Acceleration Due to Gravity and its Variation Questions in English

(D) Let $h$ be the distance above and below the surface of the earth where the weight of the body is the same. This implies the acceleration due to gravity at height $h$ $(g_h)$ must be equal to the acceleration due to gravity at depth $h$ $(g_d)$.
The formula for acceleration due to gravity at height $h$ is $g_h = g \left( 1 + \frac{h}{R} \right)^{-2} \approx g \left( 1 - \frac{2h}{R} \right)$ (for $h \ll R$). However,for a general solution,we use $g_h = \frac{g R^2}{(R+h)^2}$.
The formula for acceleration due to gravity at depth $h$ is $g_d = g \left( 1 - \frac{h}{R} \right)$.
Equating the two: $\frac{g R^2}{(R+h)^2} = g \left( 1 - \frac{h}{R} \right)$.
$\frac{1}{(1 + h/R)^2} = 1 - \frac{h}{R}$.
Let $x = \frac{h}{R}$. Then $\frac{1}{(1+x)^2} = 1 - x$.
$1 = (1-x)(1+x)^2 = (1-x)(1 + 2x + x^2) = 1 + 2x + x^2 - x - 2x^2 - x^3$.
$1 = 1 + x - x^2 - x^3$.
$x^3 + x^2 - x = 0$.
Since $x \neq 0$,we have $x^2 + x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $x$ must be positive,$x = \frac{\sqrt{5} - 1}{2}$.
Therefore,$h = \frac{\sqrt{5} - 1}{2} R = \frac{\sqrt{5} R - R}{2}$.

(A) Let $R_E$ be the radius of the Earth and $h$ be the height of the point above the surface. The distance from the center of the Earth is $r = R_E + h$.
Gravitational potential $V = -\frac{G M_E}{r} = -5.12 \times 10^7 \,J/kg$ ... $(i)$
Acceleration due to gravity $g' = \frac{G M_E}{r^2} = 6.4 \,m/s^2$ ... $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{V}{g'} = \frac{-G M_E / r}{G M_E / r^2} = -r$
$r = -\frac{V}{g'} = -\frac{-5.12 \times 10^7}{6.4} = 0.8 \times 10^7 \,m = 8000 \,km$
Since $r = R_E + h$, we have $h = r - R_E = 8000 \,km - 6400 \,km = 1600 \,km$.

(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given $h = R$,the acceleration due to gravity becomes $g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{1}{2} \right)^2 = \frac{g}{4}$.
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
Substituting $\ell = 4 \ m$ and $g' = \frac{g}{4}$,we get $T = 2\pi \sqrt{\frac{4}{g/4}} = 2\pi \sqrt{\frac{16}{g}}$.
Given $g = \pi^2 \ ms^{-2}$,we substitute this into the equation:
$T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \left( \frac{4}{\pi} \right) = 8 \ s$.

(D) The gravitational acceleration $g'$ at a height $h$ from the surface of the Earth is given by the formula: $g' = g \left(1 + \frac{h}{R}\right)^{-2}$.
Given $h = 2R$,we substitute this into the formula:
$g' = g \left(1 + \frac{2R}{R}\right)^{-2} = g(1 + 2)^{-2} = g(3)^{-2} = \frac{g}{9}$.
Given $g = 10 \,ms^{-2}$,the effective acceleration is $g' = \frac{10}{9} \,ms^{-2}$.
The gravitational force (weight) experienced by the body of mass $m = 90 \,kg$ is:
$F = m \times g' = 90 \times \frac{10}{9} = 100 \,N$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.
The acceleration due to gravity at a height $h$ is $g' = \frac{GM}{(R+h)^2}$.
Thus,$T = 2\pi \sqrt{\frac{\ell (R+h)^2}{GM}} = 2\pi \sqrt{\frac{\ell}{GM}} (R+h)$.
For height $h_1 = R$,$T_1 = 2\pi \sqrt{\frac{\ell}{GM}} (R+R) = 2\pi \sqrt{\frac{\ell}{GM}} (2R)$.
For height $h_2 = 2R$,$T_2 = 2\pi \sqrt{\frac{\ell}{GM}} (R+2R) = 2\pi \sqrt{\frac{\ell}{GM}} (3R)$.
Taking the ratio: $\frac{T_1}{T_2} = \frac{2R}{3R} = \frac{2}{3}$.
Therefore,$3T_1 = 2T_2$.

(D) At the surface of the earth, the weight of the body is given by $W_s = mg_s = 300 \,N$.
The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g_d = g_s \left(1 - \frac{d}{R}\right)$, where $g_s$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given the depth $d = R/4$, we substitute this into the formula:
$g_d = g_s \left(1 - \frac{R/4}{R}\right)$
$g_d = g_s \left(1 - \frac{1}{4}\right)$
$g_d = g_s \left(\frac{3}{4}\right)$
The weight of the body at depth $d$ is $W_d = mg_d$.
Substituting $g_d$ into the weight equation:
$W_d = m \times \left(\frac{3}{4} g_s\right)$
$W_d = \frac{3}{4} \times (mg_s)$
Since $mg_s = 300 \,N$, we have:
$W_d = \frac{3}{4} \times 300 \,N = 225 \,N$.

(C) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Let the mass and radius of the earth be $M$ and $R$ respectively. Then $g = \frac{GM}{R^2} = 9.8 \ m \ s^{-2}$.
For the given planet,the mass $M' = \frac{M}{10}$ and the radius $R' = \frac{R}{2}$ (since diameter is half,radius is also half).
The acceleration due to gravity on the planet is $g' = \frac{GM'}{R'^2}$.
Substituting the values: $g' = \frac{G(M/10)}{(R/2)^2} = \frac{GM/10}{R^2/4} = \frac{4}{10} \frac{GM}{R^2}$.
Since $\frac{GM}{R^2} = 9.8 \ m \ s^{-2}$,we have $g' = 0.4 \times 9.8 \ m \ s^{-2} = 3.92 \ m \ s^{-2}$.

(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g = \frac{GM}{r^2}$,where $r = R + h$.
Given the ratio of accelerations:
$\frac{g_p}{g_Q} = \frac{GM / r_p^2}{GM / r_Q^2} = \left( \frac{r_Q}{r_p} \right)^2$
Substituting the given values:
$\frac{36}{25} = \left( \frac{r_Q}{r_p} \right)^2$
Taking the square root on both sides:
$\frac{r_Q}{r_p} = \frac{6}{5} \implies r_Q = \frac{6}{5} r_p$
Since $r_p = R + h_p = R + R/3 = 4R/3$:
$r_Q = \frac{6}{5} \times \left( \frac{4R}{3} \right) = \frac{24R}{15} = \frac{8R}{5}$
Now,$r_Q = R + h_Q$,so:
$R + h_Q = \frac{8R}{5}$
$h_Q = \frac{8R}{5} - R = \frac{3R}{5}$

(B) Given: $R_p = \frac{R_e}{10}$,$\rho_p = \rho_e$,$\lambda = 10^{-3} \ kg \ m^{-1}$,$g_e = 10 \ m \ s^{-2}$,$R_e = 6 \times 10^6 \ m$.
Since density is the same,$M_p = \rho \cdot \frac{4}{3} \pi R_p^3 = \frac{M_e}{10^3}$.
The surface gravity of the planet is $g_p = \frac{G M_p}{R_p^2} = \frac{G (M_e / 10^3)}{(R_e / 10)^2} = \frac{G M_e}{10 R_e^2} = \frac{g_e}{10} = \frac{10}{10} = 1 \ m \ s^{-2}$.
The acceleration due to gravity at a distance $r$ from the center is $g(r) = g_p \frac{r}{R_p}$.
The wire extends from $r = R_p - \frac{R_p}{5} = \frac{4R_p}{5}$ to $r = R_p$.
The force $F$ required to hold the wire is the weight of the wire: $F = \int_{4R_p/5}^{R_p} \lambda \cdot g(r) \, dr = \int_{4R_p/5}^{R_p} \lambda \frac{g_p}{R_p} r \, dr$.
$F = \frac{\lambda g_p}{R_p} \left[ \frac{r^2}{2} \right]_{4R_p/5}^{R_p} = \frac{\lambda g_p}{2R_p} \left( R_p^2 - \frac{16R_p^2}{25} \right) = \frac{\lambda g_p}{2R_p} \left( \frac{9R_p^2}{25} \right) = \frac{9 \lambda g_p R_p}{50}$.
Substituting values: $R_p = \frac{6 \times 10^6}{10} = 6 \times 10^5 \ m$.
$F = \frac{9 \times 10^{-3} \times 1 \times 6 \times 10^5}{50} = \frac{5400}{50} = 108 \ N$.

(A) The acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$.
On the planet,the mass is $M' = 4M$ and the radius is $R' = 2R$.
The acceleration due to gravity on the planet is $g' = \frac{G(4M)}{(2R)^2} = \frac{4GM}{4R^2} = \frac{GM}{R^2} = g$.
Since the acceleration due to gravity is the same on both,the time period $T = 2\pi \sqrt{\frac{l}{g}}$ remains the same. Thus,Assertion $(A)$ is true.
The mass of the pendulum bob does not affect the time period of a simple pendulum. Thus,Reason $(R)$ is true.
However,the reason for the time period being the same is the equality of $g$ on both planets,not the mass of the pendulum. Therefore,$(R)$ is not the correct explanation of $(A)$.

(A) The acceleration due to gravity on the surface of the Earth is given by the formula:
$g = \frac{GM}{R_e^2}$
where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R_e$ is the radius of the Earth.
Given that the diameter is reduced to $1/3$ of its original value,the radius $R_e$ also reduces to $1/3$ of its original value,i.e.,$R' = R_e / 3$.
The mass $M$ remains unchanged.
The new acceleration due to gravity $g'$ is given by:
$g' = \frac{GM}{(R_e/3)^2} = \frac{GM}{R_e^2 / 9} = 9 \left( \frac{GM}{R_e^2} \right) = 9g$
Therefore,the new acceleration due to gravity is $9g$.

(B) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{\ell}{g}},$ where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity.
From this formula,it is clear that $T \propto \frac{1}{\sqrt{g}},$ which means the time period decreases as $g$ increases and increases as $g$ decreases.
At the top of a mountain,the height $h$ from the surface of the Earth is positive,and the acceleration due to gravity is given by $g_h = g_0 \left( \frac{R}{R+h} \right)^2.$
Since $g_h < g_0$ (where $g_0$ is the acceleration due to gravity at the base/surface),the value of $g$ is smaller at the top of the mountain.
Because $g$ is smaller at the top,the time period $T$ will be larger at the top of the mountain compared to the base.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A).$

(B) The weight of a body on the surface of the earth is $W = mg = 48 \ N$.
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 $h = \frac{R}{3}$,we substitute this into the formula:
$g' = g \left( \frac{R}{R + \frac{R}{3}} \right)^2 = g \left( \frac{R}{\frac{4R}{3}} \right)^2 = g \left( \frac{3}{4} \right)^2 = g \left( \frac{9}{16} \right)$.
The weight at height $h$ is $W' = mg' = mg \left( \frac{9}{16} \right)$.
Substituting $mg = 48 \ N$:
$W' = 48 \times \frac{9}{16} = 3 \times 9 = 27 \ N$.

(B) The effective acceleration due to gravity at the equator $(g_E)$ is given by $g_E = g_P - \omega^2 R$,where $g_P$ is the acceleration due to gravity at the poles,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.
Given that the weight at the equator is half the weight at the poles,we have $g_E = \frac{g_P}{2}$.
Substituting this into the equation: $\frac{g_P}{2} = g_P - \omega^2 R$.
Rearranging for $\omega$: $\omega^2 R = g_P - \frac{g_P}{2} = \frac{g_P}{2}$.
Thus,$\omega = \sqrt{\frac{g_P}{2R}}$.
Using $g_P = 9.8 \ m/s^2$ and $R = 6400 \ km = 6.4 \times 10^6 \ m$:
$\omega = \sqrt{\frac{9.8}{2 \times 6.4 \times 10^6}} = \sqrt{\frac{9.8}{12.8 \times 10^6}} = \sqrt{0.7656 \times 10^{-6}} \approx 8.75 \times 10^{-4} \ rad/s$.

(C) The variation of acceleration due to gravity $g$ with distance $r$ from the center of the Earth is given by:
$1$. For $r < R$ (inside the Earth): $g_{in} = \frac{GMr}{R^3}$. Thus, $g$ increases linearly with $r$. As $r$ increases from $0$ to $R$, $g$ increases. Therefore, statement $(b)$ is false.
$2$. At $r = 0$ (center of the Earth): $g = 0$. Thus, statement $(c)$ is true.
$3$. For $r > R$ (outside the Earth): $g_{out} = \frac{GM}{r^2}$. Thus, $g$ decreases as $r$ increases. Therefore, statement $(a)$ is true.
$4$. The effective acceleration due to gravity at the equator is $g' = g - \omega^2 R$. If the Earth stops rotating $(\omega = 0)$, then $g' = g$. Since $g - \omega^2 R < g$, the value of $g$ at the equator increases when the Earth stops rotating. Therefore, statement $(d)$ is false.
Conclusion: Statements $(a)$ and $(c)$ are true.

(B) The time of flight $t$ for a particle projected vertically upwards with initial velocity $u$ under gravity $g$ is given by the formula $t = \frac{2u}{g}$.
Since both particles are projected with the same initial velocity $u$,we have $u = \frac{g_1 t_1}{2}$ and $u = \frac{g_2 t_2}{2}$.
Equating the two expressions for $u$,we get $\frac{g_1 t_1}{2} = \frac{g_2 t_2}{2}$.
This simplifies to $g_1 t_1 = g_2 t_2$.

(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Since $G$ and $M$ are constant,$g \propto R^{-2}$. Taking the derivative,$\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta g}{g} = -2(-2 \%) = +4 \%$. Thus,$g$ increases by $4 \%$.
The rotational kinetic energy is $K = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L$ is the angular momentum. Assuming $L$ remains constant,$K \propto \frac{1}{I}$. Since $I = \frac{2}{5} MR^2$,we have $I \propto R^2$. Therefore,$K \propto R^{-2}$. Taking the derivative,$\frac{\Delta K}{K} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta K}{K} = -2(-2 \%) = +4 \%$. Thus,$K$ increases by $4 \%$. Therefore,both $g$ and $K$ increase by $4 \%$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
On the surface of the earth,$g = \frac{GM}{R^2}$,so $T = 2\pi \sqrt{\frac{l}{g}}$.
At a height $h = 2R$ above the surface,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2} = \frac{GM}{(R+2R)^2} = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} = \frac{g}{9}$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{g/9}} = 3 \times 2\pi \sqrt{\frac{l}{g}} = 3T$.
Comparing $T' = 3T$ with $T' = xT$,we get $x = 3$.

(B) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{r^2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\varrho$ and radius $r$ as $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \varrho$,we substitute this into the formula for $g$:
$g = \frac{G (\frac{4}{3} \pi r^3 \varrho)}{r^2} = \frac{4}{3} \pi G r \varrho$.
Therefore,$g \propto r \varrho$.
For planet $A$,$g_A \propto r_1 \varrho_1$.
For planet $B$,$g_B \propto r_2 \varrho_2$.
The ratio of acceleration due to gravity on $A$ to that of $B$ is $\frac{g_A}{g_B} = \frac{r_1 \varrho_1}{r_2 \varrho_2}$.

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by $g_d = g_s \left(1 - \frac{d}{R}\right)$,where $g_s$ is the acceleration due to gravity at the surface.
Given that $g_d = \frac{g_s}{n}$,we substitute this into the equation:
$\frac{g_s}{n} = g_s \left(1 - \frac{d}{R}\right)$
Dividing both sides by $g_s$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = \frac{R(n-1)}{n}$
Thus,the correct option is $A$.

(A) The weight of a body on the surface of the earth is given by $W = mg = \frac{GMm}{R^2} = 45 \ N$,where $R$ is the radius of the earth.
At a height $h = \frac{R}{2}$,the acceleration due to gravity $g'$ is given by the formula $g' = g \left( \frac{R}{R+h} \right)^2$.
Substituting $h = \frac{R}{2}$ into the formula:
$g' = g \left( \frac{R}{R + R/2} \right)^2 = g \left( \frac{R}{3R/2} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.
The weight at height $h$ is $W' = mg' = m \left( \frac{4}{9}g \right) = \frac{4}{9} W$.
Substituting $W = 45 \ N$:
$W' = \frac{4}{9} \times 45 = 4 \times 5 = 20 \ N$.

(A) The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula: $g_d = g(1 - \frac{d}{R})$.
Given that $g_d = \frac{g}{n}$,we substitute this into the equation:
$\frac{g}{n} = g(1 - \frac{d}{R})$.
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,$d = \frac{R(n-1)}{n}$.

(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.
Since $L$ is constant,$n \propto \sqrt{g}$.
On the surface of the Earth,$g_s = g$.
At a depth $d$,the acceleration due to gravity is given by $g_d = g_s \left(1 - \frac{d}{R}\right)$.
Given $d = R/3$,we have $g_d = g \left(1 - \frac{R/3}{R}\right) = g \left(1 - 1/3\right) = \frac{2}{3}g$.
The new frequency $n'$ is given by $n' = n \sqrt{\frac{g_d}{g_s}}$.
Substituting the values,$n' = n \sqrt{\frac{(2/3)g}{g}} = n \sqrt{2/3}$.

(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 velocity,and $R$ is the radius of the earth.
Given that the weight at the equator becomes $\frac{1}{6}$ of its present value,we assume the present weight is approximately $mg$ (since $\omega^2 R$ is very small compared to $g$).
Thus,$g' = \frac{1}{6} g$.
Substituting this into the equation: $\frac{1}{6} g = g - \omega^2 R$.
Rearranging for $\omega^2 R$: $\omega^2 R = g - \frac{1}{6} g = \frac{5}{6} g$.
Solving for $\omega$: $\omega = \sqrt{\frac{5g}{6R}}$.
Therefore,the required angular speed is $\sqrt{\frac{5}{6} \frac{g}{R}}$.

(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = \frac{g}{n}$,we can write: $\frac{g}{n} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$,we get: $\frac{1}{n} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides: $\frac{1}{\sqrt{n}} = \frac{R}{R+h}$.
Rearranging the equation to solve for $h$: $R+h = R \sqrt{n}$.
Therefore,$h = R \sqrt{n} - R = R(\sqrt{n}-1)$.

(B) The weight of a body at height $h$ is given by $W_h = W \left( \frac{R}{R+h} \right)^2$,where $W$ is the weight on the surface of the earth.
Given that $W_h = \frac{1}{16} W$,we have:
$\frac{1}{16} W = W \left( \frac{R}{R+h} \right)^2$
$\frac{1}{16} = \left( \frac{R}{R+h} \right)^2$
Taking the square root on both sides:
$\frac{1}{4} = \frac{R}{R+h}$
$R + h = 4R$
$h = 3R$
Therefore,the height at which the weight becomes $\frac{1}{16}$ of its surface weight is $3R$.

(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$.
Given that $g_h = \frac{g}{3}$,we substitute this into the equation:
$\frac{g}{3} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$:
$\frac{1}{3} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides:
$\frac{1}{\sqrt{3}} = \frac{R}{R+h}$.
Rearranging the terms to solve for $h$:
$R + h = \sqrt{3} R$.
$h = \sqrt{3} R - R$.
$h = (\sqrt{3} - 1) R$.

(B) The frequency of the pendulum at the surface is given by $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$.
At a depth $d$,the gravitational acceleration is given by $g_{eff} = g \left(1 - \frac{d}{R}\right)$.
For $d = \frac{R}{4}$,the effective acceleration is $g_{eff} = g \left(1 - \frac{R/4}{R}\right) = g \left(1 - \frac{1}{4}\right) = \frac{3}{4} g$.
The frequency at depth $d$ is $f_d = \frac{1}{2 \pi} \sqrt{\frac{g_{eff}}{l}} = \frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}$.
Taking the ratio of the new frequency to the original frequency $n$:
$\frac{f_d}{n} = \frac{\frac{1}{2 \pi} \sqrt{\frac{3g}{4l}}}{\frac{1}{2 \pi} \sqrt{\frac{g}{l}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Therefore,the new frequency is $f_d = \frac{\sqrt{3}}{2} n$.

(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.
At a height $h$,the acceleration due to gravity is $g_h = \frac{GM}{(R+h)^2}$.
Given that the value of $g$ is reduced by $64 \%$,the remaining value is $g_h = (100 \% - 64 \%) \text{ of } g = 36 \% \text{ of } g = 0.36g$.
Therefore,$\frac{g_h}{g} = 0.36 = \frac{36}{100}$.
Substituting the expressions for $g$ and $g_h$:
$\frac{R^2}{(R+h)^2} = \frac{36}{100}$.
Taking the square root on both sides:
$\frac{R}{R+h} = \frac{6}{10} = \frac{3}{5}$.
Cross-multiplying gives:
$5R = 3(R+h) = 3R + 3h$.
$2R = 3h$.
$h = \frac{2}{3} R$.

(B) The acceleration due to gravity at a depth $d$ is given by the formula: $g_d = g\left(1 - \frac{d}{R}\right)$.
Given that $g_d = \frac{g}{n-1}$,we substitute this into the equation:
$\frac{g}{n-1} = g\left(1 - \frac{d}{R}\right)$.
Dividing both sides by $g$:
$\frac{1}{n-1} = 1 - \frac{d}{R}$.
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n-1}$.
$\frac{d}{R} = \frac{n-1-1}{n-1} = \frac{n-2}{n-1}$.
Therefore,$d = R\left(\frac{n-2}{n-1}\right)$.

(B) The acceleration due to gravity $g$ at 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 write $g = \frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho = \frac{4}{3} \pi R \rho G$.
Given that $g$ is the same for both the earth and the planet,we have $R_e \rho_e = R_p \rho_p$.
Here,$R_e = R$ and $\rho_p = 3 \rho_e$.
Substituting these values: $R \times \rho_e = R_p \times (3 \rho_e)$.
Therefore,$R_p = \frac{R}{3}$.

(D) The weight of an object is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
At the surface of the earth $(h=0)$,the acceleration due to gravity $g$ is maximum.
As we move into a mine of depth $h_1$,the acceleration due to gravity is $g_1 = g(1 - h_1/R)$,so $W_1 = mg(1 - h_1/R) < W_2$.
As we move to the top of a mountain of height $h_3$,the acceleration due to gravity is $g_3 = g(1 - 2h_3/R)$,so $W_3 = mg(1 - 2h_3/R) < W_2$.
Since $W_2$ is the weight at the surface,it is greater than both $W_1$ and $W_3$.
Therefore,the correct relation is $W_1 < W_2 > W_3$.

(B) The acceleration due to gravity $g$ at the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$,where $\rho$ is the density,$G$ is the gravitational constant,and $R$ is the radius of the planet.
Given that the density of the planet $\rho_p = 2 \rho_e$ and the acceleration due to gravity $g_p = g_e$.
Equating the expressions for both the planet and the earth:
$\frac{4}{3} \pi \rho_p G R_p = \frac{4}{3} \pi \rho_e G R_e$
$\rho_p R_p = \rho_e R_e$
Substituting $\rho_p = 2 \rho_e$ and $R_e = R$:
$(2 \rho_e) R_p = \rho_e R$
$2 R_p = R$
$R_p = \frac{R}{2}$

(A) The effective acceleration due to gravity $g'$ at a latitude $\theta$ is given by $g' = g - R\omega^2 \cos^2 \theta$.
At the equator,the latitude $\theta = 0^\circ$,so $\cos 0^\circ = 1$. Thus,$g' = g - R\omega^2$.
Given that the weight becomes $\frac{3}{5}$ of its present value,the effective gravity $g'$ must be $\frac{3}{5}g$.
Substituting this into the equation: $\frac{3}{5}g = g - R\omega^2$.
Rearranging the terms: $R\omega^2 = g - \frac{3}{5}g = \frac{2}{5}g$.
Solving for $\omega$: $\omega^2 = \frac{2g}{5R}$,which gives $\omega = \sqrt{\frac{2g}{5R}}$.

(A) The weight of an object at the surface of the earth is $W = mg = 72 \ N$.
At a height $h$ above the surface,the acceleration due to gravity $g_h$ is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$.
Given that the height $h = \frac{R}{2}$,we substitute this into the formula:
$g_h = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9} g$.
The weight at height $h$ is $W_h = m g_h = m \left( \frac{4}{9} g \right) = \frac{4}{9} W$.
Substituting $W = 72 \ N$:
$W_h = \frac{4}{9} \times 72 = 4 \times 8 = 32 \ N$.

(C) The particle on the equator of the earth has linear speed $V = R \omega$,where $R$ is the radius of the earth and $\omega$ is the angular velocity of the earth's rotation.
At a latitude $\theta$,the particle moves in a circle of radius $r = R \cos \theta$.
The linear speed $V'$ of the particle at latitude $\theta$ is given by $V' = r \omega = (R \cos \theta) \omega$.
Substituting $\theta = 30^{\circ}$:
$V' = R \omega \cos 30^{\circ} = R \omega \left( \frac{\sqrt{3}}{2} \right)$.
Since $V = R \omega$,we have $V' = \frac{\sqrt{3}}{2} V$.

(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g_d = g(1 - \frac{d}{R})$.
Given that the depth $d = \frac{R}{2}$,we substitute this value into the equation:
$g_d = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The weight of the body at the surface is $W = mg = 300 \ N$.
The weight of the body at depth $d$ is $W_d = mg_d$.
Substituting $g_d = \frac{g}{2}$,we get $W_d = m(\frac{g}{2}) = \frac{1}{2} \times mg = \frac{1}{2} \times 300 \ N = 150 \ N$.

(B) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ of a sphere with density $\rho$ and radius $R$ is $M = \rho \cdot \frac{4}{3} \pi R^3$,we can substitute this into the formula:
$g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto R$ when the density $\rho$ is constant.
For the Earth,$g = \frac{4}{3} \pi G \rho R$.
For the planet,the radius is $R_p = 3R$ and the density is $\rho_p = \rho$.
Thus,the acceleration due to gravity on the planet is $g_n = \frac{4}{3} \pi G \rho (3R) = 3 \cdot (\frac{4}{3} \pi G \rho R) = 3g$.
Comparing $g_n = x \cdot g$ with $g_n = 3g$,we find that $x = 3$.

(D) The weight of a body at height $h$ is given by $W_h = m g_h$ and on the surface is $W = m g$.
Given that $W_h = \frac{W}{9}$,we have $m g_h = \frac{m g}{9}$,which implies $g_h = \frac{g}{9}$.
The acceleration due to gravity at height $h$ is $g_h = \frac{GM}{(R+h)^2}$ and on the surface is $g = \frac{GM}{R^2}$.
Substituting these into the equation $g_h = \frac{g}{9}$:
$\frac{GM}{(R+h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2}$
$\frac{1}{(R+h)^2} = \frac{1}{9 R^2}$
Taking the square root on both sides:
$\frac{1}{R+h} = \frac{1}{3 R}$
$R + h = 3 R$
$h = 2 R$
Therefore,the height is $2 R$.

(D) The gravitational acceleration at a depth $d$ below the Earth's surface is given by the formula: $g_d = g \left(1 - \frac{d}{R}\right)$.
Given that the acceleration due to gravity at depth $d$ is $g_d = \frac{g}{2n}$.
Substituting this into the formula:
$\frac{g}{2n} = g \left(1 - \frac{d}{R}\right)$.
Dividing both sides by $g$:
$\frac{1}{2n} = 1 - \frac{d}{R}$.
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{2n}$.
$\frac{d}{R} = \frac{2n - 1}{2n}$.
Therefore,$d = R \left(\frac{2n - 1}{2n}\right)$.

(C) The acceleration due to gravity at a depth $d$ below the earth's surface is given by the formula: $g_d = g(1 - \frac{d}{R})$.
Given that the depth $d = \frac{R}{3}$.
Substituting the value of $d$ into the formula:
$g_d = g(1 - \frac{R/3}{R})$
$g_d = g(1 - \frac{1}{3})$
$g_d = g(\frac{2}{3})$
Therefore,the acceleration due to gravity at that depth is $\frac{2g}{3}$.

(D) 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 $d$ and radius $R$ as $M = \frac{4}{3} \pi R^3 d$,we substitute this into the formula for $g$:
$g = \frac{G}{R^2} \left( \frac{4}{3} \pi R^3 d \right) = \frac{4}{3} \pi G R d$.
This shows that $g \propto R \cdot d$.
Given the ratios of radii $\frac{R_1}{R_2} = \frac{x}{y}$ and densities $\frac{d_1}{d_2} = \frac{m}{n}$,the ratio of the acceleration due to gravity is:
$\frac{g_1}{g_2} = \frac{R_1}{R_2} \times \frac{d_1}{d_2} = \frac{x}{y} \times \frac{m}{n} = \frac{xm}{yn}$.
Thus,the ratio is $xm : yn$ or $mx : ny$.

(C) The acceleration due to gravity at a depth $d$ is given by $g_d = g(1 - \frac{d}{R})$.
The acceleration due to gravity at an altitude $h$ (where $h \ll R$) is given by $g_h = g(1 - \frac{2h}{R})$.
Taking the ratio of $g_d$ to $g_h$:
$\frac{g_d}{g_h} = \frac{g(1 - \frac{d}{R})}{g(1 - \frac{2h}{R})}$
Simplifying the expression:
$\frac{g_d}{g_h} = \frac{\frac{R-d}{R}}{\frac{R-2h}{R}} = \frac{R-d}{R-2h}$.

(C) The acceleration due to gravity on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$.
Since the Earth is a sphere of radius $R$ and uniform density $\rho$,its mass $M$ can be expressed as $M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$.
Substituting the value of $M$ into the formula for $g$,we get:
$g = \frac{G}{R^2} \times (\frac{4}{3} \pi R^3 \rho)$.
Simplifying the expression,we get $g = \frac{4}{3} \pi \rho GR$.

(A) The effective acceleration due to gravity at a latitude $\phi$ is given by the formula: $g_{\phi} = g_{0} - R \omega^2 \cos^2 \phi$,where $g_{0}$ is the acceleration due to gravity at the poles (ignoring rotation).
At the equator,$\phi = 0^{\circ}$,so $g = g_{0} - R \omega^2 \cos^2 0^{\circ} = g_{0} - R \omega^2$.
At latitude $\phi = 30^{\circ}$,$g_{\phi} = g_{0} - R \omega^2 \cos^2 30^{\circ} = g_{0} - R \omega^2 \left(\frac{\sqrt{3}}{2}\right)^2 = g_{0} - \frac{3}{4} R \omega^2$.
Now,calculating the difference $|g - g_{\phi}|$:
$|g - g_{\phi}| = |(g_{0} - R \omega^2) - (g_{0} - \frac{3}{4} R \omega^2)|$
$|g - g_{\phi}| = |-\frac{1}{4} R \omega^2| = \frac{1}{4} R \omega^2$.

(B) The effective acceleration due to gravity at a latitude $\theta$ due to the rotation of the Earth is given by $g^{\prime} = g - R \omega^2 \cos^2 \theta$.
At the equator,the latitude $\theta = 0^{\circ}$,so $\cos 0^{\circ} = 1$.
Thus,the acceleration due to gravity at the equator is $g_e = g - R \omega^2$.
At the poles,the latitude $\theta = 90^{\circ}$,so $\cos 90^{\circ} = 0$.
Thus,the acceleration due to gravity at the poles is $g_p = g$.
The difference in the acceleration due to gravity between the pole and the equator is $g_p - g_e = g - (g - R \omega^2) = R \omega^2$.

(C) We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Since the mass of a planet is $M = \rho \left( \frac{4}{3} \pi R^3 \right)$,we can substitute this into the gravity formula:
$g = \frac{G}{R^2} \left( \rho \frac{4}{3} \pi R^3 \right) = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto \rho R$.
Given that $g_m = \frac{1}{6} g_e$ and the ratio of densities $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,which implies $\frac{\rho_m}{\rho_e} = \frac{3}{5}$.
Using the proportionality $g \propto \rho R$,we have:
$\frac{g_m}{g_e} = \frac{\rho_m R_m}{\rho_e R_e}$
$\frac{1}{6} = \left( \frac{3}{5} \right) \left( \frac{R_m}{R_e} \right)$
$\frac{R_m}{R_e} = \frac{1}{6} \times \frac{5}{3} = \frac{5}{18}$
Therefore,$R_m = \frac{5}{18} R_e$.

(A) Given: $\rho_A : \rho_B = 8 : 1$ and $M_A = M_B = M$.
Since density $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$,we have $\rho \propto \frac{1}{R^3}$.
Therefore,$\frac{\rho_A}{\rho_B} = \left(\frac{R_B}{R_A}\right)^3 = 8$.
Taking the cube root,$\frac{R_B}{R_A} = 2$,which means $\frac{R_A}{R_B} = \frac{1}{2}$.
The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Thus,the ratio $\frac{g_A}{g_B} = \frac{GM/R_A^2}{GM/R_B^2} = \left(\frac{R_B}{R_A}\right)^2$.
Substituting the value,$\frac{g_A}{g_B} = (2)^2 = 4$.
So,the ratio is $4:1$.