Acceleration Due to Gravity - Free MCQ Practice Test with solutions, NEET

The correct answer is Option A - A, B, C and E only

Assume Earth as a sphere of radius R with acceleration due to gravity at the surface as g.

For a point at height h above the surface, the acceleration is g_h = g · (R/(R + h))².

Putting h = R gives g_h = g · (R/(2R))² = g/4, so statement E is correct.

For a point at depth d below the surface (uniform-density model), the acceleration is g_d = g · (1 - d/R).

Putting d = R (centre) gives g_center = 0, so statement C is correct.

From g_h = g · (R/(R + h))² we see g_h decreases as h increases, so statement A is correct.

From g_d = g · (1 - d/R) we see g_d decreases as d increases, so statement B is correct; statement D is therefore incorrect in the uniform-density model.

The correct answer is Option A - R

Using the inverse-square law for gravitational acceleration, the value at height h is g_h = g \left(\dfrac{R}{R + h}\right)², where g is the acceleration at the surface and R is the Earth's radius.

Given g_h = g/4, so \left(\dfrac{R}{R + h}\right)² = \dfrac{1}{4}.

Taking the positive square root (distances are positive) gives \dfrac{R}{R + h} = \dfrac{1}{2}.

Therefore R + h = 2R, which yields h = R.

Thus the required value is h = R, so Option A is correct. (This result assumes the Earth is spherically symmetric and uses the inverse-square law.)

Correct option: C

A → III. The acceleration due to gravity at the surface is given by g = GM/R², so III matches A.

B → I. The exact acceleration at height h is g_h = GM/(R + h)² = g/(1 + h/R)². For h << R, put x = h/R and use the binomial approximation (1 + x)^-2 ≈ 1 - 2x to get g_h ≈ g(1 - 2h/R), so I matches B.

C → II. Assuming uniform density, the mass enclosed within radius R - d is (R - d)³/R³ of the total mass. Thus the acceleration at depth d is g_d = G M_enclosed /(R - d)² = g (R - d)/R = g(1 - d/R), so II matches C.

D → IV. At the centre of a spherically symmetric mass distribution the net gravitational acceleration is 0 by symmetry, so IV matches D.

Therefore the matching (A)-(III), (B)-(I), (C)-(II), (D)-(IV) is correct, which corresponds to option C.

The correct answer is Option B - 28.4 N

Weight at a point is proportional to the local acceleration due to gravity g, so we use the relation for g at height.

g' = g (R/(R + h))²

For h = R/2, the distance from the centre is R + h = 3R/2.

g' = g (R/(3R/2))² = g (2/3)² = g (4/9)

Therefore weight scales by the same factor: W' = W × 4/9

With W = 64 N, W' = 64 × 4/9 = 256/9 ≈ 28.4 N.

Thus the weight equals 28.4 N, which is Option B.

The correct answer is Option C - 3R/8

Assuming the Earth has uniform density, the acceleration at a depth d (measured below the surface) is g_d = g(1 - d/R).

The acceleration at a height h above the surface is g_h = g(R/(R + h))².

Setting these equal (since the ratio is 1:1) gives 1 - d/R = (R/(R + h))².

For d = R/4, we obtain 3/4 = (R/(R + h))².

Thus R/(R + h) = √3/2, so R + h = 2R/√3 and therefore h = R(2/√3 - 1).

Numerically h ≈ 0.1547 R, which does not equal 3R/8 or any of the provided choices; the correct value is h = R(2/√3 - 1).

The correct answer is Option A - Both Statement I and Statement II are true

For points above Earth's surface, the acceleration due to gravity is given exactly by g(h) = GM/(R + h)², where G is the gravitational constant, M is Earth's mass, R is Earth's radius and h is the height above the surface. This can be written relative to surface gravity g₀ as g(h) = g₀·(R/(R + h))², so g decreases with increasing h following the inverse-square dependence.

For heights small compared to Earth's radius (h ≪ R), the first-order approximation gives g(h) ≈ g₀·(1 - 2h/R), which shows the fractional decrease is approximately linear for small h but the exact relation remains inverse-square.

For points below the surface, using the spherical-shell result that only the mass within radius r contributes, and assuming uniform density, the enclosed mass is proportional to r³. This gives g(r) = g₀·(r/R), where r = R - d and d is depth. Substituting r yields g(d) = g₀·(1 - d/R), so g decreases linearly with depth d under the stated assumption.

Therefore, both statements are correct with the usual physical assumptions: Statement I from the inverse-square law for points outside the Earth, and Statement II from the linear behaviour of g with depth for a spherically symmetric body with uniform (or smoothly varying) density. Both statements are true.

The correct answer is Option A - 32 km

Use the relation g_h = g (R/(R + h))² for acceleration due to gravity at a height h above the surface.

(R/(R + h))² = 0.99

R/(R + h) = √0.99

1 + h/R = 1/√0.99

h = R (1/√0.99 - 1)

Substituting R = 6400 km and evaluating, √0.99 ≈ 0.994987, so 1/√0.99 ≈ 1.0050378 and 1/√0.99 - 1 ≈ 0.0050378.

Hence h ≈ 6400 km × 0.0050378 ≈ 32.24 km.

Rounded to the given options, this gives 32 km, which matches the correct option.

The correct answer is Option A - d = R/2

For a sphere of uniform density, the mass enclosed within radius r is M_enclosed = M (r^3 / R^3), where M and R are the mass and radius of the Earth.

The gravitational acceleration at distance r from the centre is g(r) = G·M_enclosed / r^2 = g·(r / R), since g = G·M / R^2 is the acceleration at the surface.

At a depth x below the surface the distance from the centre is r = R - x, so the acceleration there is g_x = g·(R - x) / R.

Setting the weight at depth to be half the surface weight gives m g_x = (1/2) m g, which reduces to (R - x) / R = 1/2.

Thus R - x = R/2 and hence x = R/2. Therefore the required depth is d = R/2.