JEE Physics Gravitation MCQs Set 15

Answer: (b) 6.4 × 10⁹ J

Explanation:
1. Escape energy \( =\tfrac12mv_e^2=\frac{GMm}{R}=mgR \).
2. \( E=100\times10\times6.4\times10^6=6.4\times10^9\ \text{J} \).

Teacher's Note:
1. Escape velocity requires the total energy to become zero.
2. Use GM = gR² to write the energy as mgR.

Answer: (a) 5 km/s

Explanation:
1. \( v_e=\sqrt{\frac{2GM}{R}}\propto\sqrt{\frac MR} \).
2. \( \frac{v_M}{v_E}=\sqrt{\frac{1/10}{1/2}}=\sqrt{0.2}\approx0.447 \).
3. \( v_M=0.447\times11.2\approx5\ \text{km/s} \).

Teacher's Note:
1. Escape velocity depends on the square root of M/R.
2. Take the ratio instead of using G.

Answer: (d) 1.5% increased

Explanation:
1. Kepler's third law: \( T^2\propto R^3\Rightarrow T\propto R^{3/2} \).
2. For a small change, \( \frac{\Delta T}{T}=\frac32\frac{\Delta R}{R}=\frac32\times1\%=1.5\% \).
3. The larger orbit has the longer period, so the period increases.

Teacher's Note:
1. Use the percentage change formula for a power law.
2. A bigger orbit means a longer period.

Answer: (c) \( \frac{\sqrt{a^2+b^2}}{m} \)

Explanation:
1. Force is the negative gradient of potential energy: \( F_x=-\frac{\partial U}{\partial x}=-a,\ F_y=-\frac{\partial U}{\partial y}=-b \).
2. Magnitude \( F=\sqrt{a^2+b^2} \), so \( |a|=\frac{\sqrt{a^2+b^2}}{m} \).

Teacher's Note:
1. Take partial derivatives for each coordinate.
2. Combine the two components with Pythagoras' theorem.

Answer: (b) G

Explanation:
1. Field \( =G(0.75)\left(\frac1{1^2}+\frac1{2^2}+\frac1{4^2}+\dots\right) \).
2. The series is a geometric progression with first term 1 and ratio 1/4: sum \( =\frac{1}{1-1/4}=\frac43 \).
3. Field \( =0.75\times\frac43G=G \) (in SI units).

Teacher's Note:
1. All fields point along the same direction, so they add.
2. Use the sum of an infinite geometric series.

Answer: (a) 9.8 m/s²

Explanation:
1. The point is now outside the shrunken earth, at the same distance from its centre as before.
2. Outside a spherical body the field is \( g=\frac{GM}{r^2} \), which depends only on the mass and the distance from the centre.
3. Mass and distance are unchanged, so g stays 9.8 m/s².

Teacher's Note:
1. Outside a sphere, the mass behaves as if concentrated at its centre.
2. The size of the earth no longer matters once the point is outside it.

Answer: (d) \( \frac{d_1v_1}{d_2} \)

Explanation:
1. Gravity is a central force, so angular momentum about the sun is conserved.
2. At the closest and farthest points the velocity is perpendicular to the radius: \( mv_1d_1=mv_2d_2 \).
3. \( v_2=\frac{d_1v_1}{d_2} \).

Teacher's Note:
1. Torque of a central force about the centre is zero.
2. At perihelion and aphelion, r and v are perpendicular.

Answer: (c) 2.3 × 10²⁷ kg

Explanation:
1. Orbit radius \( r=27\times7.4\times10^7=2.0\times10^9\ \text{m} \), period \( T=16.7\times86400=1.44\times10^6\ \text{s} \).
2. From \( T^2=\frac{4\pi^2r^3}{GM} \): \( M=\frac{4\pi^2r^3}{GT^2} \).
3. \( M=\frac{39.5\times(2.0\times10^9)^3}{6.67\times10^{-11}\times(1.44\times10^6)^2}\approx2.3\times10^{27}\ \text{kg} \).

Teacher's Note:
1. Convert days to seconds and km to metres first.
2. Use Kepler's law in the form T² = 4π²r³/GM.

Answer: (b) \( m=\frac M2 \)

Explanation:
1. \( F=\frac{Gm(M-m)}{r^2} \).
2. For a maximum \( \frac{dF}{dm}\propto M-2m=0 \), so \( m=\frac M2 \).
3. The second derivative is negative, so this is a maximum.

Teacher's Note:
1. The product m(M − m) is largest when both parts are equal.
2. Differentiate with respect to m and set it to zero.

Answer: (a) \( \sqrt{2GM_e\left(\frac1{R_e}-\frac1{R_0}\right)} \)

Explanation:
1. Energy conservation: \( -\frac{GM_em}{R_0}=-\frac{GM_em}{R_e}+\tfrac12mv^2 \).
2. \( v^2=2GM_e\left(\frac1{R_e}-\frac1{R_0}\right) \).

Teacher's Note:
1. Gravitational potential energy is negative and increases with distance.
2. The decrease in potential energy becomes kinetic energy.

Answer: (d) \( \sqrt{\frac{4G}{r}(M_e+M_m)} \)

Explanation:
1. Potential energy at the midpoint (each body is at distance r/2): \( U=-\frac{GM_em}{r/2}-\frac{GM_mm}{r/2}=-\frac{2Gm(M_e+M_m)}{r} \).
2. For escape the total energy must be zero: \( \tfrac12mv^2=\frac{2Gm(M_e+M_m)}{r} \).
3. \( v=\sqrt{\frac{4G(M_e+M_m)}{r}} \).

Teacher's Note:
1. Add the potential energies due to both bodies.
2. Escape requires zero total energy at infinity.

Answer: (c) \( \frac{GM_em}{R_e^3}x \)

Explanation:
1. Only the mass inside radius x attracts the particle: \( M'=M_e\frac{x^3}{R_e^3} \) (uniform density).
2. Force \( =\frac{GM'm}{x^2}=\frac{GM_em}{R_e^3}x \).

Teacher's Note:
1. Inside a uniform sphere the force is proportional to the distance from the centre.
2. The shell outside radius x exerts no net force.

Answer: (b) \( \frac{4\sqrt2Gm^2}{a^2} \)

Explanation:
1. Distance from the centre to each corner \( r=\frac{a}{\sqrt2} \), so \( r^2=\frac{a^2}{2} \).
2. Opposite corners (m and 3m): net force \( \frac{Gm(3m-m)}{r^2}=\frac{4Gm^2}{a^2} \) towards 3m. Opposite corners (2m and 4m): net \( \frac{Gm(4m-2m)}{r^2}=\frac{4Gm^2}{a^2} \) towards 4m.
3. The two diagonals are perpendicular: \( F=\sqrt2\times\frac{4Gm^2}{a^2}=\frac{4\sqrt2Gm^2}{a^2} \).

Teacher's Note:
1. Pair opposite corners first, then add the two perpendicular resultants.
2. The diagonals of a square are at right angles.

Answer: (b) \( x=2h \)

Explanation:
1. At height h: \( g'=g\left(1-\frac{2h}{R}\right) \), change \( =\frac{2gh}{R} \).
2. At depth x: \( g''=g\left(1-\frac xR\right) \), change \( =\frac{gx}{R} \).
3. Equate: \( x=2h \).

Teacher's Note:
1. For small h, use the binomial approximation.
2. g falls twice as fast with height as with depth.

Answer: (a) \( a=1 \)

Explanation:
1. Inside a uniform sphere \( M'=\frac43\pi r^3\rho \).
2. \( F=\frac{GM'm}{r^2}=\frac43\pi G\rho m\,r \).
3. So F varies as \( r^1 \), hence a = 1.

Teacher's Note:
1. The enclosed mass grows as r³ while the distance factor shrinks as 1/r².
2. Net effect: force proportional to r.

Answer: (a) \( v_{min}=\frac32\sqrt{\frac{5GM}{a}} \)

Explanation:
1. The body must just cross the point where the net field is zero. At distance x from the centre of the large star: \( \frac{16GM}{x^2}=\frac{GM}{(10a-x)^2}\Rightarrow x=8a \).
2. Potential per unit mass: \( V(r)=-\frac{16GM}{r}-\frac{GM}{10a-r} \). \( V(2a)=-\frac{65GM}{8a} \), \( V(8a)=-\frac{5GM}{2a} \).
3. Beyond the neutral point the smaller star pulls it in, so \( \tfrac12v^2=V(8a)-V(2a)=\frac{45GM}{8a} \).
4. \( v=\frac32\sqrt{\frac{5GM}{a}} \).

Teacher's Note:
1. The speed needed is the minimum that carries the body over the point of maximum potential, where the net force is zero.
2. Both stars contribute to the potential at every point.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Escape velocity \( v_e=\sqrt{\frac{2GM}{R}} \) contains only the mass and radius of the earth.
2. So both balls have the same escape velocity, and R explains A.

Teacher's Note:
1. The mass of the escaping body cancels out in the energy equation.
2. Escape velocity depends on the planet, not the projectile.

Answer: (c) A is true, but R is false.

Explanation:
1. At the centre the gravitational pulls from all sides cancel, so g = 0 and the weight is zero; A is true.
2. The mass of a body is constant, so R is false.

Teacher's Note:
1. g falls linearly with depth to zero at the centre.
2. Mass is a constant property of the body.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Setting \( g\left(\frac{R}{R+h}\right)^2=g\left(1-\frac hR\right) \) gives \( h/R=\frac{\sqrt5-1}{2}\approx0.62 \), so A is true.
2. R is also true, but it only says that g decreases in both directions; it does not explain why the two values are equal at this particular h.

Teacher's Note:
1. Both statements are true, but the second is a general remark.
2. Equality needs the exact solution of the equation.