Acceleration Due to Gravity and its Variation Questions in English

(B) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g' = g\left(1 - \frac{d}{R}\right)$
Given that the acceleration due to gravity at depth $d$ is $\frac{g}{4}$,we substitute $g' = \frac{g}{4}$ into the equation:
$\frac{g}{4} = g\left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{4} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{4}$
$\frac{d}{R} = \frac{3}{4}$
$d = \frac{3R}{4}$
Thus,at a depth of $\frac{3R}{4}$,the acceleration due to gravity becomes $\frac{g}{4}$.

(B) The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left( 1 - \frac{d}{R} \right)$.
Given that the acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,we have $g' = \frac{g}{n}$.
Substituting this into the formula: $\frac{g}{n} = g \left( 1 - \frac{d}{R} \right)$.
Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{d}{R}$.
Rearranging the terms to solve for $d$: $\frac{d}{R} = 1 - \frac{1}{n}$.
$\frac{d}{R} = \frac{n-1}{n}$.
Therefore,the depth $d$ is $d = R \left( \frac{n-1}{n} \right)$.

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.
For the body to be weightless,the effective gravity must be zero,so $g' = 0$.
Substituting the values: $0 = g - \omega^2 R \cos^2(60^{\circ})$.
Since $\cos(60^{\circ}) = 1/2$,we have $\cos^2(60^{\circ}) = 1/4$.
Thus,$g = \omega^2 R (1/4) \implies \omega^2 = 4g/R$.
Taking the square root,$\omega = 2 \sqrt{g/R}$.
Using $g = 9.8 \, m/s^2$ and $R = 6.4 \times 10^6 \, m$:
$\omega = 2 \sqrt{9.8 / (6.4 \times 10^6)} \approx 2 \sqrt{1.53 \times 10^{-6}} \approx 2 \times 1.237 \times 10^{-3} \approx 2.47 \times 10^{-3} \, rad/s$.
Rounding to the nearest option,$\omega = 2.5 \times 10^{-3} \, rad/s$.

(A) The gravitational force $F$ on a test mass $m$ is given by $F = mI$,where $I$ is the gravitational field intensity.
Case $1$: For points outside the sphere $(r > R)$:
The gravitational field intensity is $I = \frac{GM}{r^2}$.
Thus,$F \propto \frac{1}{r^2}$.
Therefore,$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$ for $r_1 > R$ and $r_2 > R$.
Case $2$: For points inside the sphere $(r < R)$:
The gravitational field intensity is $I = \frac{GMr}{R^3}$.
Thus,$F \propto r$.
Therefore,$\frac{F_1}{F_2} = \frac{r_1}{r_2}$ for $r_1 < R$ and $r_2 < R$.
Comparing these with the given options,option $A$ is correct.

(C) The time taken for a body to fall from a height $h$ under gravity is given by the kinematic equation $h = \frac{1}{2}gt^2$,which simplifies to $t = \sqrt{\frac{2h}{g}}$.
For Earth,$t_{earth} = \sqrt{\frac{2h}{g_{earth}}}$.
For the Moon,$t_{moon} = \sqrt{\frac{2h}{g_{moon}}}$.
Given that the acceleration due to gravity on the Moon is $g_{moon} = \frac{g_{earth}}{6}$,we can write the ratio as:
$\frac{t_{moon}}{t_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}} = \sqrt{\frac{g_{earth}}{g_{earth}/6}} = \sqrt{6}$.
Therefore,$t_{moon} = \sqrt{6}t$.

(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.
For the new planet,$M' = 2M$ and $R' = 2R$.
Thus,the new acceleration due to gravity $g'$ is:
$g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \left( \frac{GM}{R^2} \right) = \frac{g}{2}$.
The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
Therefore,$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g/2}} = \sqrt{2}$.
Given that for a seconds pendulum on Earth,$T = 2 \ s$.
So,$T' = T \times \sqrt{2} = 2\sqrt{2} \ s$.

(C) The acceleration due to gravity at a height $h$ from the surface of the Earth is given by:
$g' = \frac{g}{(1 + \frac{h}{R})^2} \quad ... (i)$
Where $g$ is the acceleration due to gravity at the surface of the Earth and $R$ is the radius of the Earth.
Multiplying both sides by the mass of the body $m$,we get the weight $W'$ at height $h$:
$W' = mg' = \frac{mg}{(1 + \frac{h}{R})^2} = \frac{W}{(1 + \frac{h}{R})^2}$
According to the question,$W' = \frac{1}{16}W$.
Substituting this into the equation:
$\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2}$
Taking the square root on both sides:
$1 + \frac{h}{R} = 4$
$\frac{h}{R} = 3$
$h = 3R$

(A) The gravitational force $F$ acting on a particle of mass $m$ near the surface of a planet of mass $M$ and radius $R = D/2$ is given by Newton's law of gravitation:
$F = \frac{GMm}{R^2} = \frac{GMm}{(D/2)^2}$
The acceleration due to gravity $g$ experienced by the particle is defined as the force per unit mass:
$g = \frac{F}{m} = \frac{GM}{(D/2)^2}$
Simplifying the expression:
$g = \frac{GM}{D^2/4} = \frac{4GM}{D^2}$
Thus,the acceleration due to gravity is $\frac{4GM}{D^2}$.

(D) The acceleration due to gravity $g$ at a distance $r$ from the centre of the Earth is given by:
For $r \le R$ (inside the Earth):
$g = \frac{4}{3} \pi \rho G r$
This shows that $g$ is directly proportional to $r$ $(g \propto r)$,which represents a straight line passing through the origin.
For $r > R$ (outside the Earth):
$g = \frac{GM}{r^2} = \frac{4}{3} \frac{\pi \rho R^3 G}{r^2}$
This shows that $g$ is inversely proportional to the square of the distance $(g \propto \frac{1}{r^2})$,which represents a hyperbolic curve.
Combining these,the graph starts from the origin,increases linearly until $r = R$,and then decreases following an inverse-square law for $r > R$. This matches the curve shown in the solution image.

(D) The gravitational potential at a height $h$ from the surface of the Earth is given by $V_h = -\frac{GM}{R+h} = -5.4 \times 10^7\, J kg^{-1}$.
The acceleration due to gravity at the same height is $g_h = \frac{GM}{(R+h)^2} = 6.0\, m s^{-2}$.
We can relate these two expressions as $g_h = \frac{GM}{(R+h)^2} = \frac{1}{R+h} \left( \frac{GM}{R+h} \right) = \frac{-V_h}{R+h}$.
Rearranging for $(R+h)$,we get $R+h = \frac{-V_h}{g_h}$.
Substituting the given values: $R+h = \frac{-(-5.4 \times 10^7)}{6.0} = \frac{5.4 \times 10^7}{6.0} = 0.9 \times 10^7 = 9.0 \times 10^6\, m$.
Converting the radius of the Earth to meters: $R = 6400\, km = 6.4 \times 10^6\, m$.
Now,$h = (R+h) - R = 9.0 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6\, m$.
Converting back to kilometers: $h = 2600\, km$.

(A) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g \left(1 + \frac{h}{R}\right)^{-2} \approx \frac{g}{(1 + h/R)^2}$.
Given $g_h = \frac{g}{4}$,we have $\frac{g}{4} = \frac{g}{(1 + h/R)^2}$.
Taking the square root on both sides,$2 = 1 + \frac{h}{R}$,which gives $\frac{h}{R} = 1$,so $h = R$.
The acceleration due to gravity at a depth $d$ below the surface is given by $g_d = g \left(1 - \frac{d}{R}\right)$.
Given $g_d = \frac{g}{4}$,we have $\frac{g}{4} = g \left(1 - \frac{d}{R}\right)$.
Dividing by $g$,$\frac{1}{4} = 1 - \frac{d}{R}$,which gives $\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}$,so $d = \frac{3R}{4}$.
Now,the ratio $\frac{h}{d} = \frac{R}{3R/4} = \frac{4}{3}$.

(C) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula:
$g_h = g \left( 1 - \frac{2h}{R_e} \right)$
where $R_e$ is the radius of the Earth.
The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g_d = g \left( 1 - \frac{d}{R_e} \right)$
According to the problem,the acceleration due to gravity at height $h$ is equal to the acceleration due to gravity at depth $d$:
$g_h = g_d$
Substituting the formulas:
$g \left( 1 - \frac{2h}{R_e} \right) = g \left( 1 - \frac{d}{R_e} \right)$
Canceling $g$ from both sides and simplifying:
$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$
$\frac{2h}{R_e} = \frac{d}{R_e}$
$d = 2h$
Given that $h = 1\, km$,we find:
$d = 2 \times 1\, km = 2\, km$

(C) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $M$ is the mass of the Earth and $R$ is the radius of the Earth.
Note that the mass of the Sun does not affect the value of $g$ on the Earth's surface.
If the universal gravitational constant $G$ becomes $10G$,then the new acceleration due to gravity $g'$ becomes $g' = \frac{(10G)M}{R^2} = 10g$.
Since $g' = 10g$,the acceleration due to gravity increases significantly.
$1$. Raindrops will fall faster because the acceleration due to gravity is higher.
$2$. Walking on the ground becomes more difficult because the effective weight of a person increases.
$3$. The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$. Since $g$ increases,$T$ will decrease.
$4$. The statement '$g$ on the Earth will not change' is incorrect because $g$ increases by a factor of $10$.

(A) Weight is defined as the force exerted on an object due to gravity,given by the formula $W = m \times g$,where $m$ is the mass and $g$ is the acceleration due to gravity. If the force of gravity disappears,$g$ becomes $0$,which results in the weight $W$ becoming $0$. However,mass is an intrinsic property of an object representing the amount of matter it contains and does not depend on gravity. Therefore,the mass remains unchanged.

(D) For scientist $A$,who goes down into a mine at depth $d$,the acceleration due to gravity is given by $g' = g(1 - d/R)$. As $d$ increases,$g'$ decreases.
For scientist $B$,who goes up in a balloon to a height $h$,the acceleration due to gravity is given by $g' = g(1 - 2h/R)$ (for $h \ll R$). As $h$ increases,$g'$ decreases.
Since the mathematical expressions for the variation of $g$ with depth and height are different,the value of $g$ measured by each scientist decreases at different rates.

(B) The acceleration due to gravity $g$ on the surface of a celestial body is given by $g = \frac{GM}{R^2}$,where $M$ is the mass and $R$ is the radius.
Given: $M_m = \frac{1}{81} M_e$ and $g_m = \frac{1}{6} g_e$.
We have the ratio: $\frac{g_m}{g_e} = \frac{M_m}{M_e} \times \left( \frac{R_e}{R_m} \right)^2$.
Substituting the values: $\frac{1}{6} = \frac{1}{81} \times \left( \frac{R_e}{R_m} \right)^2$.
Therefore,$\left( \frac{R_e}{R_m} \right)^2 = \frac{81}{6}$.
Taking the square root on both sides: $\frac{R_e}{R_m} = \sqrt{\frac{81}{6}} = \frac{9}{\sqrt{6}}$.
Thus,$R_e = \frac{9}{\sqrt{6}} R_m$.

(C) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$.
This implies that $g \propto \frac{M}{R^2}$.
According to the problem,the mass of the planet is $M_p = \frac{M_e}{2}$ and its radius is $R_p = \frac{R_e}{2}$,where $M_e$ and $R_e$ are the mass and radius of the Earth respectively.
Taking the ratio of the acceleration due to gravity on the planet $(g_p)$ to that on Earth $(g_e)$:
$\frac{g_p}{g_e} = \left( \frac{M_p}{M_e} \right) \times \left( \frac{R_e}{R_p} \right)^2 = \left( \frac{1}{2} \right) \times (2)^2 = \frac{1}{2} \times 4 = 2$.
Therefore,$g_p = 2 \times g_e = 2 \times 9.8\, m/s^2 = 19.6\, m/s^2$.

(B) The acceleration due to gravity on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
For Earth,$g_e = \frac{GM_e}{R_e^2}$.
For the other planet,the mass $M_p = 2M_e$ and the radius $R_p = 2R_e$.
Substituting these values into the formula for $g_p$:
$g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} \left( \frac{GM_e}{R_e^2} \right)$.
Since $g_e = \frac{GM_e}{R_e^2}$,we get $g_p = \frac{g_e}{2}$.

(A) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2} = 9.8 \, m/s^2$.
At a height $h$ above the surface of the earth,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given: $g = 9.8 \, m/s^2$,$R = 6400 \, km$,and $h = 480 \, km$.
Substituting these values into the formula:
$g' = 9.8 \times \left( \frac{6400}{6400 + 480} \right)^2$
$g' = 9.8 \times \left( \frac{6400}{6880} \right)^2$
$g' = 9.8 \times (0.9302)^2$
$g' = 9.8 \times 0.8653 \approx 8.48 \, m/s^2$.
Rounding to the nearest provided option,we get $8.4 \, m/s^2$.

(C) The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R}{R + h} \right)^2$.
Given that $g' = \frac{g}{2}$,we have $\frac{1}{2} = \left( \frac{R}{R + h} \right)^2$.
Taking the square root on both sides,we get $\frac{1}{\sqrt{2}} = \frac{R}{R + h}$.
This implies $R + h = R\sqrt{2}$,so $h = R(\sqrt{2} - 1)$.
Given $R = 4000 \ mile$ and $\sqrt{2} \approx 1.414$,we have $h = 4000(1.414 - 1) = 4000(0.414) = 1656 \ mile$.
Rounding to the nearest provided option,$h \approx 1600 \ mile$.

(B) The acceleration due to gravity at the surface of the earth is given by $g_0 = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $R$ is its radius.
At an altitude $h$,the acceleration due to gravity $g_h$ is given by $g_h = \frac{GM}{(R+h)^2}$.
We are given that $g_h = 25\% \text{ of } g_0$,which means $g_h = \frac{1}{4} g_0$.
Substituting the expressions for $g_h$ and $g_0$,we get $\frac{GM}{(R+h)^2} = \frac{1}{4} \frac{GM}{R^2}$.
Canceling $GM$ from both sides,we have $\frac{1}{(R+h)^2} = \frac{1}{4R^2}$.
Taking the square root of both sides,we get $\frac{1}{R+h} = \frac{1}{2R}$.
Cross-multiplying gives $2R = R + h$,which simplifies to $h = R$.

(B) The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$.
Substituting mass $M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right)$,we get:
$g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} G \pi \rho R$.
Since $G$,$\pi$,and $\frac{4}{3}$ are constants,we have $g \propto \rho R$.
To keep $g$ constant,the product $\rho R$ must remain constant.
Let the initial density be $\rho_1$ and radius be $R_1$. Let the final density be $\rho_2$ and radius be $R_2 = 5R_1$.
Then,$\rho_1 R_1 = \rho_2 R_2$.
$\rho_1 R_1 = \rho_2 (5R_1)$.
$\rho_2 = \frac{\rho_1}{5}$.
Thus,the density must be changed by a factor of $1/5$.

(A) Let the density of Earth be $d$ and its radius be $R$.
Then the density of the planet is $8d$ and its radius is $2R$.
Mass of Earth $M = \frac{4}{3} \pi R^3 d$.
Mass of the planet $M' = \frac{4}{3} \pi (2R)^3 (8d) = \frac{4}{3} \pi (8R^3) (8d) = 64 M$.
Acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$.
Acceleration due to gravity on the planet is $g' = \frac{GM'}{(2R)^2} = \frac{G(64M)}{4R^2} = 16 \frac{GM}{R^2} = 16g$.
For a body falling freely from rest,the distance $S$ covered in time $t$ is $S = \frac{1}{2}at^2$.
On Earth: $S = \frac{1}{2}g(1)^2 = \frac{g}{2}$.
On the planet: $S = \frac{1}{2}g't^2 = \frac{1}{2}(16g)t^2 = 8gt^2$.
Equating the two expressions for $S$: $\frac{g}{2} = 8gt^2$.
$t^2 = \frac{1}{16} \implies t = \frac{1}{4} = 0.25 \, \text{sec}$.
Note: The mass of the object does not affect the time of free fall.

(A) Given: $g_{h} = 1\% \text{ of } g_{s} = 0.01 g_{s}$.
The formula for acceleration due to gravity at height $h$ is $g_{h} = g_{s} \left( \frac{R}{R+h} \right)^{2}$.
Substituting the given value: $0.01 g_{s} = g_{s} \left( \frac{R}{R+h} \right)^{2}$.
Dividing both sides by $g_{s}$: $0.01 = \left( \frac{R}{R+h} \right)^{2}$.
Taking the square root on both sides: $0.1 = \frac{R}{R+h}$.
Rearranging the equation: $0.1(R + h) = R$.
$0.1R + 0.1h = R$.
$0.1h = 0.9R$.
$h = \frac{0.9}{0.1} R = 9R$.
Thus,at a height of $9R$,the acceleration due to gravity falls to $1\%$ of its value at the surface.

(B) The gravitational acceleration $g_h$ at a height $h$ from the surface of the earth is given by the formula:
$g_h = g \left( \frac{R}{R+h} \right)^2$
where $g$ is the acceleration due to gravity at the surface of the earth.
For height $h_1 = 3R$,the acceleration is:
$g_1 = g \left( \frac{R}{R+3R} \right)^2 = g \left( \frac{R}{4R} \right)^2 = g \left( \frac{1}{4} \right)^2 = \frac{g}{16}$
For height $h_2 = 4R$,the acceleration is:
$g_2 = g \left( \frac{R}{R+4R} \right)^2 = g \left( \frac{R}{5R} \right)^2 = g \left( \frac{1}{5} \right)^2 = \frac{g}{25}$
The ratio of gravitational acceleration at height $3R$ to that at height $4R$ is:
$\frac{g_1}{g_2} = \frac{g/16}{g/25} = \frac{25}{16}$
Thus,the correct option is $B$.

(A) For a point $P$ inside the planet at a distance $r_1$ from the center, the acceleration due to gravity is given by $g_{in} = g(r_1/R)$.
Given $g_{in} = g/4$, we have $r_1/R = 1/4$, so $r_1 = R/4$.
The distance of point $P$ from the surface is $d = R - r_1 = R - R/4 = 3R/4$.
For a point $Q$ outside the planet at a distance $r_2$ from the center, the acceleration due to gravity is given by $g_{out} = g(R/r_2)^2$.
Given $g_{out} = g/4$, we have $(R/r_2)^2 = 1/4$, which implies $R/r_2 = 1/2$, so $r_2 = 2R$.
The distance of point $Q$ from the surface is $h = r_2 - R = 2R - R = R$.
The maximum separation between $P$ and $Q$ is the sum of their distances from the surface: $d + h = 3R/4 + R = 7R/4$.

(D) For a point outside the Earth at a distance $r$ from the centre $(r > R)$, the acceleration due to gravity is given by $g_{out} = \frac{GM}{r^2}$. Thus, $g_{out} \propto \frac{1}{r^2}$.
For a point inside the Earth at a distance $r$ from the centre $(r < R)$, assuming uniform density $\rho$, the mass enclosed is $M' = \rho \cdot \frac{4}{3}\pi r^3$. The acceleration due to gravity is $g_{in} = \frac{GM'}{r^2} = \frac{G}{r^2} \cdot \rho \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi G \rho r$. Thus, $g_{in} \propto r$.
Therefore, both statements $(A)$ and $(C)$ are correct.

(A) The acceleration due to gravity $g$ at the surface of the Earth is given by $g = \frac{GM}{R^2}$.
Substituting the mass $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we get:
$g = \frac{G \times \rho \times \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3}\pi G \rho R$.
From this expression,we can see that the average density $\rho$ is given by $\rho = \frac{3g}{4\pi GR}$.
Since $G$,$\pi$,and $R$ (radius of the Earth) are constants,it follows that $\rho \propto g$.
Therefore,the average density of the Earth is directly proportional to $g$.

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = \frac{g}{9}$,we substitute this into the equation:
$\frac{g}{9} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$,we get: $\frac{1}{9} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides: $\frac{1}{3} = \frac{R}{R+h}$.
Cross-multiplying gives: $R + h = 3R$.
Therefore,$h = 3R - R = 2R$.

(C) For a point at a distance $y$ from the center of a uniform sphere of density $\rho$,the gravitational force $F$ acting on a body of mass $m$ is due only to the mass $M'$ contained within a sphere of radius $y$.
$M' = \text{Volume} \times \text{Density} = \left(\frac{4}{3} \pi y^3\right) \rho$.
According to Newton's law of gravitation,the force is $F = \frac{G M' m}{y^2}$.
Substituting $M'$,we get $F = \frac{G m}{y^2} \left(\frac{4}{3} \pi y^3 \rho\right) = \frac{4}{3} \pi G \rho m y$.
Since $F = ma$,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{4}{3} \pi G \rho y$.

(C) The acceleration due to gravity $g$ at the surface of a planet is given by the formula:
$g = \frac{GM}{R^2}$
where $G$ is the universal gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.
According to the problem,both planets $A$ and $B$ have the same mass $M$ and the same radius $R$.
Since $g$ depends only on the mass $M$ and the radius $R$ of the planet,and both are identical for planets $A$ and $B$,the acceleration due to gravity at the surface of both planets must be equal.
Therefore,the ratio of the acceleration due to gravity at the surface of planet $A$ to that of planet $B$ is:
$\frac{g_A}{g_B} = \frac{GM/R^2}{GM/R^2} = 1$
Thus,the correct option is $C$.

(A) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{r^2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $p$ and radius $r$ as $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 p$,we substitute this into the formula for $g$:
$g = \frac{G}{r^2} \times \left( \frac{4}{3} \pi r^3 p \right) = \frac{4}{3} \pi G r p$.
Thus,$g \propto r p$.
Therefore,the ratio of acceleration due to gravity on the two planets is $\frac{g_1}{g_2} = \frac{r_1 p_1}{r_2 p_2}$.

(A) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM_e}{R_e^2} = 9.8 \ m/s^2$.
For the planet,the mass $M_p = 3M_e$ and the radius $R_p = 3R_e$ (since the diameter is three times that of the Earth,the radius is also three times).
The acceleration due to gravity on the surface of the planet is $g_p = \frac{GM_p}{R_p^2}$.
Substituting the values: $g_p = \frac{G(3M_e)}{(3R_e)^2} = \frac{3GM_e}{9R_e^2} = \frac{1}{3} \left( \frac{GM_e}{R_e^2} \right)$.
Therefore,$g_p = \frac{g}{3} = \frac{9.8}{3} \approx 3.3 \ m/s^2$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.
At height $h$,the new time period is $T^{\prime} = 2\pi \sqrt{\frac{\ell}{g^{\prime}}}$.
Given that $T^{\prime} = \sqrt{2} T$,we have $2\pi \sqrt{\frac{\ell}{g^{\prime}}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{\ell}{g}}$.
Squaring both sides,we get $\frac{1}{g^{\prime}} = \frac{2}{g}$,which implies $g^{\prime} = \frac{g}{2}$.
The acceleration due to gravity at height $h$ is $g^{\prime} = \frac{g R^2}{(R+h)^2}$.
Equating the two expressions for $g^{\prime}$,we get $\frac{g R^2}{(R+h)^2} = \frac{g}{2}$.
This simplifies to $(R+h)^2 = 2R^2$,so $R+h = \sqrt{2}R$.
Therefore,$h = (\sqrt{2}-1)R \approx 0.414 R$.

(A) The acceleration due to gravity at height $h$ is given by $g_h = g(1 - \frac{2h}{R})$.
The fractional decrease in weight at height $h$ is $\frac{\Delta g}{g} = \frac{2h}{R} = 1\% = 0.01$.
Therefore,$\frac{h}{R} = 0.005$.
The acceleration due to gravity at depth $h$ is given by $g_d = g(1 - \frac{h}{R})$.
The fractional decrease in weight at depth $h$ is $\frac{\Delta g}{g} = \frac{h}{R} = 0.005 = 0.5\%$.
Thus,the weight of the body decreases by $0.5\%$.

(D) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R \omega^2 \cos^2 \lambda$.
For a man to feel weightlessness,the effective gravity must be zero,so $g' = 0$.
Substituting the given values,$0 = g - R \omega^2 \cos^2 60^{\circ}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $0 = g - R \omega^2 (\frac{1}{2})^2$.
$g = R \omega^2 (\frac{1}{4}) \Rightarrow \omega^2 = \frac{4g}{R} \Rightarrow \omega = 2 \sqrt{\frac{g}{R}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.
Substituting $\omega$,$T = \frac{2 \pi}{2 \sqrt{g/R}} = \pi \sqrt{\frac{R}{g}} \times 2 = 2 \pi \sqrt{\frac{R}{g}} \times 2 = 4 \pi \sqrt{\frac{R}{g}}$.

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Let $M$ and $R$ be the mass and radius of the Earth,respectively.
For the given planet,the mass $M' = M - 0.10M = 0.9M = \frac{9}{10}M$.
The radius $R' = R + 0.20R = 1.2R = \frac{12}{10}R = \frac{6}{5}R$.
The acceleration due to gravity on the planet is $g' = \frac{GM'}{(R')^2}$.
Substituting the values: $g' = \frac{G(\frac{9}{10}M)}{(\frac{6}{5}R)^2}$.
$g' = \frac{G \cdot \frac{9}{10}M}{\frac{36}{25}R^2} = \frac{9}{10} \cdot \frac{25}{36} \cdot \frac{GM}{R^2}$.
$g' = \frac{1}{2} \cdot \frac{5}{4} \cdot g = \frac{5}{8}g$.
Thus,the acceleration due to gravity on the planet is $\frac{5}{8}$ times that on the surface of the earth.

(A) The electronic charge is a fundamental physical constant and is an intrinsic property of an electron.
It does not depend on the gravitational field or the location where the measurement is performed.
Millikan's oil drop experiment determines the value of the elementary charge $e$,which is approximately $1.602 \times 10^{-19} \ C$.
Since the charge of an electron remains constant regardless of the acceleration due to gravity ($g_E$ or $g_M$),the ratio of the electronic charge on the moon to the electronic charge on the earth is $1$.

(D) For a sphere of uniform density, the acceleration due to gravity $g$ at a distance $r$ from the centre is given by:
$1$. Inside the earth $(r < R)$: $g_{in} = \frac{G M r}{R^3}$. Thus, $g \propto r$. This represents a straight line passing through the origin.
$2$. Outside the earth $(r > R)$: $g_{out} = \frac{G M}{r^2}$. Thus, $g \propto \frac{1}{r^2}$. This represents a hyperbolic curve.
Therefore, the graph starts from the origin, increases linearly until $r = R$, and then decreases following an inverse-square law. This corresponds to the figure provided in option $D$.

(C) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.
For Earth,$g = \frac{GM_e}{R_e^2}$.
For the Moon,$g_m = \frac{GM_m}{R_m^2}$.
Given: $M_e = 80 M_m$ (so $M_m = \frac{M_e}{80}$) and $R_e = 4 R_m$ (so $R_m = \frac{R_e}{4}$).
Substituting these values into the expression for $g_m$:
$g_m = \frac{G(M_e / 80)}{(R_e / 4)^2} = \frac{G M_e / 80}{R_e^2 / 16} = \frac{16}{80} \frac{G M_e}{R_e^2}$.
Since $g = \frac{G M_e}{R_e^2}$,we get $g_m = \frac{1}{5} g = \frac{g}{5}$.

(A) The mass of an object is an intrinsic property of matter and does not change regardless of its position in the universe.
As the stone moves toward the center of the Earth,the gravitational force acting on it decreases,meaning its weight $(W = mg)$ decreases.
At the center of the Earth,the gravitational acceleration $(g)$ becomes zero,so the weight and the gravitational field intensity also become zero.
Therefore,only the mass of the stone remains constant throughout the motion.

(B) For a point inside the planet at distance $r_1$ from the center,the acceleration due to gravity is given by $g(r_1) = g_0 \frac{r_1}{R}$.
Given $g(r_1) = \frac{g_0}{4}$,we have $\frac{g_0}{4} = g_0 \frac{r_1}{R}$,which gives $r_1 = \frac{R}{4}$.
For a point outside the planet at distance $r_2$ from the center,the acceleration due to gravity is given by $g(r_2) = g_0 \frac{R^2}{r_2^2}$.
Given $g(r_2) = \frac{g_0}{4}$,we have $\frac{g_0}{4} = g_0 \frac{R^2}{r_2^2}$,which implies $r_2^2 = 4R^2$,so $r_2 = 2R$.
Therefore,$r_2 - r_1 = 2R - \frac{R}{4} = \frac{7R}{4}$.

(B) Let the original mass of the asteroid be $M$ and its radius be $R$. The gravitational field at the surface is $g = \frac{GM}{R^2}$.
When a spherical part of radius $r = R/2$ is excavated,its mass $M'$ is given by $M' = \rho \cdot V' = \rho \cdot \frac{4}{3}\pi (R/2)^3 = \frac{1}{8} \rho \cdot \frac{4}{3}\pi R^3 = \frac{M}{8}$.
The gravitational field at the point on the surface just above the excavation is the vector sum of the field due to the original sphere and the field due to the removed part (treated as a negative mass).
The field due to the original sphere at the surface is $g_1 = \frac{GM}{R^2}$ (directed towards the center).
The field due to the removed sphere at its own surface is $g_2 = \frac{GM'}{r^2} = \frac{G(M/8)}{(R/2)^2} = \frac{GM/8}{R^2/4} = \frac{GM}{2R^2}$ (directed away from the center of the excavation,i.e.,outwards).
The net gravitational acceleration is $g_{net} = g_1 - g_2 = \frac{GM}{R^2} - \frac{GM}{2R^2} = \frac{GM}{2R^2}$.

(C) The gravitational acceleration $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since $M = \rho \cdot V = \rho \cdot \frac{4}{3}\pi R^3$,substituting this into the formula gives $g = \frac{G \cdot \rho \cdot \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3} \pi G \rho R$.
Thus,$g \propto \rho R$.
Given the ratio of diameters $D_1 : D_2 = 4 : 1$,the ratio of radii $R_1 : R_2 = 4 : 1$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 2$.
The ratio of gravitational accelerations is $\frac{g_1}{g_2} = \frac{\rho_1 R_1}{\rho_2 R_2} = \left(\frac{1}{2}\right) \cdot \left(\frac{4}{1}\right) = \frac{4}{2} = 2 : 1$.

(D) The effective acceleration due to gravity $g'$ at a latitude $\phi$ is given by the formula: $g' = g - \omega^2 R \cos^2 \phi$,where $\omega$ is the angular velocity of the earth and $R$ is the radius of the earth.
At the poles,the latitude $\phi = 90^\circ$,so $\cos 90^\circ = 0$. Thus,$g' = g$,which means there is no change in gravity at the poles regardless of the change in $\omega$.
At all other latitudes,as the angular velocity $\omega$ increases,the term $\omega^2 R \cos^2 \phi$ increases.
Since $g' = g - \omega^2 R \cos^2 \phi$,an increase in $\omega$ leads to a decrease in $g'$.
Consequently,the weight of an object,$W = mg'$,will decrease at all points on the earth except at the poles.

(C) The effective acceleration due to gravity at the equator (where $\theta = 0^\circ$) is given by $g' = g - \omega^2 R$.
Given that the weight becomes $\frac{3}{4} W$,the effective gravity must be $g' = \frac{3}{4} g$.
Substituting this into the equation: $\frac{3}{4} g = g - \omega^2 R$.
Rearranging for $\omega^2 R$: $\omega^2 R = g - \frac{3}{4} g = \frac{1}{4} g$.
Solving for $\omega$: $\omega = \sqrt{\frac{g}{4R}}$.
Given $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 \times 10^6 \ m$.
$\omega = \sqrt{\frac{10}{4 \times 6.4 \times 10^6}} = \sqrt{\frac{10}{25.6 \times 10^6}} = \sqrt{\frac{1}{2.56 \times 10^6}} = \frac{1}{1.6 \times 10^3} \ rad/s$.
$\omega = 0.625 \times 10^{-3} \ rad/s \approx 0.63 \times 10^{-3} \ rad/s$.

(A) The acceleration due to gravity $g$ varies with the distance $r$ from the centre of the Earth as follows:
$1$. Inside the Earth $(r < R_E)$: The acceleration due to gravity is given by $g = \frac{GM r}{R_E^3}$,which means $g \propto r$. This represents a straight line passing through the origin.
$2$. Outside the Earth $(r \geq R_E)$: The acceleration due to gravity is given by $g = \frac{GM}{r^2}$,which means $g \propto \frac{1}{r^2}$. This represents a rectangular hyperbola.
Thus,the graph shows a linear increase up to $r = R_E$ and a non-linear decrease for $r > R_E$,which is correctly depicted in the graph provided in option $A$.

(A) The acceleration due to gravity $g'$ at a distance $r$ from the center of the earth (where $r < R$,$R$ being the radius of the earth) is given by the formula:
$g' = \frac{4}{3} \pi \rho G r$
Here,$\rho$ is the uniform density of the earth and $G$ is the universal gravitational constant.
Since $\frac{4}{3}$,$\pi$,$\rho$,and $G$ are constants,we can conclude that:
$g' \propto r$
Therefore,the acceleration due to gravity inside the earth is directly proportional to the distance $r$ from the center.

(D) The acceleration due to gravity on a planet is given by $g = \frac{GM}{R^2}$.
Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_p = 3M_e$ and $D_p = 3D_e$,which implies $R_p = 3R_e$.
The ratio of gravity on the planet to the Earth is $\frac{g_p}{g_e} = \frac{M_p}{M_e} \left( \frac{R_e}{R_p} \right)^2 = 3 \left( \frac{1}{3} \right)^2 = \frac{3}{9} = \frac{1}{3}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$,so $T \propto \frac{1}{\sqrt{g}}$.
Therefore,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{3}$.
Given $T_e = 2 \ s$,we get $T_p = 2\sqrt{3} \ s$.

(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2} = 9.8\, m\,s^{-2}$.
At an altitude $h$,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2}$.
We are given $g' = 4.9\, m\,s^{-2}$,which is $\frac{g}{2}$.
So,$\frac{GM}{(R+h)^2} = \frac{1}{2} \frac{GM}{R^2}$.
Taking the reciprocal and square root on both sides: $\frac{R+h}{R} = \sqrt{2}$.
$1 + \frac{h}{R} = 1.414$.
$\frac{h}{R} = 0.414$.
$h = 0.414 \times R = 0.414 \times 6.4 \times 10^6\, m$.
$h \approx 2.65 \times 10^6\, m$.