Acceleration Due to Gravity and its Variation Questions in English

(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Let $M_e$ and $R_e$ be the mass and radius of Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.
Given: $M_e = 80 M_p$ and $D_e = 2 D_p$,which implies $R_e = 2 R_p$.
The ratio of gravity on the planet $(g_p)$ to Earth $(g_e)$ is:
$\frac{g_p}{g_e} = \frac{M_p}{M_e} \times \left( \frac{R_e}{R_p} \right)^2$
Substituting the given values:
$\frac{g_p}{9.8} = \left( \frac{1}{80} \right) \times (2)^2 = \frac{4}{80} = \frac{1}{20}$
$g_p = \frac{9.8}{20} = 0.49 \ m/s^2$.

(A) For an object on the equator to appear weightless,the effective acceleration due to gravity must be zero.
The effective gravity $g'$ at the equator is given by $g' = g - R\omega^2$.
For weightlessness,$g' = 0$,which implies $g = R\omega^2$.
Therefore,the angular speed $\omega$ is given by $\omega = \sqrt{\frac{g}{R}}$.
Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 \times 10^6\,m$.
$\omega = \sqrt{\frac{10}{6.4 \times 10^6}} = \sqrt{\frac{1}{0.64 \times 10^6}} = \sqrt{\frac{1}{64 \times 10^4}} = \frac{1}{8 \times 10^2} = \frac{1}{800} = 0.00125\,rad/s$.
Thus,$\omega = 1.25 \times 10^{-3}\,rad/s$.

(D) The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
We are given that $g' = \frac{g}{2}$.
Substituting this into the formula: $\frac{g}{2} = g \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides: $\frac{1}{\sqrt{2}} = \frac{R}{R+h}$.
Rearranging the terms: $R+h = \sqrt{2} R$.
Since $1.414 \approx \sqrt{2}$,we have $R+h = 1.414 R$.
Here,$R+h$ represents the distance from the centre of the earth. Thus,the required distance is $1.414 R$.

(B) The acceleration due to gravity $g$ on the surface of the Earth 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 substitute this into the formula for $g$:
$g = \frac{G (\rho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto \rho R$.
Let the initial density be $\rho_1$ and radius be $R_1$. Then $g_1 \propto \rho_1 R_1$.
Let the new density be $\rho_2 = 4\rho_1$ and the new radius be $R_2 = \frac{R_1}{2}$.
The new acceleration due to gravity $g_2$ is proportional to $\rho_2 R_2 = (4\rho_1) \times (\frac{R_1}{2}) = 2 \rho_1 R_1$.
Therefore,$g_2 = 2 g_1$.
Since weight $W = mg$,if $g$ doubles,the weight also doubles.

(C) The height $H$ reached by a man jumping with initial velocity $u$ is given by $H = \frac{u^2}{2g}$.
Since the initial velocity $u$ is constant,$H \propto \frac{1}{g}$,which implies $\frac{H_B}{H_A} = \frac{g_A}{g_B}$.
The acceleration due to gravity is given by $g = \frac{4}{3} \pi G \rho R$,so $g \propto \rho R$.
Given $\rho_B = \frac{1}{4} \rho_A$ and $R_B = \frac{1}{3} R_A$,we have $\frac{g_B}{g_A} = \frac{\rho_B R_B}{\rho_A R_A} = \left(\frac{1}{4}\right) \times \left(\frac{1}{3}\right) = \frac{1}{12}$.
Therefore,$\frac{H_B}{H_A} = \frac{g_A}{g_B} = 12$.
Thus,$H_B = 12 \times H_A = 12 \times 1.5 \, m = 18 \, m$.

(B) The weight of a body is given by the formula $W = m \times g$,where $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Since the mass $m$ is constant,the weight $W$ is directly proportional to the acceleration due to gravity $g$.
The value of $g$ on the surface of the Earth varies with latitude. It is given by $g = g_0 - \omega^2 R \cos^2 \phi$,where $\phi$ is the latitude.
At the equator,$\phi = 0^\circ$,so $\cos^2 \phi$ is maximum,making $g$ minimum.
At the poles,$\phi = 90^\circ$,so $\cos^2 \phi = 0$,making $g$ maximum.
Therefore,the weight of a body is maximum at the poles of the Earth.
Thus,the correct option is $B$.

(A) The acceleration due to gravity $g'$ at a height $h$ above the surface of the Earth is given by the formula:
$g' = G \frac{M}{(R + h)^2}$
Since the acceleration due to gravity on the surface of the Earth is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.
Substituting this into the expression for $g'$:
$g' = \frac{gR^2}{(R + h)^2}$
Dividing the numerator and denominator by $R^2$:
$g' = \frac{g}{(\frac{R+h}{R})^2} = \frac{g}{(1 + \frac{h}{R})^2}$
Thus,the correct option is $A$.

(C) 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 the planet can be expressed in terms of its density $d$ and volume $V$ as $M = d \times V = d \times \frac{4}{3}\pi R^3$,we substitute this into the formula for $g$.
$g = \frac{G(d \times \frac{4}{3}\pi R^3)}{R^2}$.
Simplifying the expression,we get $g = \frac{4}{3}\pi G d R$.
Since $\frac{4}{3}$,$\pi$,and $G$ are constants,it follows that $g \propto dR$.

(A) The acceleration due to gravity at a distance $r$ from the centre of the Earth (where $r < R$) is given by the formula $g' = \frac{4}{3} \pi \rho G r$,where $\rho$ is the density of the Earth and $G$ is the 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 centre.

(C) The weight of a body at the surface of the earth is $W = mg$,where $g$ is the acceleration due to gravity at the surface.
At a height $h$ above the surface,the acceleration due to gravity $g'$ is given by the formula:
$g' = g \left( \frac{R}{R + h} \right)^2$
Given that the height $h = R/2$,we substitute this 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 $W' = mg'$.
Substituting $g'$,we get $W' = m \left( \frac{4}{9}g \right) = \frac{4}{9}W$.

(A) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$.
Since mass $M = \text{Volume} \times \text{Density} = (\frac{4}{3}\pi R^3) \rho$,we can substitute this into the formula:
$g = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G R \rho$.
From this expression,we see that $g \propto \rho$ when the radius $R$ is kept constant.
If the density $\rho$ is doubled $(\rho' = 2\rho)$,the new acceleration due to gravity $g'$ will be:
$g' = 2 \times g = 2 \times 9.8\,m/s^2 = 19.6\,m/s^2$.

(D) The acceleration due to gravity $g$ at a point 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 distance from the center of the Earth.
Because the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius at the equator $(R_e)$ is greater than the radius at the poles $(R_p)$.
Since $g \propto \frac{1}{R^2}$,a smaller radius at the poles results in a higher value of acceleration due to gravity $(g_p > g_e)$.
Therefore,the correct option is $D$.

(D) The acceleration due to gravity $g'$ at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left(1 - \frac{d}{R}\right)$,where $g$ is the acceleration due to gravity at the surface,$d$ is the depth,and $R$ is the radius of the Earth.
At the center of the Earth,the depth $d$ is equal to the radius $R$ (i.e.,$d = R$).
Substituting $d = R$ into the formula: $g' = g \left(1 - \frac{R}{R}\right) = g(1 - 1) = g(0) = 0$.
Therefore,the value of acceleration due to gravity at the center of the Earth is $0 \ m/s^2$.

(A) The gravitational force on a mass $m$ at a distance $r$ from the center of the earth is given by $F = G \frac{Mm}{r^2}$.
Given the gravitational pull on $1 \,kg$ at the surface $(r = R)$ is $10 \,N$,we have $g = 10 \,m/s^2$.
The gravitational acceleration $g'$ at an orbital radius $r = 3R/2$ is given by $g' = g \left( \frac{R}{r} \right)^2$.
Substituting $r = 3R/2$,we get $g' = g \left( \frac{R}{3R/2} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.
The gravitational pull on the satellite of mass $m = 200 \,kg$ is $F' = m \times g' = 200 \times \frac{4}{9} \times 10$.
$F' = \frac{8000}{9} \approx 888.89 \,N$.
Rounding to the nearest integer,we get $889 \,N$.

(A) 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 have $g = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}\pi G \rho R$.
This implies $g \propto \rho R$,so $\frac{g_e}{g_m} = \frac{\rho_e}{\rho_m} \times \frac{R_e}{R_m}$.
Given $\frac{g_e}{g_m} = 6$ and $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,we substitute these values:
$6 = \frac{5}{3} \times \frac{R_e}{R_m}$.
Rearranging for $R_m$,we get $R_m = \frac{5}{3 \times 6} R_e = \frac{5}{18} R_e$.

(A) The acceleration due to gravity $g'$ at a height $h$ above the surface of the earth is given by the formula: $g' = g \left( \frac{R}{R + h} \right)^2$.
Here,the distance from the surface is $h = 2R$.
Substituting the value of $h$ into the formula:
$g' = g \left( \frac{R}{R + 2R} \right)^2$
$g' = g \left( \frac{R}{3R} \right)^2$
$g' = g \left( \frac{1}{3} \right)^2$
$g' = \frac{g}{9}$.

(B) The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula: $g' = g \left( 1 - \frac{d}{R} \right)$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the Earth.
Given that the effective value of acceleration due to gravity at depth $d$ is $g' = \frac{g}{4}$.
Substituting this into the formula: $\frac{g}{4} = g \left( 1 - \frac{d}{R} \right)$.
Dividing both sides by $g$: $\frac{1}{4} = 1 - \frac{d}{R}$.
Rearranging the terms: $\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,the depth is $d = \frac{3R}{4}$.

(B) The acceleration due to gravity at a height $h$ above the Earth's surface is given by $g_h = g(1 - \frac{2h}{R})$.
The fractional change in weight (or gravity) is $\frac{\Delta g}{g} = \frac{2h}{R}$.
Given $\frac{\Delta g}{g} \times 100\% = 1\%$,therefore $\frac{2h}{R} = 0.01$,which implies $\frac{h}{R} = 0.005$.
For a depth $d = h$ inside a mine,the acceleration due to gravity is $g_d = g(1 - \frac{d}{R})$.
The fractional change in weight at depth $d$ is $\frac{\Delta g}{g} = \frac{d}{R}$.
Since $d = h$,the fractional change is $\frac{h}{R} = 0.005$.
Converting to percentage: $0.005 \times 100\% = 0.5\%$.
Thus,the weight decreases by $0.5\%$.

(B) The acceleration due to gravity $g$ 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 the radius of the Earth.
Taking the natural logarithm on both sides: $\ln(g) = \ln(G) + \ln(M) - 2\ln(R)$.
Differentiating both sides to find the fractional change: $\frac{dg}{g} = \frac{dM}{M} - 2\frac{dR}{R}$.
Given that the mass $M$ decreases by $1\%$,so $\frac{dM}{M} = -0.01$.
Given that the radius $R$ decreases by $1\%$,so $\frac{dR}{R} = -0.01$.
Substituting these values into the equation: $\frac{dg}{g} = (-0.01) - 2(-0.01) = -0.01 + 0.02 = +0.01$.
This represents an increase of $1\%$ in the value of $g$.

(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{4}{3}\pi \rho GR$,where $\rho$ is the density,$G$ is the gravitational constant,and $R$ is the radius of the planet.
Given that $g_p = g_e$ and $\rho_p = 2\rho_e$,we have:
$\frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e} = 1$
Substituting the given values:
$1 = \frac{2\rho_e R_p}{\rho_e R_e}$
$1 = 2 \frac{R_p}{R_e}$
$R_p = \frac{R_e}{2} = \frac{R}{2}$
Thus,the radius of the planet is $\frac{R}{2}$.

(A) The acceleration due to gravity $g$ at 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 $\rho$ and radius $R$ as $M = \rho \times V = \rho \times \frac{4}{3}\pi R^3$,we can substitute this into the formula for $g$:
$g = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}G\pi \rho R$.
Therefore,the ratio of acceleration due to gravity for two planets is $\frac{g_1}{g_2} = \frac{\rho_1}{\rho_2} \times \frac{R_1}{R_2}$.
Given the ratios $\frac{R_1}{R_2} = \frac{2}{3}$ and $\frac{\rho_1}{\rho_2} = \frac{3}{2}$,we substitute these values:
$\frac{g_1}{g_2} = \frac{3}{2} \times \frac{2}{3} = 1$.
Thus,the ratio of acceleration due to gravity at the surface of the two planets is $1$.

(A) 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 poles,the value of $g$ is maximum $(g \approx 9.83 \ m/s^2)$,which results in the maximum weight for a given mass.
At the equator,$g$ is minimum $(g \approx 9.78 \ m/s^2)$,resulting in the minimum weight.
In a satellite,the object is in a state of weightlessness,so $g_{eff} = 0$,and the weight is $0$.
Therefore,a person will measure the highest weight $(kg-wt)$ at the poles.

(D) Let $R = 6400 \, km$ be the radius of the earth.
Acceleration due to gravity at height $h = 1600 \, km$ is given by $g_h = g \left( \frac{R}{R+h} \right)^2$.
Substituting $h = 1600 \, km$ and $R = 6400 \, km$,we get $h = R/4$.
$g_h = g \left( \frac{R}{R + R/4} \right)^2 = g \left( \frac{R}{5R/4} \right)^2 = g \left( \frac{4}{5} \right)^2 = \frac{16}{25} g$.
At depth $d$,the acceleration due to gravity is $g_d = g \left( 1 - \frac{d}{R} \right)$.
According to the problem,$g_d = \frac{1}{2} g_h$.
$g \left( 1 - \frac{d}{R} \right) = \frac{1}{2} \left( \frac{16}{25} g \right) = \frac{8}{25} g$.
$1 - \frac{d}{R} = \frac{8}{25} \Rightarrow \frac{d}{R} = 1 - \frac{8}{25} = \frac{17}{25}$.
$d = \frac{17}{25} \times 6400 \, km = 17 \times 256 \, km = 4352 \, km = 4.352 \times 10^6 \, m$.
Since this value is not among the options,the correct choice is $D$.

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.
For a body to appear weightless at the equator,the effective gravity $g'$ must be zero.
At the equator,the latitude $\lambda = 0^\circ$,so $\cos \lambda = 1$.
Substituting these values into the equation: $0 = g - \omega^2 R$.
Rearranging for $\omega$,we get $\omega = \sqrt{\frac{g}{R}}$.
Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 \times 10^6\,m$.
$\omega = \sqrt{\frac{10}{6.4 \times 10^6}} = \sqrt{\frac{1}{6.4 \times 10^5}} = \sqrt{\frac{1}{64 \times 10^4}} = \frac{1}{8 \times 10^2} = \frac{1}{800}\,rad/s$.

(B) The weight of a body on the surface of the earth is given by $W = mg = 500 \, N$.
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 $R$ is the radius of the earth.
Given that the body is halfway below the surface,the depth $d = \frac{R}{2}$.
Substituting this value into the formula:
$g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
Therefore,the weight at depth $d$ is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.
Substituting $mg = 500 \, N$:
$W' = \frac{500}{2} = 250 \, N$.

(A) The acceleration due to gravity $g$ 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 due to gravity on Earth and $g'$ be the acceleration due to gravity on the planet.
Given that $R' = 0.2 R$,we can write the ratio as:
$\frac{g'}{g} = \frac{R'}{R} = 0.2$.
Therefore,$g' = 0.2 g$.

(B) The gravitational force $F$ acting on a body of mass $m$ at the surface of the Earth is given by Newton's law of gravitation: $F = \frac{G M m}{R^2}$.
According to Newton's second law,the force is also $F = m g$,where $g$ is the acceleration due to gravity.
Equating the two expressions: $m g = \frac{G M m}{R^2}$.
Canceling $m$ from both sides,we get $g = \frac{G M}{R^2}$.
Since $G$ and $M$ are constants for the Earth,$g$ is independent of the mass of the body $m$ (which can be written as $m^0$).
Thus,$g$ is proportional to $m^0$.

(D) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Comparing the moon $(m)$ and the earth $(e)$:
$\frac{g_m}{g_e} = \frac{M_m}{M_e} \times \left( \frac{R_e}{R_m} \right)^2$.
Given $M_m = \frac{1}{9} M_e$ and $R_m = \frac{1}{2} R_e$,we have:
$\frac{g_m}{g_e} = \left( \frac{1}{9} \right) \times \left( \frac{R_e}{1/2 R_e} \right)^2 = \frac{1}{9} \times (2)^2 = \frac{4}{9}$.
Therefore,$g_m = \frac{4}{9} g_e$.
The weight of a body on the moon is $W_m = m \cdot g_m = m \cdot \left( \frac{4}{9} g_e \right) = \frac{4}{9} W_e$.
Given $W_e = 90 \ kgf$,we get $W_m = \frac{4}{9} \times 90 = 40 \ kgf$.

(A) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$.
At the equator,the latitude $\lambda = 0^\circ$,so $\cos 0^\circ = 1$.
Thus,the effective acceleration due to gravity at the equator is $g' = g - \omega^2 R$.
Here,$\omega$ is the angular velocity of the Earth's rotation and $R$ is the radius of the Earth.
If the spinning motion (angular velocity $\omega$) of the Earth increases,the term $\omega^2 R$ increases.
Since $g'$ is calculated by subtracting this term from the constant $g$,an increase in $\omega$ leads to a decrease in $g'$.
Since the weight of a body is $W = mg'$,a decrease in $g'$ results in a decrease in the weight of the body at the equator.

(B) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $M$ is the mass of the planet and $R$ is its radius.
Let the masses of the two planets be $M_1$ and $M_2$,and their radii be $R_1$ and $R_2$.
Given: $\frac{M_1}{M_2} = \frac{1}{2}$ and $\frac{R_1}{R_2} = \frac{1}{2}$.
The ratio of the acceleration due to gravity is $\frac{g_1}{g_2} = \frac{M_1}{M_2} \times (\frac{R_2}{R_1})^2$.
Substituting the given values: $\frac{g_1}{g_2} = \frac{1}{2} \times (\frac{2}{1})^2 = \frac{1}{2} \times 4 = \frac{2}{1}$.
Therefore,the ratio is $2 : 1$.

(B) The acceleration due to gravity at a latitude $\lambda$ is given by the formula:
$g' = g - R{\omega ^2}{\cos ^2}\lambda$
where $g$ is the acceleration due to gravity at the equator (ignoring the centrifugal effect) or the effective gravity at the equator where $\lambda = 0^\circ$.
At latitude $\lambda = 30^\circ$:
$g_{30} = g - R{\omega ^2}{\cos ^2}30^\circ$
Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we have $\cos^2 30^\circ = \frac{3}{4}$.
Substituting this into the equation:
$g_{30} = g - R{\omega ^2} \left( \frac{3}{4} \right)$
Therefore,the difference is:
$g - g_{30} = \frac{3}{4}{\omega ^2}R$.

(B) The acceleration due to gravity $g$ at the surface of the Earth is given by the formula:
$g = \frac{GM}{R^2}$
Where $G$ is the universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
To find the ratio of the gravitational acceleration $g$ to the gravitational constant $G$,we divide both sides by $G$:
$\frac{g}{G} = \frac{M}{R^2}$
Therefore,the correct ratio is $\frac{M}{R^2}$.

(D) For a uniform sphere of mass $M$ and radius $R$,the gravitational field $g$ at a distance $r$ from the centre is given by:
$1$. Inside the sphere $(r < R)$: $g = \frac{GMr}{R^3}$,which implies $g \propto r$.
$2$. Outside the sphere $(r > R)$: $g = \frac{GM}{r^2}$,which implies $g \propto \frac{1}{r^2}$.
Since the gravitational force $F = mg$,the ratio of forces $\frac{F_1}{F_2}$ is equal to the ratio of gravitational fields $\frac{g_1}{g_2}$.
If $r_1 < R$ and $r_2 < R$,then $\frac{F_1}{F_2} = \frac{r_1}{r_2}$. This matches option $(a)$.
If $r_1 > R$ and $r_2 > R$,then $\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$. This matches option $(b)$.
Therefore,both $(a)$ and $(b)$ are correct.

(C) 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 its radius.
Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.
Taking the logarithmic differentiation,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $1\%$,we have $\frac{\Delta R}{R} = -0.01$.
Substituting this value,$\frac{\Delta g}{g} = -2 \times (-0.01) = 0.02$.
Therefore,the acceleration due to gravity increases by $2\%$.

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - \omega^2 R \cos^2 \lambda$.
At the equator,the latitude $\lambda = 0^\circ$,so $\cos \lambda = 1$.
Thus,the effective acceleration becomes $g' = g - \omega^2 R$.
For the effective acceleration to be zero $(g' = 0)$,we have $0 = g - \omega^2 R$.
This implies $\omega^2 = \frac{g}{R}$.
Given $g = 10\,m/s^2$ and $R = 6400\,km = 6.4 \times 10^6\,m$.
$\omega = \sqrt{\frac{10}{6.4 \times 10^6}} = \sqrt{\frac{1}{6.4 \times 10^5}} = \sqrt{\frac{1}{64 \times 10^4}} = \frac{1}{8 \times 10^2} = \frac{1}{800}\,rad/s$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{l/g}$.
On the surface of the earth,the acceleration due to gravity is $g$. Thus,$T_1 = 2\pi \sqrt{l/g}$.
At a height $h = R$ above the earth's surface,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Substituting $h = R$,we get $g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{R}{2R} \right)^2 = \frac{g}{4}$.
The time period at height $R$ is $T_2 = 2\pi \sqrt{l/g'} = 2\pi \sqrt{l/(g/4)} = 2 \times 2\pi \sqrt{l/g}$.
Therefore,$T_2/T_1 = 2$.

(C) For a point inside the Earth $(r < R)$, the acceleration due to gravity is given by $g_{in} = \frac{4}{3} \pi G \rho r$, which means $g \propto r$. This represents a straight line passing through the origin.
For a point outside the Earth $(r \geq R)$, the acceleration due to gravity is given by $g_{out} = \frac{GM}{r^2}$, which means $g \propto \frac{1}{r^2}$. This represents a curve that decreases as $r$ increases.
Therefore, the graph shows a linear increase up to $r = R$ and then a non-linear decrease for $r > R$, which corresponds to the graph provided in option $C$.

(C) When a stone is dropped into a hole bored along the diameter of the earth,the gravitational force acting on the stone at a distance $r$ from the centre is given by $F = -(\frac{GMm}{R^3})r$.
Since the force is proportional to the displacement $r$ and directed towards the centre,the stone undergoes simple harmonic motion ($S$.$H$.$M$.) about the centre of the earth.
The time period of this oscillation is given by $T = 2\pi \sqrt{\frac{R}{g}}$,where $R$ is the radius of the earth and $g$ is the acceleration due to gravity at the surface.

(A) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
As we go inside a mine (below the Earth's surface),the effective acceleration due to gravity $g$ decreases.
Since $T \propto \frac{1}{\sqrt{g}}$,a decrease in the value of $g$ leads to an increase in the time period $T$.
Therefore,the period of oscillation inside a mine will be greater than $T$.

(B) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Given: $M_p = 2M_e$ and $R_p = 2R_e$.
Therefore,the ratio of gravity on earth to the planet is:
$\frac{g_e}{g_p} = \frac{M_e}{M_p} \times \left(\frac{R_p}{R_e}\right)^2 = \frac{1}{2} \times (2)^2 = \frac{4}{2} = 2$.
So,$g_p = \frac{g_e}{2}$.
The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.
For a second's pendulum on earth,$T_e = 2 \, \text{s}$.
Thus,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{2}$.
$T_p = T_e \times \sqrt{2} = 2\sqrt{2} \, \text{s}$.

(A) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$,where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.
As we move from the equator to the pole,the value of $g$ increases due to the Earth's shape (it is flattened at the poles).
Since $T \propto \frac{1}{\sqrt{g}}$,an increase in the value of $g$ results in a decrease in the time period $T$.
Therefore,the period of the simple pendulum decreases.

(B) The frequency of a simple pendulum is given by the formula $n = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$,which implies $n \propto \sqrt{g}$.
When a pendulum is taken deep into a mine at a depth $d$,the acceleration due to gravity $g'$ is given by $g' = g \left( 1 - \frac{d}{R} \right)$,where $R$ is the radius of the Earth.
Since $g' < g$,the value of acceleration due to gravity decreases as we go deep into the mine.
Because the frequency $n$ is directly proportional to the square root of $g$,a decrease in $g$ leads to a decrease in the frequency of the pendulum.

(A) 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 have $g = \frac{G}{R^2} \times \rho \times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \rho GR$.
Therefore,$g \propto \rho R$.
Taking the ratio for the Moon and Earth: $\frac{g_m}{g_e} = \frac{\rho_m}{\rho_e} \times \frac{R_m}{R_e}$.
Given $\frac{g_m}{g_e} = \frac{1}{6}$ and $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,we have $\frac{\rho_m}{\rho_e} = \frac{3}{5}$.
Substituting these values: $\frac{1}{6} = \frac{3}{5} \times \frac{R_m}{R_e}$.
Solving for $R_m$: $\frac{R_m}{R_e} = \frac{1}{6} \times \frac{5}{3} = \frac{5}{18}$.
Thus,$R_m = \frac{5}{18}R_e$.

(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 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 Moon: $g_m = \frac{GM_m}{R_m^2}$.
Given that $M_m = \frac{M_e}{80}$ and $R_m = \frac{R_e}{4}$.
Taking the ratio of the two accelerations:
$\frac{g_m}{g_e} = \frac{M_m}{M_e} \times \left( \frac{R_e}{R_m} \right)^2$
Substituting the given values:
$\frac{g_m}{g_e} = \left( \frac{1}{80} \right) \times \left( \frac{4}{1} \right)^2 = \frac{1}{80} \times 16 = \frac{16}{80} = \frac{1}{5}$.
Therefore,$g_m = \frac{g_e}{5} = \frac{g}{5}$.

(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Comparing the gravity on the planet $(g_p)$ and Earth $(g_e)$:
$\frac{g_p}{g_e} = \frac{M_p}{M_e} \times \left( \frac{R_e}{R_p} \right)^2$
Given $M_p = \frac{1}{10} M_e$ and $R_p = \frac{1}{3} R_e$:
$\frac{g_p}{g_e} = \frac{1}{10} \times \left( \frac{3}{1} \right)^2 = \frac{9}{10}$.
The maximum height reached is $H = \frac{u^2}{2g}$,which implies $H \propto \frac{1}{g}$.
Therefore,$\frac{H_p}{H_e} = \frac{g_e}{g_p} = \frac{10}{9}$.
$H_p = \frac{10}{9} \times H_e = \frac{10}{9} \times 90 = 100\,m$.

(D) The acceleration due to gravity $g$ 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 substitute this into the formula:
$g = \frac{G (\frac{4}{3} \pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G R \rho$.
Thus,$g \propto R \rho$.
Therefore,the ratio of the acceleration due to gravity for two planets is $\frac{g_1}{g_2} = \frac{R_1 \rho_1}{R_2 \rho_2}$,which can be written as $g_1 : g_2 = R_1 \rho_1 : R_2 \rho_2$.

(A) 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 distance from the surface is $h = 2R$.
Substituting the value of $h$ into the formula:
$g' = g \left( \frac{R}{R + 2R} \right)^2$
$g' = g \left( \frac{R}{3R} \right)^2$
$g' = g \left( \frac{1}{3} \right)^2$
$g' = \frac{g}{9}$.
Therefore,the acceleration due to gravity at a distance of $2R$ from the surface is $\frac{g}{9}$.

(B) The acceleration due to gravity at a height $h$ above the surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = 1\% \text{ of } g$,so $g' = \frac{g}{100}$.
Substituting this into the formula: $\frac{g}{100} = g \left( \frac{R}{R+h} \right)^2$.
Taking the square root on both sides: $\frac{1}{10} = \frac{R}{R+h}$.
Cross-multiplying gives: $R + h = 10R$.
Therefore,$h = 9R$.

(C) The weight of an object at height $h$ is given by $W' = W \left( \frac{R}{R+h} \right)^2$.
Given $W = 72 \, N$ and $h = R/2$.
Substituting the values:
$W' = 72 \left( \frac{R}{R + R/2} \right)^2$
$W' = 72 \left( \frac{R}{3R/2} \right)^2$
$W' = 72 \left( \frac{2}{3} \right)^2$
$W' = 72 \times \frac{4}{9}$
$W' = 8 \times 4 = 32 \, N$.

(B) The change in acceleration due to gravity at height $h$ is given by $\frac{\Delta g_h}{g} \approx \frac{2h}{R}$.
Given that the weight decreases by $1\%$,we have $\frac{2h}{R} \times 100\% = 1\%$,which implies $\frac{h}{R} = 0.005$.
At a depth $d = h$,the change in acceleration due to gravity is given by $\frac{\Delta g_d}{g} = \frac{d}{R} = \frac{h}{R}$.
Substituting the value of $\frac{h}{R}$,we get $\frac{\Delta g_d}{g} = 0.005$.
Therefore,the percentage decrease in weight at depth $h$ is $0.005 \times 100\% = 0.5\%$.