Q.The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the gravitational force acting on a body due to earth at a height equal to twice the radius of the earth is 30 N, then the mass of the body is (Acceleration due to gravity on the surface of the earth $=10\ ms^{-2}$) (A) 27 kg (B) 54 kg (C) 30 kg (D) 60 kg

   ›Reveal solutionSolution

   This tests the inverse-square falloff of gravity with distance from Earth's center — find $g$ at height $2R$ (i.e. at $3R$ from the center), then use $F=mg$ to get the mass. Answer: (A) 27 kg.

   Concept and Intuition

   Newton's law of gravitation gives $g(r)=\dfrac{GM}{r^2}$ for a point (or distance $r\ge R$ from a spherical Earth's center). At the surface, $g_0=\dfrac{GM}{R^2}$. At any other distance $r$ from the center, $g(r)=g_0\left(\dfrac{R}{r}\right)^2$ — gravity weakens with the square of the distance ratio, not linearly. Crucially, "height $h$ above the surface" means the distance from the center is $R+h$, not $h$ itself — a very common point of confusion.

   Step-by-Step Solution

   1. Distance from center at height $h=2R$: $r=R+h=R+2R=3R$.
   2. Gravitational acceleration there:

   $g=g_0\left(\frac{R}{3R}\right)^2=\frac{g_0}{9}=\frac{10}{9}\ m/s^2.$

   3. Given force: $F=mg=30\ N$, so

   $m=\frac{F}{g}=\frac{30}{10/9}=30\times\frac{9}{10}=27\ kg.$

   Common Mistakes

   • Using $r=2R$ instead of $r=3R$ (forgetting to add the Earth's own radius $R$ to the height $2R$ above the surface).
   • Using a linear falloff ($g=g_0/2$ for $h=2R$, treating $g\propto 1/r$) instead of the correct inverse-square law.

   ✓Final answer

   The correct option is (A) — 27 kg.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.For a planet having uniform density, the Gravitational field inside the planet varies with the distance from the centre as (A) $1/r$ (B) $1/r^2$ (C) $r$ (D) Constant

   ›Reveal solutionSolution

   Inside a uniform sphere the field grows linearly with distance from the centre because only the mass within radius $r$ pulls on a test point there, and that mass itself scales as $r^3$.

   Concept and Intuition

   By the shell theorem, a uniform spherical shell exerts zero gravitational field at any point strictly inside it. So for a point at radius $r$ inside a uniform-density planet, only the sphere of radius $r$ (and mass) contributes to the field there; every shell outside radius $r$ cancels out. This is exactly analogous to the electric field inside a uniformly charged sphere.

   Step-by-Step Solution

   1. Let the planet have uniform density $\rho$ and radius $R$. Consider a point at distance $r < R$ from the centre.
   2. Mass enclosed within radius $r$: $M_{enc} = \rho \times \dfrac{4}{3}\pi r^3$.
   3. Only this enclosed mass contributes to the gravitational field at radius $r$ (shells outside contribute nothing, by the shell theorem):

   $g(r) = \frac{G M_{enc}}{r^2} = \frac{G \rho \frac{4}{3}\pi r^3}{r^2} = \frac{4}{3}\pi G \rho\, r.$

   4. Since $\rho$, $G$ are constants, $g(r) \propto r$ — the field increases linearly from zero at the centre to a maximum at the surface ($r=R$), then falls off as $1/r^2$ outside.

   Common Mistakes

   • Applying $g \propto 1/r^2$ (the outside-the-planet law) to points inside the planet — that law only holds for $r \ge R$, treating the whole mass as concentrated at the centre.
   • Forgetting that mass enclosed itself depends on $r$ (as $r^3$), which is what converts the $1/r^2$ into a net $r^1$ dependence inside.

   ✓Final answer

   The correct option is (C) — $g \propto r$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.When 'g' is acceleration due to gravity on earth, the gain in potential energy of an object of mass $m$ raised from the surface of earth to a height equal to the radius of earth 'R' is given by (A) $\frac{mgR}{2}$ (B) $\frac{mgR}{4}$ (C) $mgR$ (D) $2mgR$

   ›Reveal solutionSolution

   For large heights (comparable to Earth's radius), the simple $mgh$ formula fails; the correct gain in PE when raised to height $R$ (one Earth radius) works out to $\frac{mgR}{2}$.

   Concept and Intuition

   The formula $\Delta U = mgh$ is only valid for heights small compared to Earth's radius, where gravity is essentially constant. When raised to a height as large as Earth's radius itself, gravity weakens significantly with altitude, so we must use the more general gravitational potential energy expression $U(r) = -\dfrac{GMm}{r}$ and account for this variation.

   Step-by-Step Solution

   1. Gravitational PE at Earth's surface (distance $R$ from center): $U_1 = -\dfrac{GMm}{R}$.
   2. Gravitational PE at height $h=R$ above the surface (distance $2R$ from center): $U_2 = -\dfrac{GMm}{2R}$.
   3. Gain in PE: $\Delta U = U_2 - U_1 = -\dfrac{GMm}{2R} + \dfrac{GMm}{R} = \dfrac{GMm}{2R}$.
   4. Using $g = \dfrac{GM}{R^2} \Rightarrow GM = gR^2$: $\Delta U = \dfrac{gR^2 m}{2R} = \dfrac{mgR}{2}$.
   5. This matches the general formula $\Delta U = \dfrac{mgh}{1+h/R}$ at $h=R$: $\dfrac{mgR}{1+1}=\dfrac{mgR}{2}$.

   Common Mistakes

   • Blindly using $\Delta U = mgh = mgR$ (option C), ignoring that gravity decreases substantially over a height equal to Earth's radius.
   • Sign errors when subtracting the two (negative) potential energy values.

   ✓Final answer

   The correct option is (A) — $\frac{mgR}{2}$.

   ANSWER: A

4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The weight of a body decreases by 1% when it is raised to a small height 'h' above the earth's surface. If the same body is taken to a depth 'h' in a mine, then its weight is (A) Decreases by 0.5% (B) Decreases by 2.0% (C) Increases by 0.5% (D) Increases by 1.0%

   ›Reveal solutionSolution

   Gravity falls off twice as fast with height as it does with depth (for small $h$), so a 1% decrease at height $h$ corresponds to only a 0.5% decrease at the same depth $h$ — option (A).

   Concept and Intuition

   Above the Earth's surface, gravity decreases according to the inverse-square law, which for small heights $h\ll R$ approximates to a linear decrease with slope $2/R$: $g_h\approx g(1-2h/R)$. Below the surface, only the mass enclosed within radius $(R-h)$ contributes (uniform density assumption), giving a linear decrease with slope $1/R$: $g_d=g(1-h/R)$. The height formula's coefficient is exactly double the depth formula's coefficient — this factor-of-two relationship is the crux of the problem.

   Step-by-Step Solution

   1. Approximate gravity at small height $h$: $g_h=\dfrac{g}{(1+h/R)^2}\approx g\left(1-\dfrac{2h}{R}\right)$ for $h\ll R$.
   2. Given: weight (hence $g$) decreases by 1% at height $h$, so $\dfrac{2h}{R}=0.01 \Rightarrow \dfrac{h}{R}=0.005$.
   3. Gravity at depth $h$ inside the Earth: $g_d=g\left(1-\dfrac{h}{R}\right)$.
   4. Substitute $\dfrac{h}{R}=0.005$: $g_d=g(1-0.005)=0.995g$.
   5. This is a decrease of $0.5\%$ from the surface value $g$.

   Common Mistakes

   • Using the same formula (or the same numeric percentage) for both height and depth — the two expressions have different coefficients (factor of 2 vs factor of 1) and must not be conflated.
   • Forgetting the height-approximation formula requires $h\ll R$, which is implicitly assumed since the height/depth here produce only small (~1%) changes.

   ✓Final answer

   The correct option is (A) — Decreases by 0.5%.

   ANSWER: A

5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Considering earth as a sphere of uniform density, how much would a body weighs halfway down the centre of earth if it is weighed 80 N on the surface (A) 70 N (B) 40 N (C) 60 N (D) 50 N

   ›Reveal solutionSolution

   Gravity inside a uniform sphere scales linearly with distance from the centre, so halfway down (r = R/2) the weight is exactly half the surface weight: 40 N.

   Concept and Intuition

   For a uniformly dense sphere like an idealised Earth, only the mass enclosed within radius $r$ contributes to gravity at that radius (the shell theorem — mass outside contributes zero net force). Since enclosed mass scales as $r^3$ and gravitational acceleration involves $M_{enclosed}/r^2$, the two effects combine to make $g(r)\propto r$ inside the sphere — a simple linear relationship.

   Step-by-Step Solution

   1. At the surface ($r=R$), $g(R) = g_{surface}$, and the weight is $80\ \text{N}$.
   2. Inside a uniform sphere, $g(r) = g_{surface}\cdot\dfrac{r}{R}$ for $r\le R$.
   3. "Halfway down to the centre" means $r = R/2$:

   $g(R/2) = g_{surface}\times\frac12$

   4. So the weight there is also halved: $W = 80\times\dfrac12 = 40\ \text{N}$.

   Common Mistakes

   • Using the inverse-square law (as if outside the Earth) instead of the linear relationship that holds inside a uniform sphere.
   • Confusing "halfway down to the centre" (r = R/2) with "half the radius remaining" or other distance interpretations.

   ✓Final answer

   The correct option is (B) — 40 N.

   ANSWER: B

6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If a body is thrown vertically upwards from a height of 0.5 R (R – radius of the earth) with a velocity equal to the escape velocity a body from the surface of the earth, then the velocity of the body when it escapes from the gravitational influence of the earth is (g – acceleration due to gravity on the surface of the earth) (A) $\sqrt{2gR}$ (B) $\sqrt{gR}$ (C) $\sqrt{\dfrac{2gR}{3}}$ (D) $\sqrt{\dfrac{2gR}{5}}$

   ›Reveal solutionSolution

   This tests energy conservation for a body launched with surface escape speed from an elevated point, finding its residual speed at infinity.

   Concept and Intuition

   Escape velocity from the surface is defined by $\tfrac12 v_e^2 = \dfrac{GM}{R}$ (just enough KE to reach infinity with zero leftover speed, when launched from the surface). Here, the body is launched with this same speed $v_e = \sqrt{2gR}$, but from a higher point (distance $1.5R$ from the centre, i.e., height $0.5R$ above the surface), where the gravitational potential energy is less negative (weaker binding). So this launch speed is now more than enough to escape from that starting point, and the body retains some residual speed as it reaches infinity. We find that residual speed using conservation of total mechanical energy between the launch point and infinity.

   Step-by-Step Solution

   1. Launch point: distance from Earth's centre $= R + 0.5R = 1.5R$; launch speed $v_e = \sqrt{2gR}$ (equal to the surface escape velocity, using $GM = gR^2$).
   2. Total mechanical energy at launch: $E_i = \tfrac12 v_e^2 - \dfrac{GM}{1.5R} = \tfrac12(2gR) - \dfrac{gR^2}{1.5R} = gR - \dfrac{2}{3}gR$.
   3. At infinity, PE $\to 0$, so $E_f = \tfrac12 v_f^2$.
   4. Energy conservation: $\tfrac12 v_f^2 = gR - \dfrac23 gR = \dfrac13 gR$.
   5. $v_f^2 = \dfrac{2}{3}gR \Rightarrow v_f = \sqrt{\dfrac{2gR}{3}}$.

   Common Mistakes

   • Using $R$ instead of $1.5R$ for the potential energy at the launch point — the body starts above the surface, so the PE term uses the actual distance from the centre.
   • Assuming the residual velocity is zero just because the launch speed equals "the" escape velocity — that's only true when launched from the surface itself; launching the same speed from higher up leaves a surplus.

   ✓Final answer

   The correct option is (C) — $\sqrt{\dfrac{2gR}{3}}$.

   ANSWER: C

7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.The height at which the weight of a person becomes 25% of his weight on the surface of the earth is ($R_E$ is radius of the earth) (A) $\dfrac{R_E}{2}$ (B) $\dfrac{R_E}{4}$ (C) $R_E$ (D) $2R_E$

   ›Reveal solutionSolution

   Weight becomes 25% of surface weight when $g$ drops to a quarter of its surface value, which by the inverse-square law of gravity happens at a height equal to Earth's radius, $h=R_E$.

   Concept and Intuition

   The acceleration due to gravity at a height $h$ above the Earth's surface is $g_h = g\left(\dfrac{R_E}{R_E+h}\right)^2$ (inverse-square law, treating Earth's mass as concentrated at its centre). Since weight is proportional to $g$ at a given location (for fixed mass), the weight ratio equals the $g$ ratio.

   Step-by-Step Solution

   1. Set up the ratio: $\dfrac{W_h}{W_0} = \dfrac{g_h}{g} = \left(\dfrac{R_E}{R_E+h}\right)^2 = 0.25$.
   2. Take the square root: $\dfrac{R_E}{R_E+h} = 0.5$.
   3. Solve: $R_E = 0.5(R_E+h) \Rightarrow 2R_E = R_E+h \Rightarrow h = R_E$.

   Common Mistakes

   • Forgetting to take the square root (working directly with the squared ratio) or using the wrong (linear, not inverse-square) variation of $g$ with height, which is only valid for $h \ll R_E$.

   ✓Final answer

   The correct option is (C) — $R_E$.

   ANSWER: C

8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The time period of a simple pendulum on the surface of the earth is T. If the pendulum is taken to a height equal to half of the radius of the earth, then its time period is (A) $\dfrac{T}{2}$ (B) $\dfrac{3T}{2}$ (C) $2T$ (D) $3T$

   ›Reveal solutionSolution

   This tests how a simple pendulum's period changes with altitude through the variation of $g$ with height; the answer is (B) $\dfrac{3T}{2}$.

   Concept and Intuition

   The period of a simple pendulum is $T = 2\pi\sqrt{l/g}$. Since $l$ is unchanged, any change in period comes purely from the change in the local acceleration due to gravity, which decreases with height above the Earth's surface as $g_h = g\dfrac{R^2}{(R+h)^2}$.

   Step-by-Step Solution

   1. At height $h = R/2$: $g_h = g\dfrac{R^2}{(R+R/2)^2} = g\dfrac{R^2}{(3R/2)^2} = g\dfrac{R^2}{\tfrac{9}{4}R^2} = \dfrac{4g}{9}$.
   2. Since $T \propto \dfrac{1}{\sqrt{g}}$, the new period is $T' = T\sqrt{\dfrac{g}{g_h}} = T\sqrt{\dfrac{g}{4g/9}} = T\sqrt{\dfrac{9}{4}} = \dfrac{3T}{2}$.

   Common Mistakes

   • Using $g_h = g(1 - 2h/R)$ (the small-height approximation) instead of the exact inverse-square law, which is inaccurate for $h$ comparable to $R$.
   • Forgetting the square root when relating $T$ to $g$.

   ✓Final answer

   The correct option is (B) — $\dfrac{3T}{2}$.

   ANSWER: B

9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The acceleration due to gravity at a height of $(\sqrt{2}-1)R$ from the surface of the earth is (Acceleration due to gravity on the surface of the earth = $10\ \mathrm{m\,s^{-2}}$ and R is radius of the earth) (A) $2.5\ \mathrm{m\,s^{-2}}$ (B) $7.5\ \mathrm{m\,s^{-2}}$ (C) $5\ \mathrm{m\,s^{-2}}$ (D) $10\ \mathrm{m\,s^{-2}}$

   ›Reveal solutionSolution

   The chosen height $(\sqrt2-1)R$ is deliberately engineered so $R+h=\sqrt2\,R$, making $(R+h)^2=2R^2$ and $g_h=g/2=5\ \mathrm{m\,s^{-2}}$ — option (C).

   Concept and Intuition

   Gravity outside a uniform sphere behaves as if all its mass were concentrated at the centre, so $g$ falls off as the inverse square of the distance from the centre, $g(r)=\dfrac{GM}{r^2}$. At the surface, $r=R$ gives $g$; at height $h$ above the surface, $r=R+h$, so $g_h=g\left(\dfrac{R}{R+h}\right)^2$. The specific height in this problem is chosen precisely so that $R+h$ becomes a clean multiple of $R$ (namely $\sqrt2\,R$), which is the standard trick behind such questions.

   Step-by-Step Solution

   1. Formula: $g_h=g\left(\dfrac{R}{R+h}\right)^2$.
   2. Given $h=(\sqrt2-1)R$, so $R+h=R+(\sqrt2-1)R=\sqrt2\,R$.
   3. Then $\left(\dfrac{R}{R+h}\right)^2=\left(\dfrac{R}{\sqrt2R}\right)^2=\left(\dfrac{1}{\sqrt2}\right)^2=\dfrac12$.
   4. So $g_h=g\times\dfrac12=\dfrac{10}{2}=5\ \mathrm{m\,s^{-2}}$.

   Common Mistakes

   • Using the depth formula ($g_d=g(1-d/R)$, valid only below the surface) instead of the height formula (valid above the surface) — this is a height, so the inverse-square form must be used.
   • Arithmetic slip in simplifying $(\sqrt2-1)R+R=\sqrt2R$ — some students mistakenly leave an extra $R$ term instead of combining correctly.

   ✓Final answer

   The correct option is (C) — $5\ \mathrm{m\,s^{-2}}$.

   ANSWER: C

10. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.The depth 'd' at which the value of acceleration due to gravity becomes $\dfrac{1}{n}$ times the value at the surface (Radius of earth – R) (A) $\dfrac{R}{n}$ (B) $R\left(\dfrac{n-1}{n}\right)$ (C) $R\left(\dfrac{n}{n+1}\right)$ (D) $\dfrac{R}{n^2}$

    ›Reveal solutionSolution

    Rearranging the standard depth-variation-of-g formula gives $d=R\left(\dfrac{n-1}{n}\right)$.

    Concept and Intuition

    Inside the Earth (assuming uniform density), gravitational acceleration decreases linearly with depth because only the mass within radius $(R-d)$ contributes: $g_d = g\left(1-\dfrac{d}{R}\right)$.

    Step-by-Step Solution

    1. Given $g_d = \dfrac{g}{n}$.
    2. $g\left(1-\dfrac{d}{R}\right) = \dfrac{g}{n} \Rightarrow 1-\dfrac{d}{R}=\dfrac1n$.
    3. $\dfrac{d}{R} = 1-\dfrac1n = \dfrac{n-1}{n}$.
    4. $d = R\left(\dfrac{n-1}{n}\right)$.

    Common Mistakes

    • Confusing the depth formula (linear in $d$) with the altitude/height formula (which involves $R/(R+h)$, an inverse-square-like fall-off).

    ✓Final answer

    The correct option is (B) — $R\left(\dfrac{n-1}{n}\right)$.

    ANSWER: B

11. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.The acceleration due to gravity at a height of 6400 km from the surface of the earth is 2.5 ms$^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km) (A) $1.11$ ms$^{-2}$ (B) $1.5$ ms$^{-2}$ (C) $2.22$ ms$^{-2}$ (D) $1.25$ ms$^{-2}$

    ›Reveal solutionSolution

    Using the inverse-square variation of gravity with distance from Earth's center, the first data point pins down surface gravity $g_0=10\ \text{ms}^{-2}$; then $g$ at twice the radius above the surface follows directly. Answer: $1.11\ \text{ms}^{-2}$.

    Concept and Intuition

    Above the Earth's surface, gravitational acceleration falls off as $g(h) = g_0\left(\dfrac{R}{R+h}\right)^2$, since gravity depends on the distance from the center, $R+h$, not the height above the surface directly.

    Step-by-Step Solution

    1. At $h_1 = 6400$ km $=R$: distance from center $=R+h_1=2R$. So $g(h_1) = g_0\left(\dfrac{R}{2R}\right)^2=\dfrac{g_0}{4}$.
    2. Given $g(h_1)=2.5\ \text{ms}^{-2}$: $\dfrac{g_0}{4}=2.5 \Rightarrow g_0=10\ \text{ms}^{-2}$.
    3. At $h_2=12800$ km $=2R$: distance from center $=R+h_2=3R$. So $g(h_2)=g_0\left(\dfrac{R}{3R}\right)^2=\dfrac{g_0}{9}=\dfrac{10}{9}\approx1.11\ \text{ms}^{-2}$.

    Common Mistakes

    • Using $h$ (height above surface) in place of $R+h$ (distance from Earth's center) in the inverse-square law.
    • Assuming $g$ halves for doubling the height, rather than applying the correct inverse-square scaling.

    ✓Final answer

    The correct option is (A) — $1.11\ \text{ms}^{-2}$.

    ANSWER: A

12. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.What is the height from the surface of earth, where acceleration due to gravity will be $\dfrac{1}{4}$ of that of the earth? ($R_E = 6400\ \mathrm{km}$) (A) $6400\ \mathrm{km}$ (B) $3200\ \mathrm{km}$ (C) $1600\ \mathrm{km}$ (D) $640\ \mathrm{km}$

    ›Reveal solutionSolution

    Gravity falls with the inverse square of distance from Earth's centre; requiring it to drop to one-quarter its surface value gives $h = R_E = 6400\ \text{km}$.

    Concept and Intuition

    Acceleration due to gravity outside a spherical Earth behaves as if all its mass were at the centre: $g_h = \dfrac{GM}{(R+h)^2}$. Since $g = \dfrac{GM}{R^2}$ at the surface, $g_h/g = \left(\dfrac{R}{R+h}\right)^2$ — a purely geometric inverse-square dependence on distance from the centre.

    Step-by-Step Solution

    1. Write $g_h = \dfrac{gR^2}{(R+h)^2}$.
    2. We need $g_h = \dfrac{g}{4}$, so $\dfrac{R^2}{(R+h)^2} = \dfrac{1}{4}$.
    3. Taking square roots: $\dfrac{R}{R+h} = \dfrac{1}{2} \implies R+h = 2R \implies h = R$.
    4. With $R_E = 6400\ \text{km}$, the height is $h = 6400\ \text{km}$.

    Common Mistakes

    • Using the flat-earth approximation $g_h \approx g(1-2h/R)$, which is only valid for $h \ll R$ — here $h$ is comparable to $R$, so the exact inverse-square formula must be used.

    ✓Final answer

    The correct option is (A) — $6400\ \mathrm{km}$.

    ANSWER: A