Acceleration Due to Gravity and its Variation Questions in English

(A) Statement $I$ is true because Newton's Law of Universal Gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Statement $II$ is true because the acceleration due to gravity $g$ at the center of the Earth is zero. Since weight is defined as $W = mg$,if $g = 0$,then the weight $W$ of the person at the center of the Earth is also zero.
Therefore,both statements are true.

(D) seconds pendulum has a time period $T = 2 \ s$.
The formula for the time period of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g'}}$.
At a height $h = 2R$ from the surface,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^{2} = g \left( \frac{R}{R+2R} \right)^{2} = g \left( \frac{1}{3} \right)^{2} = \frac{g}{9}$.
Given $g = \pi^{2} \ m/s^{2}$,so $g' = \frac{\pi^{2}}{9} \ m/s^{2}$.
Substituting the values into the time period formula: $2 = 2 \pi \sqrt{\frac{L}{\pi^{2}/9}}$.
$1 = \pi \sqrt{\frac{9L}{\pi^{2}}} = \pi \cdot \frac{3\sqrt{L}}{\pi} = 3\sqrt{L}$.
$\sqrt{L} = \frac{1}{3} \Rightarrow L = \frac{1}{9} \ m$.

(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g' = \frac{GM}{r^2}$.
Given the distance from the center is $r = \frac{5}{4}R$,the acceleration due to gravity at this height is:
$g' = \frac{GM}{(\frac{5}{4}R)^2} = \frac{GM}{\frac{25}{16}R^2} = \frac{16}{25} \left( \frac{GM}{R^2} \right) = \frac{16}{25}g$.
Since weight $W = mg$,the new weight $W' = m g' = \frac{16}{25}mg = \frac{16}{25}W$.
The decrease in weight is $\Delta W = W - W' = W - \frac{16}{25}W = \frac{9}{25}W$.
The percentage decrease in weight is $\frac{\Delta W}{W} \times 100 = \frac{9/25 W}{W} \times 100 = \frac{9}{25} \times 100 = 36\%$.
Thus,the percentage decrease in the weight of the object is $36\%$.

(A) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g' = g(1 - \frac{2h}{R})$,where $R$ is the radius of the Earth and $h << R$.
The change in acceleration due to gravity is $\Delta g = g - g' = g(\frac{2h}{R})$.
The fractional decrease in weight (which is proportional to the change in $g$) is given by $\frac{\Delta g}{g} = \frac{2h}{R}$.
To find the percentage decrease,we multiply by $100$:
$\text{Percentage decrease} = \frac{\Delta g}{g} \times 100 = \frac{2h}{R} \times 100$.
Substituting the given values $h = 32 \ km$ and $R = 6400 \ km$:
$\text{Percentage decrease} = 2 \times \frac{32}{6400} \times 100 = 2 \times \frac{1}{200} \times 100 = 1 \%$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.
The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R_E}{R_E + h} \right)^2$,which implies $\sqrt{\frac{g'}{g}} = \frac{R_E}{R_E + h}$.
Given the time periods $T_1 = 4\,s$ and $T_2 = 6\,s$,we have the ratio:
$\frac{T_1}{T_2} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.
Substituting the values:
$\frac{4}{6} = \frac{6400 + h}{6400}$.
$\frac{2}{3} = 1 + \frac{h}{6400}$.
$\frac{h}{6400} = \frac{2}{3} - 1 = -\frac{1}{3}$.
Wait,re-evaluating the ratio: Since $T \propto \frac{1}{\sqrt{g}}$,then $\frac{T_2}{T_1} = \sqrt{\frac{g}{g'}} = \frac{R_E + h}{R_E}$.
$\frac{6}{4} = \frac{6400 + h}{6400}$.
$1.5 = 1 + \frac{h}{6400}$.
$0.5 = \frac{h}{6400}$.
$h = 0.5 \times 6400 = 3200\,km$.

(D) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.
Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.
Taking the natural logarithm and differentiating,we get the relative error formula: $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $2 \%$,we have $\frac{\Delta R}{R} = -0.02$.
Substituting this into the error formula: $\frac{\Delta g}{g} = -2 \times (-0.02) = 0.04$.
To express this as a percentage,multiply by $100$: $\frac{\Delta g}{g} \times 100 = 4 \%$.
Since the result is positive,the acceleration due to gravity will increase by $4 \%$.

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R_e})$ for $h \ll R_e$.
The acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g_d = g(1 - \frac{d}{R_e})$.
According to the problem,$g_h = g_d$,so:
$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$
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 $d = \alpha h$,we compare the two expressions:
$\alpha h = 2h$
$\alpha = 2$.

(B) The acceleration due to gravity $g$ decreases as we move away from the earth's surface,given by $g(h) = g_0(1 - 2h/R_e)$.
Let $C$ be the centre of mass of the cabin. The acceleration of any object at a height $h$ is $a = g(h)$.
Since $A$ is above $C$ and $B$ is below $C$,their heights are $h_A > h_C > h_B$.
Consequently,the gravitational accelerations satisfy $a_B > a_C > a_A$.
When viewed from the frame of the cabin (which has acceleration $a_C$),the relative accelerations are:
$a_{A,rel} = a_A - a_C < 0$ (meaning $A$ accelerates upward relative to the cabin).
$a_{B,rel} = a_B - a_C > 0$ (meaning $B$ accelerates downward relative to the cabin).
Thus,$A$ moves slowly upward and $B$ moves slowly downward relative to the cabin. Therefore,option $(b)$ is correct.

(A) Due to the rotation of the planet,the effective acceleration due to gravity at latitude $\lambda$ is given by $g^{\prime} = g - \omega^2 R \cos^2 \lambda$.
At the equator,$\lambda = 0^{\circ}$,so $g_e = g - \omega^2 R$.
At a latitude of $60^{\circ}$,$\lambda = 60^{\circ}$,so $g_{60} = g - \omega^2 R \cos^2(60^{\circ}) = g - \frac{\omega^2 R}{4}$.
Given that the weight at the equator is $f$ times the weight at $60^{\circ}$,we have $g_e = f \cdot g_{60}$.
Substituting the expressions: $g - \omega^2 R = f \left( g - \frac{\omega^2 R}{4} \right)$.
Rearranging gives $g(1 - f) = \omega^2 R \left( 1 - \frac{f}{4} \right) = \frac{\omega^2 R}{4} (4 - f)$.
Using $g = \frac{GM}{R^2}$,$M = \frac{4}{3} \pi R^3 \rho$,and $\omega = \frac{2\pi}{T}$,we substitute:
$\frac{G}{R^2} \left( \frac{4}{3} \pi R^3 \rho \right) (1 - f) = \frac{4 \pi^2}{T^2} R \left( \frac{4 - f}{4} \right)$.
$\frac{4}{3} \pi G R \rho (1 - f) = \frac{\pi^2 R}{T^2} (4 - f)$.
Solving for density $\rho$: $\rho = \left( \frac{4 - f}{1 - f} \right) \cdot \frac{3 \pi}{4 G T^2}$.

(D) The value of acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R})$,where $R$ is the radius of the earth.
The value of acceleration due to gravity at a depth $d$ below the surface is given by $g_d = g(1 - \frac{d}{R})$.
According to the problem,$g_h = g_d$.
Therefore,$g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$.
Simplifying the equation,we get $\frac{2h}{R} = \frac{d}{R}$,which implies $d = 2h$.
Given that $h = 10 \,km$,we have $d = 2 \times 10 \,km = 20 \,km$.

(D) The orbital radius of the ball is $r = R + h$,where $R$ is the radius of the Earth $(6400 \, km)$ and $h$ is the height $(9 \, km)$.
Since $h \ll R$,we have $r \approx R$.
The acceleration of an object in a circular orbit is given by the centripetal acceleration $a = v^2 / r$.
Substituting the orbital velocity $v = \sqrt{GM/r}$,we get $a = (GM/r) / r = GM/r^2$.
Since $r \approx R$,the acceleration $a \approx GM/R^2$.
By definition,the acceleration due to gravity near the Earth's surface is $g = GM/R^2$.
Therefore,the acceleration of the ball in orbit is nearly equal to $g$.

(B) Given,mass of the second planet $M_2 = 8 M_1$,where $M_1$ is the mass of the first planet.
Since the density $\rho$ is the same for both planets,we have $M = \frac{4}{3} \pi R^3 \rho$.
Thus,$M \propto R^3$,which implies $R \propto M^{1/3}$.
Therefore,$\frac{R_2}{R_1} = \left(\frac{M_2}{M_1}\right)^{1/3} = (8)^{1/3} = 2$,so $R_2 = 2 R_1$.
The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
The ratio of the acceleration due to gravity of the second planet to that of the first planet is:
$\frac{g_2}{g_1} = \frac{M_2}{M_1} \times \left(\frac{R_1}{R_2}\right)^2$
Substituting the values,we get $\frac{g_2}{g_1} = 8 \times \left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2$.
Thus,the ratio is $2$.

(D) Given that,the weight of the body at the surface of the earth is $W = mg = 144 \, N$.
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 $h = 3R$,we substitute this into the equation:
$g' = g \left( \frac{R}{R + 3R} \right)^2 = g \left( \frac{R}{4R} \right)^2 = g \left( \frac{1}{4} \right)^2 = \frac{g}{16}$
The weight of the body at height $h$ is $W' = mg'$.
$W' = m \left( \frac{g}{16} \right) = \frac{W}{16}$
Substituting the value of $W = 144 \, N$:
$W' = \frac{144}{16} = 9 \, N$.
Therefore,the weight of the body at a height of $3R$ is $9 \, N$.

(B) The weight of a body at the Earth's surface is given by $W = \frac{GMm}{r^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the body,and $r$ is the radius of the Earth.
Since $G, M,$ and $m$ are constant,$W \propto \frac{1}{r^2}$.
Given that the radius contracts by $0.1 \%$,the new radius is $r' = r(1 - 0.001) = 0.999r$.
The new weight $W'$ is given by $W' = \frac{GMm}{(0.999r)^2} = \frac{W}{(0.999)^2}$.
Using the binomial approximation $(1-x)^{-n} \approx 1+nx$ for small $x$,we have $W' = W(1 - 0.001)^{-2} \approx W(1 + 2 \times 0.001) = W(1 + 0.002)$.
Thus,$W' = W + 0.002W$,which represents an increase of $0.002 \times 100 = 0.2 \%$.
Therefore,the weight increases by $0.2 \%$.

(B) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Given that the mass of the Earth decreases by $25 \%$,the new mass $M' = M - 0.25M = 0.75M = \frac{3}{4}M$.
The radius of the Earth increases by $50 \%$,so the new radius $R' = R + 0.50R = 1.5R = \frac{3}{2}R$.
The new acceleration due to gravity $g'$ is:
$g' = \frac{GM'}{(R')^2} = \frac{G(0.75M)}{(1.5R)^2} = \frac{0.75GM}{2.25R^2} = \frac{0.75}{2.25} g = \frac{1}{3} g \approx 0.333g$.
The decrease in acceleration due to gravity is $\Delta g = g - g' = g - 0.333g = 0.667g$.
Expressed as a percentage,the decrease is $0.667 \times 100 \% \approx 67 \%$.
Therefore,the acceleration due to gravity at its surface decreases by nearly $67 \%$.

(B) The acceleration due to gravity $g$ at the surface of a body of radius $R$ and density $\rho$ is given by:
$g = \frac{GM}{R^2} = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3} \pi G \rho R$
Given that $g_p = g_e$ and $\rho_p = 1.5 \rho_e = \frac{3}{2} \rho_e$.
Equating the gravity expressions for the planet and the Earth:
$\frac{4}{3} \pi G \rho_p R_p = \frac{4}{3} \pi G \rho_e R_e$
$\rho_p R_p = \rho_e R_e$
Substituting $\rho_p = \frac{3}{2} \rho_e$ and $R_e = R$:
$(\frac{3}{2} \rho_e) R_p = \rho_e R$
$R_p = \frac{2}{3} R$
Thus,the radius of the planet is $\frac{2}{3} R$.

(A) The weight of a body at height $h$ is given by $W_h = W \left(1 - \frac{2h}{R}\right)$,where $W = mg$ is the weight at the surface.
Percentage decrease in weight at height $h$ is $\frac{W - W_h}{W} \times 100 = \frac{2h}{R} \times 100 = 1.5 \%$.
From this,we get $\frac{h}{R} = \frac{1.5}{2 \times 100} = 0.75 \%$.
The weight of a body at depth $h$ is given by $W_d = W \left(1 - \frac{h}{R}\right)$.
Percentage decrease in weight at depth $h$ is $\frac{W - W_d}{W} \times 100 = \frac{h}{R} \times 100$.
Substituting the value of $\frac{h}{R}$,we get $\frac{h}{R} \times 100 = 0.75 \%$.
Thus,the weight decreases by $0.75 \%$.

(D) For an object on the equator to feel weightless,the effective acceleration due to gravity must be zero.
The effective acceleration due to gravity at the equator is given by $g_{eff} = g - \omega^2 R$,where $g$ is the acceleration due to gravity,$\omega$ is the angular velocity of the earth,and $R$ is the radius of the earth.
Setting $g_{eff} = 0$,we get $g - \omega^2 R = 0$,which implies $\omega^2 R = g$.
Thus,$\omega = \sqrt{\frac{g}{R}}$.
The duration of the day (time period $T$) is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{R}{g}}$.
Substituting the values $R \approx 6.4 \times 10^6 \, m$ and $g \approx 9.8 \, m/s^2$:
$T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6}{9.8}} \approx 2 \times 3.14 \times 808 \, s \approx 5074 \, s$.
Converting to hours: $T \approx \frac{5074}{3600} \, hr \approx 1.41 \, hr$.

(D) The acceleration due to gravity at a depth $d$ inside the Earth is given by $g_d = \frac{GMd}{R^3} = \beta$,where $d < R$.
From this,we can express the gravitational constant and mass product as $GM = \frac{\beta R^3}{d}$.
The acceleration due to gravity at a height $d$ above the surface of the Earth is given by $g_h = \frac{GM}{(R+d)^2}$.
Substituting the value of $GM$ from the first equation into the second,we get $g_h = \frac{\beta R^3}{d(R+d)^2}$.

(D) The weight of a body on the surface of the earth is given by $W = mg = \frac{GMm}{R^2} = 72 \, N$.
At a height $h = 2R$ from the surface,the distance from the center of the earth is $r = R + h = R + 2R = 3R$.
The weight at this height is $W' = mg' = \frac{GMm}{r^2} = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$.
Substituting the value of $W$ from the first equation,we get $W' = \frac{1}{9} \times \left( \frac{GMm}{R^2} \right) = \frac{72}{9} = 8 \, N$.

(C) The acceleration due to gravity $(g)$ at the surface of a planet is given by the formula: $g = \frac{GM}{R^2}$.
Here,$M$ is the mass of the planet and $R$ is its radius.
Since the planet is a sphere,its mass can be expressed in terms of its density $(\rho)$ and volume $(V)$: $M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right)$.
Substituting this into the expression for $g$:
$g = \frac{G}{R^2} \times \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi G R \rho$.
Since $G$,$\pi$,and $\rho$ are constants for both planets,we find that $g \propto R$.
Therefore,the acceleration due to gravity is directly proportional to the radius of the planet. Thus,it would be greater on the larger planet.

(C) The gravitational acceleration $g$ is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is its radius.
Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.
Taking the logarithmic derivative,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $1.5 \%$,we have $\frac{\Delta R}{R} = -1.5 \% = -0.015$.
Substituting this value into the equation:
$\frac{\Delta g}{g} = -2 \times (-1.5 \%) = +3 \%$.
Therefore,the gravitational acceleration increases by $3 \%$.

(B) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since mass $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we can substitute this into the formula:
$g = \frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto \rho R$.
Given $\rho_p = 2 \rho_e$ and $R_p = 1.5 R_e$,where $p$ denotes the planet and $e$ denotes the earth:
$\frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e} = \frac{2 \rho_e \times 1.5 R_e}{\rho_e R_e} = 2 \times 1.5 = 3$.
Therefore,the acceleration due to gravity on the surface of the planet is $3$ times that on the surface of the earth.

(B) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula $g' = g(1 - \frac{2h}{R_e})$,where $R_e$ is the radius of the Earth.
Given that the value of $g$ decreases by $2 \%$,the new value $g'$ is $98 \%$ of $g$,so $g' = 0.98g$.
Substituting this into the formula: $0.98g = g(1 - \frac{2h}{R_e})$.
Dividing both sides by $g$: $0.98 = 1 - \frac{2h}{R_e}$.
Rearranging the terms: $\frac{2h}{R_e} = 1 - 0.98 = 0.02$.
Solving for $h$: $h = 0.01 \times R_e$.
Given $R_e = 6400 \, km$,we have $h = 0.01 \times 6400 \, km = 64 \, km$.

(A) The weight of a man at the equator is given by $w_{eq} = m(g - \omega^2 R)$,where $m$ is the mass,$g$ is the acceleration due to gravity,$\omega$ is the angular velocity of the Earth,and $R$ is the radius of the Earth.
At the poles,the effect of rotation is zero,so the weight is $w_p = mg$.
The change in weight is $\Delta w = w_p - w_{eq} = mg - m(g - \omega^2 R) = m\omega^2 R$.
The fractional change in weight is $\frac{\Delta w}{w_{eq}} = \frac{m\omega^2 R}{m(g - \omega^2 R)} = \frac{\omega^2 R}{g - \omega^2 R}$.
Given $\omega^2 R \approx 0.0337 \, m/s^2$ and $g \approx 9.81 \, m/s^2$,we have:
$\frac{\Delta w}{w_{eq}} = \frac{0.0337}{9.81 - 0.0337} \approx \frac{0.0337}{9.7763} \approx 0.003447$.
Converting to percentage: $0.003447 \times 100 = 0.3447 \%$.
Thus,the weight increases by approximately $0.34 \%$.

(B) The apparent weight $w'$ of an object of mass $m$ at the equator is given by the formula: $w' = mg - m \omega^2 R$,where $g$ is the acceleration due to gravity,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.
When the Earth is rotating,the effective weight is reduced by the centrifugal force $m \omega^2 R$.
If the Earth suddenly stops rotating,the angular speed $\omega$ becomes $0$.
Consequently,the new apparent weight $w''$ becomes $w'' = mg$.
Comparing the two,the weight increases by the magnitude of the centrifugal force,which is $m \omega^2 R$.
Therefore,the weight of the object increases by $m \omega^2 R$.

(C) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula:
$g_h = g \left[ \frac{R}{R+h} \right]^2$
Given:
$g = 9.8 \, m/s^2$
$R = 6400 \, km$
$h = 480 \, km$
Substituting the values into the formula:
$g_h = 9.8 \left[ \frac{6400}{6400 + 480} \right]^2$
$g_h = 9.8 \left[ \frac{6400}{6880} \right]^2$
$g_h = 9.8 \left[ \frac{40}{43} \right]^2$
$g_h = 9.8 \times (0.9302)^2$
$g_h = 9.8 \times 0.8653$
$g_h \approx 8.48 \, m/s^2$
Rounding to the nearest value,we get $8.5 \, m/s^2$.

(B) The acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R_e})$.
The change in the value of $g$ at height $h$ is $\Delta g_h = g - g_h = g(\frac{2h}{R_e})$.
The acceleration due to gravity at a depth $x$ below the surface is given by $g_x = g(1 - \frac{x}{R_e})$.
The change in the value of $g$ at depth $x$ is $\Delta g_x = g - g_x = g(\frac{x}{R_e})$.
According to the problem,the change in $g$ is the same in both cases:
$\Delta g_h = \Delta g_x$
$g(\frac{2h}{R_e}) = g(\frac{x}{R_e})$
$x = 2h$.

(A) The effective acceleration due to gravity at the equator is given by $g' = g - \omega^2 R$,where $g$ is the acceleration due to gravity at the poles (or without rotation),$\omega$ is the angular speed,and $R$ is the radius of the Earth.
The weight of a person at the equator is $W' = m g'$.
Given that the new weight is $\frac{3}{5}$ of the present weight,we have $m g' = \frac{3}{5} m g$.
Substituting the expression for $g'$:
$m(g - \omega^2 R) = \frac{3}{5} m g$
Divide both sides by $m$:
$g - \omega^2 R = \frac{3}{5} g$
Rearranging the terms to solve for $\omega^2 R$:
$\omega^2 R = g - \frac{3}{5} g = \frac{2}{5} g$
Solving for $\omega$:
$\omega^2 = \frac{2 g}{5 R}$
$\omega = \sqrt{\frac{2 g}{5 R}}$

(D) To ensure safety,the velocity attained upon reaching the ground must be the same on both the Earth and the planet.
Let $g_1 = 9.8 \, m/s^2$ be the acceleration due to gravity on Earth and $h_1 = 3 \, m$ be the safe height on Earth.
Let $g_2 = 1.96 \, m/s^2$ be the acceleration due to gravity on the planet and $h_2$ be the corresponding safe height on the planet.
The velocity $v$ attained after falling from height $h$ is given by $v = \sqrt{2gh}$.
Equating the velocities for safety: $\sqrt{2 g_1 h_1} = \sqrt{2 g_2 h_2}$.
Squaring both sides: $g_1 h_1 = g_2 h_2$.
Substituting the values: $9.8 \times 3 = 1.96 \times h_2$.
$h_2 = \frac{9.8 \times 3}{1.96} = 5 \times 3 = 15 \, m$.
Thus,the corresponding height on the planet is $15 \, m$.

(A) Let $M_e$ be the mass of the Earth and $R$ be the radius of the Earth.
$1$. For a point inside the Earth at a distance $r$ from the center $(r < R)$, the acceleration due to gravity is given by:
$g_{in} = \frac{G M_e r}{R^3}$
This implies $g \propto r$, which is a linear relationship passing through the origin.
$2$. For a point outside the Earth at a distance $r$ from the center $(r \geq R)$, the acceleration due to gravity is given by:
$g_{out} = \frac{G M_e}{r^2}$
This implies $g \propto \frac{1}{r^2}$, which is a hyperbolic decay curve.
At the surface $(r = R)$, the value of $g$ is maximum, $g_{max} = \frac{G M_e}{R^2}$.
Comparing this with the given options, the graph that shows a linear increase for $r < R$ and a $1/r^2$ decrease for $r > R$ is represented by the first graph (Graph $A$).

(C) 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 weightless,the effective gravity must be zero,so $g' = 0$.
Substituting the given values: $0 = g - R\omega^2 \cos^2 60^{\circ}$.
Since $\cos 60^{\circ} = 1/2$,we have $\cos^2 60^{\circ} = 1/4$.
Thus,$0 = g - R\omega^2 (1/4)$,which implies $R\omega^2 / 4 = g$.
Solving for $\omega$,we get $\omega^2 = 4g/R$,so $\omega = 2\sqrt{g/R}$.
The duration of the day $T$ is related to angular velocity $\omega$ by $T = 2\pi / \omega$.
Substituting $\omega$,we get $T = 2\pi / (2\sqrt{g/R}) = \pi \sqrt{R/g}$.

(D) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula $g' = g \left( 1 + \frac{h}{R_e} \right)^{-2}$.
The weight of the body at the surface is $W = mg = 18 \, N$.
The weight at height $h$ is $W' = mg' = \frac{mg}{(1 + h/R_e)^2}$.
Given $h = 3200 \, km$ and $R_e = 6400 \, km$,the ratio $\frac{h}{R_e} = \frac{3200}{6400} = \frac{1}{2} = 0.5$.
Substituting these values into the weight formula:
$W' = \frac{18}{(1 + 0.5)^2} = \frac{18}{(1.5)^2} = \frac{18}{2.25} = 8 \, N$.

(C) Statement $I$ is correct: The acceleration due to gravity $g$ at a height $h$ above the surface is given by $g_h = g(1 + h/R)^{-2}$,which decreases as $h$ increases. At a depth $d$ below the surface,it is given by $g_d = g(1 - d/R)$,which decreases as $d$ increases.
Statement $II$ is incorrect: For a height $h$ and depth $d$ where $h = d$,the values are $g_h = g(1 + h/R)^{-2}$ and $g_d = g(1 - h/R)$. Using binomial approximation for small $h$,$g_h \approx g(1 - 2h/R)$,while $g_d = g(1 - h/R)$. Since $(1 - 2h/R) \neq (1 - h/R)$,the values are not the same.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
As we move to Mount Everest (higher altitude),the value of $g$ decreases.
Since $T \propto \frac{1}{\sqrt{g}}$,a decrease in $g$ leads to an increase in the time period $T$.
An increase in the time period means the clock takes more time to complete one oscillation,which implies the clock becomes slow,not fast.
Therefore,Assertion $A$ is incorrect.
The value of $g$ at altitude $h$ is $g' = g(1 - \frac{2h}{R_e})$,which is indeed less than $g$ at the surface. Thus,Reason $R$ is correct.
Hence,$A$ is incorrect but $R$ is correct.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,where $g$ is the acceleration due to gravity at the surface.
At a height $h = R$ above the surface,the acceleration due to gravity $g'$ is given by $g' = \frac{g}{(1 + h/R)^2}$.
Substituting $h = R$,we get $g' = \frac{g}{(1 + R/R)^2} = \frac{g}{(2)^2} = \frac{g}{4}$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{\ell}{g'}} = 2\pi \sqrt{\frac{\ell}{g/4}} = 2 \times 2\pi \sqrt{\frac{\ell}{g}} = 2T$.
Given that the new time period is $xT$,we have $xT = 2T$,which implies $x = 2$.

(B) When a particle is at a distance $x$ from the center of the Earth, the gravitational force acting on it is directed towards the center and is given by $F = -\frac{GMm}{R^3}x$.
Since $F = ma$, the acceleration $a$ is given by $a = -\frac{GM}{R^3}x$.
Using the relation $g = \frac{GM}{R^2}$, we can write the acceleration as $a = -\frac{g}{R}x$.
Comparing this with the standard equation for simple harmonic motion, $a = -\omega^2 x$, we get $\omega^2 = \frac{g}{R}$, or $\omega = \sqrt{\frac{g}{R}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$.
Substituting the given values: $R = 6400 \, km = 6400 \times 10^3 \, m$ and $g = 10 \, m/s^2$.
$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} = 2 \times 3.14 \times \sqrt{640000} = 2 \times 3.14 \times 800 \, s$.
$T = 5024 \, s$.
Converting to minutes: $T = \frac{5024}{60} \approx 83.73 \, minutes$, which is approximately $1$ hour and $24$ minutes.

(D) The acceleration due to gravity at a depth $d$ is given by $g_d = g_0 \left(1 - \frac{d}{R}\right)$,where $g_0 = \frac{GM}{R^2}$.
The acceleration due to gravity at a height $h = 3R$ is given by $g_h = \frac{GM}{(R+h)^2} = \frac{GM}{(R+3R)^2} = \frac{GM}{(4R)^2} = \frac{GM}{16R^2} = \frac{g_0}{16}$.
According to the problem,$g_d = 4 \times g_h$.
Substituting the expressions: $g_0 \left(1 - \frac{d}{R}\right) = 4 \times \left(\frac{g_0}{16}\right)$.
$1 - \frac{d}{R} = \frac{1}{4}$.
$\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}$.
$d = \frac{3}{4} \times R = \frac{3}{4} \times 6400 \ km = 4800 \ km$.

(B) The weight of a body at the surface of the Earth is given by $W = mg = G \frac{Mm}{R^2}$,where $M$ is the mass of the Earth and $R$ is the radius of the Earth.
At a height $h$ above the surface,the weight $W'$ is given by $W' = mg' = G \frac{Mm}{(R+h)^2}$.
Given that $h = 9R$,we substitute this into the expression for $W'$:
$W' = G \frac{Mm}{(R + 9R)^2} = G \frac{Mm}{(10R)^2} = G \frac{Mm}{100R^2}$.
Since $W = G \frac{Mm}{R^2}$,we can write $W' = \frac{W}{100}$.
Therefore,the weight of the body at that height is $W/100$.

(C) Statement-$I$ is true because the Earth is not a perfect sphere (it is an oblate spheroid) and it rotates about its axis. The acceleration due to gravity $g$ varies with latitude $\phi$ as $g_{\phi} = g - \omega^2 R_e \cos^2 \phi$. Thus,it is different at different places.
Statement-$II$ is false because the acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g_d = g(1 - d/R_e)$. As depth $d$ increases,the term $(1 - d/R_e)$ decreases,meaning $g_d$ decreases as we go deeper into the Earth.

(D) The escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.
Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G \rho \frac{4}{3}\pi R^3}{R}} = \sqrt{\frac{8}{3}G\pi\rho} \cdot R$.
Thus,$V_e \propto \rho^{1/2} R$.
Given $\frac{V_{eA}}{V_{eB}} = \frac{1}{2}$ and $\frac{R_A}{R_B} = \frac{1}{3}$.
From $\frac{V_{eA}}{V_{eB}} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{R_A}{R_B}$,we get $\frac{1}{2} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{1}{3}$,which implies $\sqrt{\frac{\rho_A}{\rho_B}} = \frac{3}{2}$,so $\frac{\rho_A}{\rho_B} = \frac{9}{4}$.
The acceleration due to gravity is $g = \frac{GM}{R^2} = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}G\pi\rho R$.
Therefore,$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B} = \left(\frac{\rho_A}{\rho_B}\right) \left(\frac{R_A}{R_B}\right) = \left(\frac{9}{4}\right) \left(\frac{1}{3}\right) = \frac{3}{4}$.

(C) The mass of a planet is given by $M = \rho \times \frac{4}{3} \pi R^3$,where $\rho$ is the density and $R$ is the radius.
Since the density $\rho$ is constant,$M \propto R^3$,which implies $R \propto M^{1/3}$.
The weight of an object is $W = mg$,where the acceleration due to gravity $g = \frac{GM}{R^2}$.
Substituting $R \propto M^{1/3}$ into the expression for $g$,we get $g \propto \frac{M}{(M^{1/3})^2} = \frac{M}{M^{2/3}} = M^{1/3}$.
Since the weight $W \propto g$,we have $W \propto M^{1/3}$.
Given that the mass of the planet is double the mass of the earth $(M_p = 2M_e)$,the new weight $W'$ will be $W' = (2)^{1/3} W$.

(B) The weight of a body on the surface of the earth is given by $W = mg = 100\,N$.
The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given $h = \frac{R}{4}$,we substitute this into the formula:
$g' = g \left( \frac{R}{R + R/4} \right)^2 = g \left( \frac{R}{5R/4} \right)^2 = g \left( \frac{4}{5} \right)^2 = \frac{16}{25}g$.
The weight at height $h$ is $W' = mg' = m \left( \frac{16}{25}g \right) = \frac{16}{25} \times W$.
Substituting $W = 100\,N$:
$W' = \frac{16}{25} \times 100 = 16 \times 4 = 64\,N$.

(D) The weight of a body on the surface of the earth is given by $W = mg = 400\,N$.
At a depth $d$ below the surface of the earth,the acceleration due to gravity $g'$ is given by the formula $g' = g(1 - \frac{d}{R})$,where $R$ is the radius of the earth.
Given that the depth $d = \frac{R}{2}$,we substitute this into the formula:
$g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The new weight $W'$ of the body at this depth is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.
Since $mg = 400\,N$,we have $W' = \frac{400}{2} = 200\,N$.

(A) For a point at a height $h$ above the surface of the earth,the acceleration due to gravity is given by:
$g(h) = \frac{GM}{(R+h)^2}$
where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.
We can rewrite this expression as:
$g(h) = \frac{GM}{R^2(1 + \frac{h}{R})^2}$
$g(h) = \frac{GM}{R^2} (1 + \frac{h}{R})^{-2}$
Since $g = \frac{GM}{R^2}$ is the acceleration due to gravity at the surface of the earth,we have:
$g(h) = g (1 + \frac{h}{R})^{-2}$
Given the condition $h \ll R$,we can use the binomial approximation $(1 + x)^n \approx 1 + nx$ for $|x| \ll 1$:
$(1 + \frac{h}{R})^{-2} \approx 1 - \frac{2h}{R}$
Substituting this back into the equation:
$g(h) \approx g (1 - \frac{2h}{R})$
Thus,the acceleration due to gravity at height $h$ is $g^{\prime} = g(1 - \frac{2h}{R})$.

(D) The weight of a body on the surface of the Earth is given by $W = mg = 200 \, N$.
The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula $g' = g(1 - d/R)$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.
Given the depth $d = R/2$,we substitute this into the formula:
$g' = g(1 - (R/2)/R) = g(1 - 1/2) = g/2$.
The weight of the body at depth $d$ is $W' = mg' = m(g/2) = (mg)/2$.
Since $mg = 200 \, N$,we have $W' = 200/2 = 100 \, N$.

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2 \lambda$.
Statement $I$ is true because the rotation of the earth introduces a centrifugal force,which modifies the effective acceleration due to gravity.
Statement $II$ is false. At the poles,$\lambda = 90^{\circ}$,so $\cos 90^{\circ} = 0$,meaning the effect is zero (minimum). At the equator,$\lambda = 0^{\circ}$,so $\cos 0^{\circ} = 1$,meaning the effect is $R\omega^2$,which is the maximum reduction in $g$.
Therefore,Statement $I$ is true and Statement $II$ is false.

(C) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we can write:
$g = \frac{G}{R^2} \times \rho \times \frac{4}{3} \pi R^3 = \left( \frac{4}{3} \pi G \right) \rho R$.
Therefore,$g \propto \rho R$.
For planet $A$: $g_A \propto \rho \times R$.
For planet $B$: $g_B \propto \frac{\rho}{3} \times 4R = \frac{4}{3} \rho R$.
Taking the ratio:
$\frac{g_A}{g_B} = \frac{\rho R}{\frac{4}{3} \rho R} = \frac{1}{4/3} = \frac{3}{4}$.
Thus,the ratio $(g_A : g_B)$ is $3:4$.

(C) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (\frac{4}{3} \pi R^3)$,we have $g = \frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho = \frac{4}{3} \pi G \rho R$.
For planet $A$: $g_A = \frac{4}{3} \pi G \rho R$.
For planet $B$: $g_B = \frac{4}{3} \pi G (\frac{\rho}{2}) (1.5 R) = \frac{4}{3} \pi G \rho R \times (0.5 \times 1.5) = \frac{4}{3} \pi G \rho R \times 0.75$.
Therefore,the ratio $\frac{g_B}{g_A} = \frac{0.75}{1} = \frac{3}{4}$.

(D) The acceleration due to gravity on the surface of Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of Earth,and $R$ is the radius of Earth.
Since $M$ is constant,we have $g \propto \frac{1}{R^2}$.
Let the initial radius be $R_1 = R$ and the final radius be $R_2 = \frac{R}{2}$.
Then,the ratio of the new acceleration $g_2$ to the initial acceleration $g_1$ is given by:
$\frac{g_2}{g_1} = \frac{R_1^2}{R_2^2} = \frac{R^2}{(R/2)^2} = \frac{R^2}{R^2/4} = 4$.
Therefore,$g_2 = 4g_1 = 4g$.