Class 11 Physics Chapter 7: Gravitation MCQs with answers

Answer: (a) 49 F

F ∝ m₁m₂/r². The product of masses becomes 2 × 2 = 4 times and r² becomes 9 times, so F′ = (4/9)F. 2/3 F treats the distance as entering to the first power; 9/4 F inverts the ratio.

Answer: (b) [M⁻¹ L³ T⁻²]

G = Fr²/(m₁m₂), so [G] = [M L T⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²]. [M L² T⁻²] is the dimension of energy.

Answer: (d) 5 cm

For every point on the ellipse the sum of its distances from F₁ and F₂ equals the string length, 10 cm. At either end of the major axis this sum equals the length of the major axis, 2a, so 2a = 10 cm and a = 5 cm. 4 cm is the semi-minor axis, √(5² − 3²); 3 cm is half the distance between the foci.

Answer: (b) on the surface of the Earth

Inside a uniform Earth g grows in proportion to r and is zero at the centre; outside it falls as 1/r². So g is greatest at r = R_E, on the surface, and decreases whether one goes up or down.

Answer: (c) gas molecules formed on its surface can have speeds larger than the Moon's escape speed of about 2.3 km s⁻¹

The escape speed from the Moon is only about 2.3 km s⁻¹, about five times smaller than the Earth's, because both g and the radius are smaller there. Gas molecules moving faster than this escape the Moon's pull.

Answer: (d) J kg⁻¹

Gravitational potential is the potential energy of unit mass, so its unit is J kg⁻¹. J is the unit of potential energy, N kg⁻¹ of gravitational intensity, and N m² kg⁻² of G.

Answer: (a) the astronaut and the satellite are both in free fall towards the Earth

Gravity at the orbit is not small, and gravitational shielding is not possible. Both the astronaut and the satellite fall freely towards the Earth with the same acceleration, so the floor presses nothing on the astronaut.

Answer: (c) 4

At equilibrium G(Mm/d²)L = τθ, so θ ∝ 1/d². Halving d makes θ four times larger. 2 treats the force as varying with 1/d.

Answer: (d) −5.0 × 10⁹ J

In a circular orbit P.E. : K.E. : E = −2 : 1 : −1, so E = −K.E. = −5.0 × 10⁹ J. −1.0 × 10¹⁰ J is the potential energy, not the total.

Answer: (c) the value of the potential energy of a body at a given height

Choosing the zero is choosing the arbitrary constant in V = −Gm₁m₂/r + constant. Only the value at a point changes; differences of potential energy, the work done between two points, and the force are unaltered.