JEE Physics Gravitation MCQs Set 24

Answer: (c) \( 1.67\times10^{-9}(\hat i+\hat j) \) N

Explanation:
1. Each neighbour is 0.2 m away. Force magnitude \( = Gm^2/r^2 = 6.67\times10^{-11}/0.04 = 1.67\times10^{-9} \) N.
2. One force points along +y and the other along +x, so the resultant is \( 1.67\times10^{-9}(\hat i+\hat j) \) N.

Teacher's Note:
1. Resolve the two attractions along the axes.
2. Both point away from the origin towards the other masses.

Answer: (c) \( 4\sqrt2\,Gm^2/a^2 \)

Explanation:
1. Each corner is at distance \( a/\sqrt2 \), so \( r^2 = a^2/2 \). Force from mass M is \( 2GmM/a^2 \).
2. Opposite masses (m, 3m) give a net \( 2Gm(2m)/a^2 = 4Gm^2/a^2 \).
3. Opposite masses (2m, 4m) also give \( 4Gm^2/a^2 \), at right angles.
4. Resultant \( = 4\sqrt2\,Gm^2/a^2 \).

Teacher's Note:
1. Combine opposite pairs first.
2. The two net forces are perpendicular.

Answer: (b) against the gravitation of earth in onward journey

Explanation:
1. Escaping the earth's strong field needs the largest energy.
2. The moon's gravity is only about one sixth of the earth's, so the other stages need less fuel.

Teacher's Note:
1. The earth is much more massive than the moon.
2. The biggest fuel cost is leaving the earth.

Answer: (a) 1 : 1

Explanation:
1. By Newton's third law the two gravitational forces are equal in magnitude.
2. Ratio = 1 : 1.

Teacher's Note:
1. Action and reaction are always equal.
2. The masses do not change the ratio.

Answer: (d) 9D/10

Explanation:
1. Let the point be at x from the earth: \( GM_e/x^2 = GM_m/(D-x)^2 \).
2. \( x/(D-x) = \sqrt{81} = 9 \).
3. \( x = 9D/10 \).

Teacher's Note:
1. The point lies closer to the lighter body.
2. Take the square root of the mass ratio.

Answer: (d) 0.4

Explanation:
1. \( g\propto M/R^2 \).
2. \( g_M/g_E = 0.1/(0.5)^2 = 0.4 \).

Teacher's Note:
1. Radius enters squared.
2. Diameter ratio equals radius ratio.

Answer: (a) \( \sqrt6 \) seconds

Explanation:
1. \( h = \tfrac12gt^2 \), so \( t\propto1/\sqrt g \).
2. \( t_{moon} = 1\times\sqrt6 = \sqrt6 \) s.

Teacher's Note:
1. Time of fall varies as the inverse square root of g.

Answer: (a) 6400 km

Explanation:
1. g' = g/4, so \( (R/(R+h))^2 = 1/4 \).
2. \( R+h = 2R \), giving \( h = R = 6400 \) km.

Teacher's Note:
1. A decrease by 75% leaves one quarter.
2. One quarter means twice the distance from the centre.

Answer: (b) 0.99 g

Explanation:
1. For small h, \( g' = g(1 - 2h/R) = g(1 - 64/6400) \).
2. \( = 0.99g \).

Teacher's Note:
1. Use the small-height approximation.

Answer: (b) g increases linearly from zero at the centre up to r = R, then decreases as 1/r² for r > R

Explanation:
1. Inside a uniform sphere \( g = GMr/R^3 \), which is proportional to r.
2. Outside \( g = GM/r^2 \).
3. So g rises linearly to a maximum at the surface and then falls as \( 1/r^2 \).

Teacher's Note:
1. The field inside depends only on the mass within radius r.
2. The maximum value is at the surface.

Answer: (b) \( \dfrac{g}{G}R_E^2 \)

Explanation:
1. At the surface \( g = GM/R_E^2 \).
2. \( M = gR_E^2/G \).

Teacher's Note:
1. Rearrange the surface formula for g.
2. Only the earth's radius and g need measuring.

Answer: (d) \( h = (\sqrt2-1)R \)

Explanation:
1. \( (R/(R+h))^2 = 1/2 \).
2. \( R+h = \sqrt2R \), so \( h = (\sqrt2-1)R \).

Teacher's Note:
1. Take the square root before solving.
2. h is measured from the surface.

Answer: (c) increase

Explanation:
1. Rotation reduces the effective g by \( \omega^2R\cos^2\lambda \).
2. Without rotation this term vanishes, so g increases (except at the poles where it was unchanged).

Teacher's Note:
1. The centrifugal effect opposes gravity.

Answer: (d) will fall down with acceleration g/6

Explanation:
1. The moon has no atmosphere, so there is no buoyant force.
2. The balloon falls freely with the moon's g = g/6.

Teacher's Note:
1. Balloons rise only because of buoyancy in a fluid.

Answer: (a) increase by 2%

Explanation:
1. \( g\propto1/R^2 \), so \( \Delta g/g = 2\times1\% = 2\% \).
2. A smaller radius gives a larger g: increase by 2%.

Teacher's Note:
1. Fractional change doubles for an inverse square law.

Answer: (a) decrease by 0.5%

Explanation:
1. Above the surface the fall is \( 2h/R = 1\% \), so \( h/R = 0.5\% \).
2. At depth h the fall is \( h/R = 0.5\% \).
3. Weight decreases by 0.5%.

Teacher's Note:
1. Depth gives half the fractional decrease of the same height.

Answer: (d) S₂ and S₄

Explanation:
1. A physical (beam) balance compares masses, so g cancels and the reading is the same at the pole and equator: S₂ is true.
2. A spring balance measures weight mg, and g is larger at the pole: S₄ is true.

Teacher's Note:
1. Beam balance compares masses.
2. Spring balance reads weight.

Answer: (c) \( 4GM_p/D_p^2 \)

Explanation:
1. Radius \( R = D_p/2 \).
2. \( g = GM_p/R^2 = 4GM_p/D_p^2 \).

Teacher's Note:
1. Acceleration does not depend on the falling mass m.

Answer: (b) 2 : 1

Explanation:
1. Inside: \( g_{R/2} = g/2 \).
2. Outside: \( g_{2R} = g/4 \).
3. Ratio = 2 : 1.

Teacher's Note:
1. Use the linear law inside and the inverse square law outside.

Answer: (d) g on the Earth will not change

Explanation:
1. \( g = GM/R^2 \) depends on the product GM.
2. The product stays the same, so g is unchanged and the other three effects do not occur.

Teacher's Note:
1. Only the product GM matters for g.

Answer: (b) \( -4.8\times10^7 \)

Explanation:
1. Outside the earth \( g = GM/r^2 \) and \( V = -GM/r = -gr \).
2. \( V = -6.0\times8\times10^6 = -4.8\times10^7 \) J/kg.

Teacher's Note:
1. Potential is negative for an attractive field.
2. Use V = −g r outside a sphere.

Answer: (b) \( -3GM/a \)

Explanation:
1. Shell contributes \( -GM/a \) everywhere inside.
2. The particle contributes \( -GM/(a/2) = -2GM/a \).
3. Total = \( -3GM/a \).

Teacher's Note:
1. Potentials add as scalars.
2. Inside a shell the potential is constant.

Answer: (a) \( V = -GM/R \) (constant) for r < R and \( V = -GM/r \) for r > R

Explanation:
1. Inside the shell the field is zero, so the potential stays equal to its surface value \( -GM/R \).
2. Outside it equals that of a point mass, \( -GM/r \).

Teacher's Note:
1. Zero field means constant potential.
2. The potential is continuous at the shell.

Answer: (b) 2600 km

Explanation:
1. \( V/g = -r \), so \( r = 5.4\times10^7/6.0 = 9\times10^6 \) m = 9000 km.
2. \( h = 9000 - 6400 = 2600 \) km.

Teacher's Note:
1. V divided by g gives the distance from the centre.

Answer: (c) \( -GM_mm\left(\dfrac{81}{R}+\dfrac1r\right) \)

Explanation:
1. \( U = -GM_em/R - GM_mm/r \).
2. Substitute \( M_e = 81M_m \): \( U = -GM_mm(81/R + 1/r) \).

Teacher's Note:
1. Potential energies of the two attractions add.