( \\sqrt{\\dfrac{Gm}{4R}} \)

Explanation:
1. Separation is \( 2R \), so \( \\dfrac{mv^2}{R}=\\dfrac{Gm^2}{4R^2} \).
2. Hence \( v=\\sqrt{\\dfrac{Gm}{4R}} \).

Teacher's Note:
1. The circle's radius is the distance of each particle from the centre of mass.
2. The force uses the full separation of 2R.

Answer: (b) \( \\dfrac{5GmM}{6R} \)

Explanation:
1. Orbit radius \( 3R \). Energy in orbit \( =-\\dfrac{GMm}{2(3R)}=-\\dfrac{GMm}{6R} \).
2. Energy at the surface at rest \( =-\\dfrac{GMm}{R} \).
3. Energy required \( =-\\dfrac{GMm}{6R}+\\dfrac{GMm}{R}=\\dfrac{5GMm}{6R} \).

Teacher's Note:
1. The satellite must also be given orbital kinetic energy, not just lifted.
2. The orbit radius is altitude plus planet radius.

Answer: (a) \( -\\dfrac{GM}{R} \)

Explanation:
1. Full sphere at a point \( R/2 \) from the centre: \( V_1=-\\dfrac{GM}{2R^3}\\left(3R^2-\\dfrac{R^2}{4}\\right)=-\\dfrac{11GM}{8R} \).
2. The removed sphere (mass \( M/8 \), radius \( R/2 \)) at its own centre: \( V_2=-\\dfrac32\\dfrac{G(M/8)}{R/2}=-\\dfrac{3GM}{8R} \).
3. Remaining potential \( =V_1-V_2=-\\dfrac{11GM}{8R}+\\dfrac{3GM}{8R}=-\\dfrac{GM}{R} \).

Teacher's Note:
1. Subtract the potential of the removed sphere from that of the full sphere.
2. Inside a uniform sphere the potential is \( -\\dfrac{GM(3R^2-r^2)}{2R^3} \).

Answer: (c) \( \\dfrac E{16} \)

Explanation:
1. Volume ratio 64 gives radius ratio \( R_e:R_m=4:1 \).
2. Energy \( =\\dfrac{GMm}{R}\\propto\\rho R^2 \) for equal density, so it falls by \( 4^2=16 \).

Teacher's Note:
1. For equal densities, the escape energy per unit mass is proportional to the square of the radius.
2. Cube root of 64 is 4.

Answer: (b) 16 km/s

Explanation:
1. \( v^2=v_i^2+2GM\\left(\\dfrac1R-\\dfrac1{10R}\\right)=v_i^2+0.9\\,v_e^2 \).
2. \( v^2=144+0.9\\times125.4=256.9 \), so \( v\\approx16\\text{ km/s} \).

Teacher's Note:
1. Use the energy gained by falling from 10R to R, with 2GM/R = v_e^2.
2. Square the escape speed to get 125.4.