Acceleration Due to Gravity and its Variation Questions in English

(C) The weight of a body at the surface of the earth is $W = mg_0$,where $g_0 = \frac{GM}{R^2}$.
At a height $h = \frac{R}{2}$,the acceleration due to gravity $g'$ is given by:
$g' = \frac{GM}{(R+h)^2} = \frac{GM}{(R + R/2)^2} = \frac{GM}{(3R/2)^2} = \frac{GM}{9R^2/4} = \frac{4}{9} \left( \frac{GM}{R^2} \right) = \frac{4}{9} g_0$.
Therefore,the weight at height $h$ is $W' = mg' = m \left( \frac{4}{9} g_0 \right) = \frac{4}{9} W$.

(B) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2}$,where $G$ is the universal gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.
At a depth $h$ below the surface,the acceleration due to gravity $g'$ is given by the formula $g' = g(1 - \frac{h}{R})$.
According to the problem,$g' = \frac{g}{n}$.
Substituting this into the formula: $\frac{g}{n} = g(1 - \frac{h}{R})$.
Dividing both sides by $g$: $\frac{1}{n} = 1 - \frac{h}{R}$.
Rearranging the terms to solve for $h$: $\frac{h}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.
Therefore,$h = \frac{R(n-1)}{n}$.

(B) The acceleration due to gravity at a latitude $\lambda$ due to the rotation of the Earth is given by the formula:
$g^{\prime} = g - R_{E} \omega^2 \cos^2 \lambda$
At the equator,the latitude $\lambda = 0^{\circ}$. Since $\cos(0^{\circ}) = 1$,we have:
$g_E = g - R_{E} \omega^2 (1)^2 = g - R_{E} \omega^2$
At the poles,the latitude $\lambda = 90^{\circ}$. Since $\cos(90^{\circ}) = 0$,we have:
$g_P = g - R_{E} \omega^2 (0)^2 = g$
Now,calculating the difference $(g_P - g_E)$:
$g_P - g_E = g - (g - R_E \omega^2) = R_E \omega^2$

(B) The value of acceleration due to gravity decreases as we go below the surface of the earth. The weight $(W)$ of a body is defined as the product of its mass $(m)$ and the acceleration due to gravity $(g)$, i.e., $W = mg$.
At a depth $h$ below the surface of the earth, the acceleration due to gravity $g'$ is given by the formula:
$g' = g \left(1 - \frac{h}{R}\right)$
Multiplying both sides by mass $m$, we get the weight at depth $h$:
$W' = W \left(1 - \frac{h}{R}\right)$
Given that $W' = 250 \,N$, $W = 500 \,N$, and $R = 6400 \,km$:
$250 = 500 \left(1 - \frac{h}{6400}\right)$
$0.5 = 1 - \frac{h}{6400}$
$\frac{h}{6400} = 0.5$
$h = 0.5 \times 6400 = 3200 \,km$.

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

(A) The acceleration due to gravity at a depth $d$ below the surface of the Earth is given by the formula:
$g^{\prime} = g \left(1 - \frac{d}{R}\right)$
Given that the acceleration due to gravity at depth $d$ is $\frac{g}{n}$,we substitute this into the equation:
$\frac{g}{n} = g \left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = \frac{R(n-1)}{n}$

(A) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g' = g\left(1 - \frac{d}{R}\right)$.
According to the problem,the value of acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,so $g' = \frac{g}{n}$.
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 $d$: $\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n}$.
Therefore,$d = R\left(\frac{n-1}{n}\right)$.

(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 $g' = \frac{g}{3}$,we substitute this into the equation:
$\frac{g}{3} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by '$g$',we get:
$\frac{1}{3} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides:
$\frac{1}{\sqrt{3}} = \frac{R}{R+h}$.
Cross-multiplying gives:
$R + h = \sqrt{3} R$.
Rearranging for '$h$':
$h = \sqrt{3} R - R = (\sqrt{3} - 1) R$.

(C) The acceleration due to gravity at a distance $r$ from the center of the earth is given by $g' = \frac{GM}{r^2}$.
In the first case,the distance from the surface is $R$,so the distance from the center is $r_1 = R + R = 2R$. Given $g_1 = \frac{g}{4}$.
In the second case,the distance from the surface is $\frac{R}{2}$,so the distance from the center is $r_2 = R + \frac{R}{2} = \frac{3R}{2}$.
Using the ratio: $\frac{g_2}{g_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{2R}{3R/2}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$.
Therefore,$g_2 = \frac{16}{9} \times g_1 = \frac{16}{9} \times \frac{g}{4} = \frac{4g}{9}$.

(C) The acceleration due to gravity $g'$ 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 the height $h = R$,we substitute this into the formula:
$g' = g \left( \frac{R}{R+R} \right)^2$
$g' = g \left( \frac{R}{2R} \right)^2$
$g' = g \left( \frac{1}{2} \right)^2$
$g' = \frac{g}{4}$
Therefore,the gravitational acceleration at a height $R$ above the earth's surface is $\frac{g}{4}$.

(B) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula:
$g' = g(1 - \frac{d}{R})$
where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.
Given that $g' = 60\% \text{ of } g$,we have:
$g' = 0.6g$
Substituting this into the formula:
$0.6g = g(1 - \frac{d}{R})$
$0.6 = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - 0.6$
$\frac{d}{R} = 0.4$
$\frac{d}{R} = \frac{4}{10} = \frac{2}{5}$
Therefore,$d = \frac{2}{5}R$.

(D) The acceleration due to gravity $g$ at a latitude $\phi$ is given by the formula: $g_{\phi} = g - \omega^2 R \cos^2 \phi$,where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the earth,and $R$ is the radius of the earth.
At the equator,$\phi = 0^{\circ}$,so $\cos 0^{\circ} = 1$,which gives $g_{eq} = g - \omega^2 R$ (minimum value).
At the poles,$\phi = 90^{\circ}$,so $\cos 90^{\circ} = 0$,which gives $g_{pole} = g$ (maximum value).
Additionally,the earth is an oblate spheroid,meaning the radius at the equator is greater than the radius at the poles $(R_{eq} > R_{pole})$.
Since $g = \frac{GM}{R^2}$,a smaller radius at the poles results in a higher value of $g$.
Therefore,as we move from the equator to the poles,the value of acceleration due to gravity increases.

(A) The acceleration due to gravity at the surface of the earth is given by $g = \frac{GM}{R^2}$.
At a height '$h$' above the surface,the acceleration due to gravity '$g_h$' is given by $g_h = \frac{GM}{(R+h)^2}$.
Given that $g_h = \frac{g}{3}$,we substitute the expressions:
$\frac{g}{3} = \frac{GM}{(R+h)^2}$.
Substituting $g = \frac{GM}{R^2}$ into the equation:
$\frac{1}{3} \left( \frac{GM}{R^2} \right) = \frac{GM}{(R+h)^2}$.
$\frac{1}{3R^2} = \frac{1}{(R+h)^2}$.
Taking the square root on both sides:
$\frac{1}{\sqrt{3}R} = \frac{1}{R+h}$.
$R+h = \sqrt{3}R$.
$h = \sqrt{3}R - R$.
$h = R(\sqrt{3}-1)$.

(D) Given,the distance of the bottom of the hole from the surface of the Earth is $d = \frac{R_e}{2}$,where $R_e$ is the radius of the Earth.
On the surface of the Earth,the weight of the body is $W = mg = 300 \ N$.
The acceleration due to gravity at a depth $d$ below the surface is given by the formula $g' = g(1 - \frac{d}{R_e})$.
Substituting $d = \frac{R_e}{2}$ into the formula:
$g' = g(1 - \frac{R_e/2}{R_e}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The weight of the body at the bottom of the hole is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.
Substituting $mg = 300 \ N$:
$W' = \frac{300}{2} = 150 \ N$.

(D) When a body is suspended by a string at a position $P$ with latitude $\lambda$,the body rotates with the angular velocity $\omega$ of the Earth. The effective force acting on the body is the gravitational force minus the centripetal force required for circular motion.
The tension $T$ in the string is given by:
$T = mg - mr\omega^2 \cos \lambda$
Since $g = \frac{GM}{R^2}$ and the radius of the circular path at latitude $\lambda$ is $r = R \cos \lambda$,we substitute these into the equation:
$T = m \left( \frac{GM}{R^2} \right) - m(R \cos \lambda) \omega^2 \cos \lambda$
$T = \frac{GMm}{R^2} - mR\omega^2 \cos^2 \lambda$
At the equator,the latitude $\lambda = 0^{\circ}$.
Substituting $\lambda = 0^{\circ}$ and $\cos 0^{\circ} = 1$:
$T = \frac{GMm}{R^2} - mR\omega^2 (1)^2$
$T = \frac{GMm}{R^2} - mR\omega^2$

(B) The value of $g$ at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.
Thus,the change in $g$ is $\Delta g_h = g - g_h = g(\frac{2h}{R})$.
The value of $g$ at a depth $x$ below the surface of the Earth is given by $g_x = g(1 - \frac{x}{R})$.
Thus,the change in $g$ is $\Delta g_x = g - g_x = g(\frac{x}{R})$.
Given that the change in $g$ is the same at height $h$ and depth $x$,we equate the two expressions:
$g(\frac{2h}{R}) = g(\frac{x}{R})$.
Solving for $x$,we get $x = 2h$.

(C) 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 $\frac{1}{n}$ 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 $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = R\left(\frac{n-1}{n}\right)$

(A) The formula for gravitational acceleration at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = \frac{g}{4}$,we substitute this into the equation:
$\frac{g}{4} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$,we get $\frac{1}{4} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides,we obtain $\frac{1}{2} = \frac{R}{R+h}$.
Cross-multiplying gives $R + h = 2R$.
Therefore,$h = 2R - R = R$.

(A) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula:
$g^{\prime} = g \left( \frac{R}{R+h} \right)^{2}$
Given that the height $h = nR$,we substitute this into the equation:
$g^{\prime} = g \left( \frac{R}{R+nR} \right)^{2}$
$g^{\prime} = g \left( \frac{R}{R(1+n)} \right)^{2}$
$g^{\prime} = g \left( \frac{1}{1+n} \right)^{2}$
Now,we find the ratio of the acceleration due to gravity on the surface $(g)$ to that at the altitude $(g^{\prime})$:
$\frac{g}{g^{\prime}} = \frac{g}{g \left( \frac{1}{1+n} \right)^{2}}$
$\frac{g}{g^{\prime}} = (1+n)^{2}$
Therefore,the ratio is $(n+1)^{2}$.

(C) The mass of the Earth $M$ can be expressed in terms of its volume $V$ and mean density $\rho$ as:
$M = V \rho = \frac{4}{3} \pi R^3 \rho$ ... $(i)$
The acceleration due to gravity $g$ on the surface of the Earth is given by:
$g = \frac{GM}{R^2}$ ... $(ii)$
Substituting the expression for $M$ from equation $(i)$ into equation $(ii)$:
$g = \frac{G}{R^2} \times \left( \frac{4}{3} \pi R^3 \rho \right)$
$g = \frac{4}{3} \pi R G \rho$
Rearranging the equation to solve for the mean density $\rho$:
$\rho = \frac{3 g}{4 \pi R G}$

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$.
Substituting the mass of the Earth $M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3$,where $\rho$ is the average density and $R$ is the radius of the Earth:
$g = \frac{G}{R^2} \cdot \left( \frac{4}{3} \pi R^3 \rho \right) = \frac{4}{3} \pi G R \rho$.
Since $G$,$\pi$,and $R$ are constants for the Earth,we have $g \propto \rho$.
Therefore,the average density $\rho$ is directly proportional to the acceleration due to gravity $g$.

(A) The gravitational acceleration $g$ on the surface of a planet is given by the formula: $g = \frac{4}{3} \pi G R \rho$, where $G$ is the gravitational constant, $R$ is the radius, and $\rho$ is the density.
Since the density $\rho$ is the same for both the planet and the earth, we have $g \propto R$.
Let $g$ be the acceleration on earth and $g^{\prime}$ be the acceleration on the planet.
Given $R^{\prime} = 0.2 R$, we have:
$\frac{g^{\prime}}{g} = \frac{R^{\prime}}{R} = 0.2$
Therefore, $g^{\prime} = 0.2 g$.

(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula: $g = \frac{GM}{R^2}$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we can substitute this into the formula for $g$:
$g = \frac{G (\rho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi \rho G R$.
From this expression,it is clear that $g \propto \rho$ when the radius $R$ is kept constant.
Therefore,$\frac{g_2}{g_1} = \frac{\rho_2}{\rho_1}$.
Given that the density is doubled,$\rho_2 = 2\rho_1$.
Thus,$g_2 = 2 \times g_1 = 2 \times 9.8 \ m/s^2 = 19.6 \ m/s^2$.

(A) The acceleration due to gravity on Earth is given by $g = \frac{GM}{R^2}$.
For the planet, the mass is $M' = 4M$ and the radius is $R' = R$.
Therefore, the acceleration due to gravity on the planet is $g' = \frac{GM'}{R'^2} = \frac{G(4M)}{R^2} = 4g$.
Given $g = 10 \,ms^{-2}$, the acceleration due to gravity on the planet is $g' = 4 \times 10 = 40 \,ms^{-2}$.
The work done in lifting a body of mass $m = 5 \,kg$ through a height $h = 2 \,m$ is given by $W = m g' h$.
Substituting the values, $W = 5 \,kg \times 40 \,ms^{-2} \times 2 \,m = 400 \,J$.

(C) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - R\omega^2$.
For the weight of the body to be zero,the effective gravity must be zero,which means $g' = 0$.
Therefore,$g - R\omega^2 = 0$.
This implies $R\omega^2 = g$.
Solving for $\omega$:
$\omega^2 = \frac{g}{R}$
$\omega = \sqrt{\frac{g}{R}}$
Given $g = 10 \,ms^{-2}$ and $R = 6400 \,km = 6.4 \times 10^6 \,m$.
$\omega = \sqrt{\frac{10}{6.4 \times 10^6}}$
$\omega = \sqrt{\frac{10}{64 \times 10^5}} = \sqrt{\frac{1}{64 \times 10^4}}$
$\omega = \frac{1}{8 \times 10^2} = \frac{1}{800} \,rad/s$.

(B) The acceleration due to gravity $g$ is given by $g = \frac{GM}{R^2}$.
Let $g_m, M_m, R_m$ be the gravity,mass,and radius of the moon,and $g_e, M_e, R_e$ be those of the earth.
Given: $g_m = \frac{1}{6} g_e$ and $M_m = \frac{1}{8} M_e$.
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 given values: $\frac{1}{6} = \frac{1}{8} \times \left(\frac{R_e}{R_m}\right)^2$.
Rearranging gives: $\left(\frac{R_e}{R_m}\right)^2 = \frac{8}{6} = \frac{4}{3}$.
Taking the square root: $\frac{R_e}{R_m} = \sqrt{\frac{4}{3}}$.
Thus,the radius of the earth is $\sqrt{4/3}$ times the radius of the moon.

(B) 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 $\frac{3}{5}$ of its initial value,we have $g' = \frac{3}{5}g$.
Substituting this into the equation: $\frac{3}{5}g = g - \omega^2 R$.
Rearranging the terms: $\omega^2 R = g - \frac{3}{5}g = \frac{2}{5}g$.
Thus,$\omega = \sqrt{\frac{2g}{5R}}$.
Given $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 \times 10^6 \ m$.
$\omega = \sqrt{\frac{2 \times 10}{5 \times 6.4 \times 10^6}} = \sqrt{\frac{20}{32 \times 10^6}} = \sqrt{\frac{1}{1.6 \times 10^6}} = \sqrt{0.625 \times 10^{-6}} \approx 0.791 \times 10^{-3} \ rad/s = 7.91 \times 10^{-4} \ rad/s$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
At a height $h = R$ from the earth's surface,the new time period is $T' = xT$.
The acceleration due to gravity at height $h$ is $g_h = \frac{GM}{(R+h)^2}$.
Since $h = R$,$g_h = \frac{GM}{(R+R)^2} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{g}{4}$.
Now,the ratio of time periods is $\frac{T'}{T} = \frac{2\pi \sqrt{l/g_h}}{2\pi \sqrt{l/g}} = \sqrt{\frac{g}{g_h}}$.
Substituting $g_h = \frac{g}{4}$,we get $x = \sqrt{\frac{g}{g/4}} = \sqrt{4} = 2$.

(B) The frequency of a simple pendulum is given by $f = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At the surface of the earth,$f = F = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At a depth $d$ below the surface,the effective acceleration due to gravity is $g_d = g(1 - \frac{d}{R})$.
Given $d = \frac{R}{3}$,the effective gravity becomes $g_d = g(1 - \frac{R/3}{R}) = g(1 - \frac{1}{3}) = \frac{2g}{3}$.
The frequency at depth $d$ is $f_d = \frac{1}{2\pi} \sqrt{\frac{g_d}{l}} = \frac{1}{2\pi} \sqrt{\frac{2g}{3l}}$.
Comparing this with the frequency at the surface:
$f_d = \sqrt{\frac{2}{3}} \times \left( \frac{1}{2\pi} \sqrt{\frac{g}{l}} \right) = \sqrt{\frac{2}{3}} F$.
Since $\sqrt{\frac{2}{3}} = \sqrt{\frac{1}{1.5}} = \frac{1}{\sqrt{1.5}}$,the frequency is $\frac{F}{\sqrt{1.5}}$.

(D) Given that the acceleration due to gravity on planet $A$ is $9$ times that on planet $B$:
$g_{A} = 9g_{B}$ $(i)$
Using the third equation of motion, $v^2 = u^2 + 2gh$. Since the initial velocity $u$ is the same for the jump on both planets and the final velocity $v$ at the maximum height is $0$, we have $u^2 = 2gh$, or $h = \frac{u^2}{2g}$.
Since $u$ is constant for the same person,
$h \propto \frac{1}{g}$
Therefore, $\frac{h_{B}}{h_{A}} = \frac{g_{A}}{g_{B}}$
Substituting the given values:
$\frac{h_{B}}{2} = \frac{9g_{B}}{g_{B}} = 9$
$h_{B} = 9 \times 2 = 18 \,m$.

(A) The gravitational acceleration on the surface of a planet with mass $M$ and radius $R$ is given by $g = \frac{GM}{R^2}$.
For the given planet,the mass $M_p = 2M_e$ and the radius $R_p = 2R_e$.
Therefore,the gravitational acceleration on the planet $g_p$ is:
$g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} g_e$.
The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
For a second's pendulum on Earth,$T_e = 2 \ s$ with acceleration $g_e$.
On the planet,the new time period $T_p$ is:
$T_p = 2\pi \sqrt{\frac{l}{g_p}} = 2\pi \sqrt{\frac{l}{g_e/2}} = \sqrt{2} \times (2\pi \sqrt{\frac{l}{g_e}}) = \sqrt{2} \times T_e$.
Substituting $T_e = 2 \ s$,we get $T_p = 2\sqrt{2} \ s$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.
For a seconds pendulum on Earth,$T = 2 \text{ s}$.
The acceleration due to gravity on a planet is $g' = \frac{GM'}{R'^2}$.
Given $M' = 3M$ and $R' = 3R$,we have:
$\frac{g'}{g} = \frac{M'}{M} \cdot \frac{R^2}{R'^2} = 3 \cdot \frac{1}{3^2} = \frac{3}{9} = \frac{1}{3}$.
The time period on the planet is $T' = 2\pi \sqrt{\frac{L}{g'}}$.
Thus,$\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{3}$.
$T' = T \cdot \sqrt{3} = 2 \sqrt{3} \text{ seconds}$.

(A) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At the surface of the Earth,$n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$.
At a depth $d$ below the surface,the acceleration due to gravity $g'$ is given by $g' = g(1 - \frac{d}{R})$.
Given $d = \frac{R}{2}$,we have $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The new frequency $n'$ at depth $d$ is $n' = \frac{1}{2\pi} \sqrt{\frac{g'}{l}}$.
Substituting $g' = \frac{g}{2}$,we get $n' = \frac{1}{2\pi} \sqrt{\frac{g/2}{l}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2\pi} \sqrt{\frac{g}{l}} \right)$.
Therefore,$n' = \frac{n}{\sqrt{2}}$.

(D) The acceleration due to gravity $ g_d $ at a depth $ d $ below the Earth's surface is given by the formula:
$ g_d = g \left( 1 - \frac{d}{R} \right) $
Where $ g $ is the acceleration due to gravity at the surface $( 9.8 \,ms^{-2} )$,$ d $ is the depth $( 1600 \,km )$,and $ R $ is the radius of the Earth $( 6400 \,km )$.
Substituting the values:
$ g_d = 9.8 \left( 1 - \frac{1600}{6400} \right) $
$ g_d = 9.8 \left( 1 - \frac{1}{4} \right) $
$ g_d = 9.8 \times \frac{3}{4} $
$ g_d = 7.35 \,ms^{-2} $
Therefore,the acceleration due to gravity at a depth of $ 1600 \,km $ is $ 7.35 \,ms^{-2} $.

(A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g_h = \frac{g}{(1 + \frac{h}{R})^2}$.
Given that the height $h = \frac{R}{2}$,where $R$ is the radius of the Earth.
Substituting the value of $h$ into the formula:
$g_h = \frac{g}{(1 + \frac{R/2}{R})^2} = \frac{g}{(1 + \frac{1}{2})^2} = \frac{g}{(\frac{3}{2})^2}$.
$g_h = \frac{g}{9/4} = \frac{4}{9}g$.
Using $g = 9.8 \ m/s^2$:
$g_h = \frac{4}{9} \times 9.8 \approx 4.355 \ m/s^2 \approx 4.4 \ m/s^2$.

(D) The acceleration due to gravity $g$ at a distance $x$ from the centre of the Earth is given by:
$1$. Inside the Earth $(x < R)$: $g = \frac{GMx}{R^3}$,which implies $g \propto x$. This is a linear relationship.
$2$. Outside the Earth $(x \ge R)$: $g = \frac{GM}{x^2}$,which implies $g \propto \frac{1}{x^2}$. This is an inverse square relationship.
Thus,the graph starts from the origin $(0,0)$,increases linearly until $x = R$,and then decreases following an inverse square law for $x > R$.

(A) The acceleration due to gravity on the surface of 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 $g_e = \frac{GM_e}{R_e^2}$ be the acceleration due to gravity on Earth.
For the strange planet,we are given $g_p = 2g_e$.
Checking option $A$: If $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{2}$,then $g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{GM_e/2}{R_e^2/4} = 2 \frac{GM_e}{R_e^2} = 2g_e$.
Thus,option $A$ is correct.

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

(B) The acceleration due to gravity '$g$' at a distance '$r$' from the center of the Earth is given by:
$1$. Inside the Earth $(r < R)$: $g = \frac{GMr}{R^3}$,which implies $g \propto r$. This is a linear relationship.
$2$. Outside the Earth $(r \geq R)$: $g = \frac{GM}{r^2}$,which implies $g \propto \frac{1}{r^2}$. This is a non-linear,inverse-square decay.
Therefore,the graph shows a linear increase from the center $(r=0)$ to the surface $(r=R)$,and a non-linear decrease for distances greater than the radius of the Earth $(r > R)$. Graph $B$ correctly represents this behavior.

(B) The value of acceleration due to gravity at height $h = 10 \,km$ is given by $g_h = g(1 - \frac{2h}{R_e}) = x$,where $R_e$ is the radius of the Earth.
At a depth $d$ below the surface,the acceleration due to gravity is given by $g_d = g(1 - \frac{d}{R_e}) = x$.
Equating the two expressions for $x$:
$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$
$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$
$\frac{2h}{R_e} = \frac{d}{R_e}$
$d = 2h$.
Given $h = 10 \,km$,we get $d = 2 \times 10 \,km = 20 \,km$.

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
On the surface of the earth,$T = 2\pi \sqrt{\frac{l}{g}}$,where $g = \frac{GM}{R^2}$.
At a height $h = R/2$,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Substituting $h = R/2$,we get $g' = g \left( \frac{R}{R + R/2} \right)^2 = g \left( \frac{R}{3R/2} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.
The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{(4/9)g}} = \frac{3}{2} \left( 2\pi \sqrt{\frac{l}{g}} \right) = \frac{3}{2}T$.

(C) The acceleration due to gravity $g'$ at a height $h$ from the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R+h} \right)^2$,where $g$ is the acceleration due to gravity on the surface of the earth and $R$ is the radius of the earth.
Given: $h = (\sqrt{2}-1)R$ and $g = 10 \ m \ s^{-2}$.
Substituting the value of $h$ in the formula:
$g' = g \left( \frac{R}{R + (\sqrt{2}-1)R} \right)^2$
$g' = g \left( \frac{R}{R + \sqrt{2}R - R} \right)^2$
$g' = g \left( \frac{R}{\sqrt{2}R} \right)^2$
$g' = g \left( \frac{1}{\sqrt{2}} \right)^2$
$g' = g \times \frac{1}{2}$
$g' = \frac{10}{2} = 5 \ m \ s^{-2}$.

(A) Given: $g_1 = 2.5 \,ms^{-2}$ at height $h_1 = 6400 \,km$. Radius of earth $R = 6400 \,km$.
The acceleration due to gravity at height $h$ is given by $g = \frac{GM}{(R+h)^2}$.
This implies $g \propto \frac{1}{(R+h)^2}$.
For height $h_1 = 6400 \,km$, distance from center is $r_1 = R + h_1 = 6400 + 6400 = 12800 \,km = 2R$.
For height $h_2 = 12800 \,km$, distance from center is $r_2 = R + h_2 = 6400 + 12800 = 19200 \,km = 3R$.
Taking the ratio: $\frac{g_2}{g_1} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{2R}{3R} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.
Therefore, $g_2 = \frac{4}{9} \times g_1 = \frac{4}{9} \times 2.5 = \frac{10}{9} \approx 1.11 \,ms^{-2}$.

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula $g_h = g \left( \frac{R_E}{R_E + h} \right)^2$,where $g$ is the acceleration due to gravity at the surface and $R_E$ is the radius of the earth.
Given that $g_h = \frac{g}{4}$,we substitute this into the equation:
$\frac{g}{4} = g \left( \frac{R_E}{R_E + h} \right)^2$
$\frac{1}{4} = \left( \frac{R_E}{R_E + h} \right)^2$
Taking the square root of both sides:
$\frac{1}{2} = \frac{R_E}{R_E + h}$
$R_E + h = 2R_E$
$h = R_E$
Given $R_E = 6400 \ km$,therefore $h = 6400 \ km$.

(C) The acceleration due to gravity $g'$ at a height $h$ above the Earth's surface is given by the formula:
$g' = g \left( \frac{R_E}{R_E + h} \right)^2$
Given that the height $h = 4 R_E$ and the acceleration due to gravity on the surface $g = 10 \,ms^{-2}$.
Substituting the values into the formula:
$g' = 10 \left( \frac{R_E}{R_E + 4 R_E} \right)^2$
$g' = 10 \left( \frac{R_E}{5 R_E} \right)^2$
$g' = 10 \left( \frac{1}{5} \right)^2$
$g' = 10 \times \frac{1}{25} = \frac{10}{25} = 0.4 \,ms^{-2}$

(A) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
Since the Earth is an oblate spheroid,its radius $R$ is minimum at the poles and maximum at the equator. Because $g \propto \frac{1}{R^2}$,$g$ is greater at the poles.
As we move to a height $h$ above the surface,the acceleration due to gravity $g^{\prime}$ is given by $g^{\prime} = \frac{g}{(1 + h/R)^2}$. As $h$ increases,$g^{\prime}$ decreases.
At the center of the Earth,the effective acceleration due to gravity is zero.
Therefore,statements $A$ and $B$ are correct.

(C) The gravitational acceleration $g$ at a distance $r$ from the centre of a uniform solid sphere of mass $M$ and radius $R$ (where $r \ge R$) is given by $g = \frac{GM}{r^2}$.
On the surface,$r = R$,so $a_o = \frac{GM}{R^2}$.
We want to find the distance $r$ where the acceleration becomes $a = \frac{a_o}{4}$.
Substituting the expressions,we get $\frac{GM}{r^2} = \frac{1}{4} \left( \frac{GM}{R^2} \right)$.
Canceling $GM$ from both sides,we get $\frac{1}{r^2} = \frac{1}{4R^2}$.
Taking the square root of both sides,$r^2 = 4R^2$,which gives $r = 2R$.
Thus,the distance from the centre of the sphere is $2R$.

(C) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since density $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$,we can express the radius $R$ as $R = \left( \frac{3M}{4\pi \rho} \right)^{1/3}$.
Substituting $R$ into the formula for $g$:
$g = \frac{GM}{(\frac{3M}{4\pi \rho})^{2/3}} = G M^{1/3} (\frac{4\pi \rho}{3})^{2/3}$.
This shows that $g \propto M^{1/3} \rho^{2/3}$.
Given for the planet: $M^{\prime} = 8M$ and $\rho^{\prime} = 27\rho$.
Taking the ratio:
$\frac{g^{\prime}}{g} = \left( \frac{M^{\prime}}{M} \right)^{1/3} \left( \frac{\rho^{\prime}}{\rho} \right)^{2/3}$.
$\frac{g^{\prime}}{g} = (8)^{1/3} \times (27)^{2/3} = 2 \times (3^3)^{2/3} = 2 \times 3^2 = 2 \times 9 = 18$.
Therefore,$g^{\prime} = 18g$.