Acceleration Due to Gravity and its Variation Questions in English

(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.
Given: $\frac{W_{earth}}{W_{planet}} = \frac{9}{4} = \frac{g_{earth}}{g_{planet}}$.
We know that $g = \frac{GM}{R^2}$. Since density $\rho$ is constant,$M = \rho \cdot \frac{4}{3}\pi R^3$,so $g = \frac{G \cdot \rho \cdot \frac{4}{3}\pi R^3}{R^2} = G \cdot \rho \cdot \frac{4}{3}\pi R \propto R$.
However,the problem states the mass of the planet is $\frac{1}{9}$ of the Earth's mass. Let $M_p = \frac{M_e}{9}$.
Using $g = \frac{GM}{R^2}$,we have $\frac{g_e}{g_p} = \frac{M_e}{M_p} \cdot \frac{R_p^2}{R_e^2} = 9 \cdot \frac{R_p^2}{R^2}$.
Given $\frac{g_e}{g_p} = \frac{9}{4}$,so $\frac{9}{4} = 9 \cdot \frac{R_p^2}{R^2}$.
$\frac{1}{4} = \frac{R_p^2}{R^2} \implies \frac{R_p}{R} = \frac{1}{2}$.
Therefore,$R_p = \frac{R}{2}$.

(B) When a particle of mass $m$ is at a distance $r$ from the center of the Earth,the gravitational force acting on it is due only to the mass of the Earth contained within a sphere of radius $r$. Let this mass be $M'$.
The volume of this inner sphere is $V' = \frac{4}{3} \pi r^3$.
Since the density $\rho$ is uniform,the mass $M'$ is given by $M' = \rho V' = \rho \left( \frac{4}{3} \pi r^3 \right)$.
The gravitational force $F$ on the particle of mass $m$ at distance $r$ is given by Newton's law of gravitation:
$F = \frac{G M' m}{r^2}$
Substituting the value of $M'$:
$F = \frac{G m}{r^2} \left( \rho \cdot \frac{4}{3} \pi r^3 \right)$
Simplifying the expression:
$F = \left( \frac{4}{3} \pi G \rho \right) m r$

(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 acceleration due to gravity at a depth $d$ below the earth's surface is given by $g_d = g(1 - \frac{d}{R})$.
Given that $g_h = g_d$ for $d = 200 \ km$,we equate the two expressions:
$g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$
Canceling $g$ and $1$ from both sides:
$-\frac{2h}{R} = -\frac{d}{R}$
This simplifies to:
$2h = d$
$h = \frac{d}{2}$
Substituting $d = 200 \ km$:
$h = \frac{200}{2} = 100 \ km$.

(B) The variation of acceleration due to gravity $(g)$ is given by:
$1$. For $r > R$ (outside the Earth): $g = \frac{GM}{r^2}$. Thus,$g$ decreases as $r$ increases.
$2$. For $r < R$ (inside the Earth): $g = \frac{GMr}{R^3}$. Thus,$g$ decreases as we move towards the centre ($r$ decreases).
$3$. At the centre of the Earth $(r = 0)$: $g = 0$.
$4$. Effect of rotation: The effective gravity is $g' = g - \omega^2 R \cos^2 \lambda$. If the Earth stops rotating $(\omega = 0)$,then $g' = g$. Since $g$ was previously $g - \omega^2 R \cos^2 \lambda$,the value of $g$ increases (except at the poles where $\cos \lambda = 0$). Therefore,statement $(d)$ is false.
Thus,statements $(a), (b),$ and $(c)$ are true.

(B) The acceleration due to gravity $g$ varies with the distance $d$ from the centre of the Earth as follows:
$1$. Inside the Earth $(d < R)$:
The acceleration due to gravity at a distance $d$ from the centre is given by $g_d = \frac{GMd}{R^3} = g \frac{d}{R}$,where $g$ is the acceleration due to gravity at the surface. This shows that $g_d \propto d$,which is a linear relationship.
$2$. At the surface of the Earth $(d = R)$:
The acceleration due to gravity is maximum,$g = \frac{GM}{R^2}$.
$3$. Outside the Earth $(d > R)$:
The acceleration due to gravity at a distance $d$ from the centre is given by $g_d = \frac{GM}{d^2}$. This shows that $g_d \propto \frac{1}{d^2}$,which is an inverse square relationship.
Combining these,the graph starts from the origin $(d=0, g=0)$,increases linearly to a maximum at $d=R$,and then decreases following an inverse square curve for $d > R$. This corresponds to the graph shown in option $B$.

(A) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - \omega^2 R$,where $g$ 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 man's weight at the equator is $(3/5)$ of his actual weight,we have $W' = (3/5)W$.
Since $W = mg$,this implies $mg' = (3/5)mg$,which simplifies to $g' = (3/5)g$.
Substituting this into the equation for effective gravity: $(3/5)g = g - \omega^2 R$.
Rearranging the terms: $\omega^2 R = g - (3/5)g = (2/5)g$.
Solving for $\omega$: $\omega^2 = (2/5)(g/R)$,therefore $\omega = \sqrt{(2/5)(g/R)}$.

(C) The weight of a body is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.
Due to the rotation of the Earth and its shape (oblate spheroid),the radius of the Earth at the equator is greater than at the poles.
Since $g = \frac{GM}{R^2}$,the value of $g$ is minimum at the equator and maximum at the poles.
Therefore,when a body is moved from the pole to the equator,the value of $g$ decreases,which causes the weight of the body to decrease.

(C) 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 $M_m$ and $R_m$ be the mass and radius of the moon.
Given: $M_m = \frac{M_e}{8}$ and $g_m = \frac{g_e}{6}$.
Taking the ratio: $\frac{g_m}{g_e} = \frac{G M_m / R_m^2}{G M_e / R_e^2} = \frac{M_m}{M_e} \times \left( \frac{R_e}{R_m} \right)^2$.
Substituting the given values: $\frac{1}{6} = \frac{1}{8} \times \left( \frac{R_e}{R_m} \right)^2$.
Rearranging gives: $\left( \frac{R_e}{R_m} \right)^2 = \frac{8}{6}$.
Therefore,$R_e = \sqrt{\frac{8}{6}} R_m$.

(A) 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,$R$ is the radius of the earth,and $d$ is the depth.
Given that the body is halfway down to the center of the earth,the depth $d = \frac{R}{2}$.
Substituting this value into the formula,we get $g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.
The weight of the body at depth $d$ is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.
Since the weight at the surface is $W = mg = 250\,N$,the weight at half the depth is $W' = \frac{250}{2} = 125\,N$.

(B) For a point inside the earth $(r < R)$,the acceleration due to gravity is given by $g' = \frac{G M r}{R^3}$,which implies $g' \propto r$. Thus,the graph is a straight line passing through the origin.
For a point outside the earth $(r > R)$,the acceleration due to gravity is given by $g' = \frac{G M}{r^2}$,which implies $g' \propto \frac{1}{r^2}$. Thus,the graph is a curve showing an inverse square relationship.
At the surface of the earth $(r = R)$,the value of $g$ is maximum. Combining these two regions,the correct variation is shown in figure $(ii)$.

(D) The acceleration due to gravity $g$ on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times \frac{4}{3} \pi R^3$,we can write $g = \frac{G}{R^2} \times \rho \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \rho G R$.
This shows that $g \propto R$ when the density $\rho$ is constant.
Given that the new planet has the same density as Earth but is $3$ times bigger in size (radius),let $R^{\prime} = 3R$.
Therefore,$\frac{g^{\prime}}{g} = \frac{R^{\prime}}{R} = \frac{3R}{R} = 3$.
Thus,$g^{\prime} = 3g$.

(C) The acceleration due to gravity on the Earth is $g = \frac{GM_e}{R_e^2}$.
Given,the mass of the planet $M_p = \frac{1}{9} M_e$ and the radius of the planet $R_p = \frac{1}{2} R_e$.
The acceleration due to gravity on the planet is $g_p = \frac{GM_p}{R_p^2}$.
Substituting the values,$g_p = \frac{G(M_e/9)}{(R_e/2)^2} = \frac{G M_e / 9}{R_e^2 / 4} = \frac{4}{9} \frac{GM_e}{R_e^2} = \frac{4}{9} g$.
The weight of the body on the Earth is $W_e = mg = 9 \ N$.
The weight of the body on the planet is $W_p = mg_p = m(\frac{4}{9} g) = \frac{4}{9} (mg) = \frac{4}{9} \times 9 \ N = 4 \ N$.

(D) 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 the weight of the body at height $h$ is $\frac{1}{9}$ of its weight at the surface,we have $g' = \frac{g}{9}$.
Substituting this into the formula: $\frac{g}{9} = g \left(1 + \frac{h}{R}\right)^{-2}$.
This simplifies to: $\frac{1}{9} = \left(1 + \frac{h}{R}\right)^{-2}$.
Taking the square root of both sides: $\frac{1}{3} = \left(1 + \frac{h}{R}\right)^{-1}$.
Inverting both sides: $3 = 1 + \frac{h}{R}$.
Therefore,$\frac{h}{R} = 2$,which gives $h = 2R$.

(A) For a particle of mass $m$ at a distance $r$ $(r < R_e)$ from the center of the Earth,the gravitational force is due only to the mass $M'$ of the sphere of radius $r$.
The mass $M'$ of the sphere of radius $r$ is given by: $M' = \rho \cdot V = \left( \frac{M_e}{\frac{4}{3} \pi R_e^3} \right) \cdot \left( \frac{4}{3} \pi r^3 \right) = M_e \frac{r^3}{R_e^3}$.
The gravitational force $F$ on the particle is: $F = \frac{G M' m}{r^2}$.
Substituting the value of $M'$: $F = \frac{G (M_e \frac{r^3}{R_e^3}) m}{r^2} = \frac{G M_e m r^3}{R_e^3 r^2} = \frac{G M_e m}{R_e^3} r$.

(B) The acceleration due to gravity on the surface of the earth is $g = \frac{GM}{R^2}$.
For the planet,the mass $M_p = \frac{M}{7}$ and the radius $R_p = \frac{R}{2}$.
The acceleration due to gravity on the planet is $g_p = \frac{GM_p}{R_p^2} = \frac{G(M/7)}{(R/2)^2} = \frac{GM/7}{R^2/4} = \frac{4}{7} \frac{GM}{R^2} = \frac{4}{7}g$.
The weight of the body on the planet is $W_p = m \cdot g_p = m \cdot (\frac{4}{7}g) = \frac{4}{7} \cdot (mg)$.
Given that the weight on earth $W_e = mg = 700\,gm\,wt$.
Therefore,$W_p = \frac{4}{7} \times 700 = 400\,gm\,wt$.

(C) The effective acceleration due to gravity at the equator is given by $g' = g - \omega'^2 R$, where $\omega'$ is the new angular velocity.
For the weight of a body at the equator to be zero, the effective gravity must be zero, so $g - \omega'^2 R = 0$.
This implies $\omega' = \sqrt{g/R}$.
Given the present angular velocity $\omega = \sqrt{g/R_e}$ is not the standard way to relate, we use the condition $g = \omega_0^2 R$ where $\omega_0$ is the required angular velocity.
We know $g = 9.8 \ m/s^2$ and $R = 6.4 \times 10^6 \ m$.
Calculating $\omega_0 = \sqrt{9.8 / (6.4 \times 10^6)} \approx 1.24 \times 10^{-3} \ rad/s$.
The present angular velocity of the Earth is $\omega = 2\pi / T$, where $T = 86400 \ s$.
$\omega \approx 7.27 \times 10^{-5} \ rad/s$.
Ratio $\frac{\omega_0}{\omega} = \frac{1.24 \times 10^{-3}}{7.27 \times 10^{-5}} \approx 17$.
Thus, the required angular velocity is $17\omega$.

(C) The effective weight $R$ of a body of mass $m$ at a latitude $\lambda$ on the rotating earth is given by $R = mg - m\omega^2 R_e \cos^2 \lambda$,where $\omega$ is the angular velocity of the earth and $R_e$ is the radius of the earth.
At the equator,$\lambda = 0^\circ$,so $\cos \lambda = 1$. Thus,$R = mg - m\omega^2 R_e$.
At the poles,$\lambda = 90^\circ$,so $\cos \lambda = 0$. Thus,$R = mg$.
If the earth stops rotating,$\omega = 0$.
At the poles,the weight remains $R = mg$ (no change).
At the equator,the weight becomes $R = mg$,which is an increase compared to the rotating case $(mg - m\omega^2 R_e)$.
Therefore,the weight of a body will increase at the equator.

(B) 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 universal gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.
To find the mass $M$,we rearrange the formula:
$M = \frac{gR^2}{G}$
Thus,the correct formula is $g \frac{R^2}{G}$.

(A) Given: Radius of Earth $R_e = 6400 \; km$,Radius of Mars $R_m = 3200 \; km$.
Mass of Earth $M_e = 10 M_m$,Weight on Earth $W_e = 200 \; N$.
The weight of an object is given by $W = mg$,where $g = \frac{GM}{R^2}$.
Therefore,the ratio of weights is $\frac{W_m}{W_e} = \frac{m g_m}{m g_e} = \frac{M_m}{M_e} \times \left(\frac{R_e}{R_m}\right)^2$.
Substituting the values: $\frac{W_m}{200} = \frac{1}{10} \times \left(\frac{6400}{3200}\right)^2$.
$\frac{W_m}{200} = \frac{1}{10} \times (2)^2 = \frac{4}{10} = 0.4$.
$W_m = 200 \times 0.4 = 80 \; N$.

(A) Let the distance fallen be $h$. The time taken to fall through distance $h$ under gravity $g'$ is $T = \sqrt{\frac{2h}{g'}}$ and the final velocity is $V = \sqrt{2g'h}$.
For the first planet $(g_1 = 2g)$: $T_1 = \sqrt{\frac{2h}{2g}} = \sqrt{\frac{h}{g}}$ and $V_1 = \sqrt{2(2g)h} = 2\sqrt{gh}$.
For the second planet $(g_2 = 8g)$: $T_2 = \sqrt{\frac{2h}{8g}} = \frac{1}{2}\sqrt{\frac{h}{g}}$ and $V_2 = \sqrt{2(8g)h} = 4\sqrt{gh}$.
Given $t = T_1 - T_2 = \sqrt{\frac{h}{g}} - \frac{1}{2}\sqrt{\frac{h}{g}} = \frac{1}{2}\sqrt{\frac{h}{g}}$.
Given $v = V_2 - V_1 = 4\sqrt{gh} - 2\sqrt{gh} = 2\sqrt{gh}$.
From the time equation,$\sqrt{h} = 2t\sqrt{g}$.
Substituting this into the velocity equation: $v = 2\sqrt{g}(2t\sqrt{g}) = 4gt$.

(A) The acceleration due to gravity $g$ is given by $g = G \cdot \frac{4}{3} \pi R \rho$.
Given $\rho' = \frac{2}{3} \rho$ and $R' = \frac{1}{4} R$,the acceleration due to gravity on the moon $g'$ is:
$g' = G \cdot \frac{4}{3} \pi R' \rho' = G \cdot \frac{4}{3} \pi \left(\frac{R}{4}\right) \left(\frac{2}{3} \rho\right) = \frac{1}{6} g$.
$(I)$ Maximum height $h_{\max} = \frac{u^2}{2g}$. Since the initial velocity $u$ is the same,$h_{\max} \propto \frac{1}{g}$.
Therefore,$h' g' = h g \implies h' = h \left(\frac{g}{g'}\right) = 0.5 \times 6 = 3\, m$.
$(II)$ The time of flight $t = \frac{2u}{g}$. Since $u$ is constant,$t \propto \frac{1}{g}$.
Therefore,$\frac{t'}{t} = \frac{g}{g'} = \frac{g}{g/6} = 6$.
Thus,the ratio of time on the moon to the earth is $6 : 1$.

(C) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.
Let the two pans be at heights $h_1$ and $h_2$ such that $h_1 - h_2 = h$. The weights measured are $W_1 = mg_1$ and $W_2 = mg_2$.
The difference in weight (error) is $\Delta W = mg_1 - mg_2 = mg(1 - \frac{2h_1}{R}) - mg(1 - \frac{2h_2}{R}) = mg(\frac{2h_2}{R} - \frac{2h_1}{R}) = \frac{2mg}{R}(h_2 - h_1)$.
Since $h_1 - h_2 = h$,we have $\Delta W = \frac{2mgh}{R}$.
Substituting $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3 \rho$,we get $g = \frac{G}{R^2} \cdot \frac{4}{3}\pi R^3 \rho = \frac{4}{3}\pi G R \rho$.
Substituting this into the error formula: $\Delta W = \frac{2mh}{R} \cdot (\frac{4}{3}\pi G R \rho) = \frac{8}{3}\pi G \rho mh$.

(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.
Taking the natural logarithm on both sides,we get $\ln g = \ln G + \ln M - 2 \ln R$.
Differentiating both sides,we obtain the relative error formula: $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $1\%$,we have $\frac{\Delta R}{R} = -0.01$.
Substituting this value,we get $\frac{\Delta g}{g} = -2(-0.01) = 0.02$,which is $2\%$.
Since the result is positive,the acceleration due to gravity increases by $2\%$.

(A) For a uniform sphere of mass $M$ and radius $R$:
$1$. Inside the sphere $(r < R)$,the gravitational field intensity $E$ (or force per unit mass) is given by $E = \frac{GMr}{R^3}$. Thus,$E \propto r$.
Therefore,if $r_1 < R$ and $r_2 < R$,then $\frac{F_1}{F_2} = \frac{r_1}{r_2}$.
$2$. Outside the sphere $(r > R)$,the gravitational field intensity is given by $E = \frac{GM}{r^2}$. Thus,$E \propto \frac{1}{r^2}$.
Therefore,if $r_1 > R$ and $r_2 > R$,then $\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$.
Comparing these with the given options,option $A$ is correct.

(D) In a free fall,the apparent weight of a body becomes zero. This is because the body and the surface it is on (or the frame of reference) are both accelerating downwards at $g$. The normal force $N$ acting on the body is given by $N = m(g - a)$. Since $a = g$ in free fall,$N = m(g - g) = 0$. Thus,the body experiences weightlessness.
However,the acceleration due to gravity acting on a body in free fall is $g = 9.8 \; m/s^2$,which is definitely not zero. Therefore,the $Assertion$ is correct,but the $Reason$ is incorrect.

(C) Let the tennis ball bounce with an initial velocity $u$. It will rise to a height $h$ where its final velocity becomes $0$. Using the equation of motion $v^2 - u^2 = 2as$,we get:
$0^2 - u^2 = 2(-g')h$,where $g'$ is the acceleration due to gravity on the hill.
Therefore,$h = \frac{u^2}{2g'}$.
Since the acceleration due to gravity on the hill $(g')$ is less than that on the surface of the earth $(g)$ due to the increase in altitude,the height $h$ will be greater on the hill.
Thus,the $Assertion$ is correct,but the $Reason$ is incorrect because $g'$ is actually less than $g$.

(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2 \lambda$,where $g$ is the acceleration due to gravity if the earth were stationary,$R$ is the radius of the earth,and $\omega$ is the angular velocity of the earth.
The increase in the value of $g$ when the earth stops rotating is $\Delta g = g - g' = R\omega^2 \cos^2 \lambda$.
Given:
$R = 6400 \times 10^3 \ m = 6.4 \times 10^6 \ m$
$\lambda = 45^{\circ} \implies \cos^2 45^{\circ} = (1/\sqrt{2})^2 = 0.5$
$\omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{24 \times 3600} \approx 7.27 \times 10^{-5} \ rad/sec$
Substituting the values:
$\Delta g = (6.4 \times 10^6) \times (7.27 \times 10^{-5})^2 \times 0.5$
$\Delta g = (6.4 \times 10^6) \times (52.85 \times 10^{-10}) \times 0.5$
$\Delta g \approx 0.0169 \ m/sec^2$
Converting to $C.G.S.$ units $(cm/sec^2)$:
$1 \ m/sec^2 = 100 \ cm/sec^2$
$\Delta g = 0.0169 \times 100 = 1.69 \ cm/sec^2 \approx 1.68 \ cm/sec^2$.

(A) For a body on the equator to appear weightless,the effective gravity must be zero.
The condition for weightlessness due to the rotation of the Earth is given by $g - \omega^2 R = 0$.
Therefore,$\omega^2 R = g$,which implies $\omega = \sqrt{g / R}$.
Given $g = 10\, m/s^2$ and $R = 6.4 \times 10^3\, km = 6.4 \times 10^6\, m$.
Substituting the values: $\omega = \sqrt{\frac{10}{6.4 \times 10^6}}$.
$\omega = \sqrt{\frac{10}{6.4} \times 10^{-6}} = \sqrt{1.5625 \times 10^{-6}}$.
$\omega = 1.25 \times 10^{-3}\, rad/s$.

(D) The variation of acceleration due to gravity $g$ with altitude $h$ (where $h \ll R$) is given by: $g_h = g(1 - \frac{2h}{R})$.
The variation of acceleration due to gravity $g$ with depth $d$ (where $d \ll R$) is given by: $g_d = g(1 - \frac{d}{R})$.
According to the problem,the change in the value of $g$ at height $h$ is the same as the change at depth $d$. This implies that the values of $g$ at these points are equal,so $g_h = g_d$.
Equating the two expressions: $g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$.
Simplifying the equation: $1 - \frac{2h}{R} = 1 - \frac{d}{R}$.
Subtracting $1$ from both sides: $-\frac{2h}{R} = -\frac{d}{R}$.
Multiplying by $-R$: $d = 2h$.

(C) In a free fall,the effective weight of a body becomes zero. This is because the normal force acting on the object is zero,as the object and the surface are accelerating downwards at the same rate $g$. Thus,the object experiences weightlessness.
The acceleration due to gravity acting on an object in free fall is $g = 9.8 \ m/s^2$,which is non-zero. Therefore,the $Assertion$ is correct,but the $Reason$ is incorrect.

(D) 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 that the body is halfway down to the center,the depth $d = R/2$.
Substituting this value into the formula:
$g' = g(1 - (R/2)/R) = g(1 - 1/2) = g/2$.
Since weight $W = mg$,the new weight $W'$ at depth $d$ is $W' = mg' = m(g/2) = W/2$.
Given the weight at the surface $W = 200 \; N$,the weight at depth $d = R/2$ is $W' = 200/2 = 100 \; N$.

(D) The weight of the box at the North Pole is $W_p = mg = 196 \; N$. Given $g = 10 \; m/s^2$,the mass $m = 196 / 10 = 19.6 \; kg$.
At the equator,the effective acceleration due to gravity is $g' = g - \omega^2 R$,where $\omega$ is the angular velocity of the Earth.
The weight at the equator is $W_e = m(g - \omega^2 R) = mg - m\omega^2 R$.
The angular velocity $\omega = \frac{2\pi}{T}$,where $T = 24 \times 3600 \; s$.
Substituting the values: $W_e = 196 - 19.6 \times \left( \frac{2\pi}{24 \times 3600} \right)^2 \times 6400 \times 10^3$.
Calculating the term $m\omega^2 R = 19.6 \times (7.27 \times 10^{-5})^2 \times 6.4 \times 10^6 \approx 19.6 \times 0.0337 \approx 0.66 \; N$.
Therefore,$W_e = 196 - 0.66 = 195.34 \; N$.
The closest option is $195.32 \; N$.

(A) Decreases,$(b)$ Decreases,$(c)$ Mass of the body,$(d)$ More.
$(a)$ Acceleration due to gravity at height $h$ is $g_{h} = g(1 - 2h/R_{e})$. Thus,it decreases with increasing altitude.
$(b)$ Acceleration due to gravity at depth $d$ is $g_{d} = g(1 - d/R_{e})$. Thus,it decreases with increasing depth.
$(c)$ Since $g = GM/R^{2}$,the acceleration due to gravity is independent of the mass of the body $(m)$.
$(d)$ The potential energy difference is $\Delta U = -GmM(1/r_{2} - 1/r_{1})$. The formula $mg(r_{2}-r_{1})$ is only an approximation for small distances near the surface. Thus,the former is more accurate.

(C) Weight of the body on the surface of the earth,$W = mg = 63 \; N$.
The acceleration due to gravity at a height $h$ from the Earth's surface is given by the formula:
$g' = g \left( 1 + \frac{h}{R_e} \right)^{-2}$
Given that the height $h = \frac{R_e}{2}$,we substitute this into the formula:
$g' = g \left( 1 + \frac{R_e/2}{R_e} \right)^{-2} = g \left( 1 + \frac{1}{2} \right)^{-2} = g \left( \frac{3}{2} \right)^{-2} = g \left( \frac{2}{3} \right)^2 = \frac{4}{9} g$
The gravitational force (weight) at height $h$ is $W' = mg'$.
$W' = m \left( \frac{4}{9} g \right) = \frac{4}{9} (mg)$
Substituting the value of $W = 63 \; N$:
$W' = \frac{4}{9} \times 63 = 4 \times 7 = 28 \; N$.

(D) The weight of a body of mass $m$ at the Earth's surface is given by $W = mg = 250 \; N$.
The acceleration due to gravity at a depth $d$ below the Earth's surface is given by the formula $g' = g(1 - \frac{d}{R_e})$,where $R_e$ is the radius of the Earth.
Given that the body is halfway down to the center,the depth is $d = \frac{R_e}{2}$.
Substituting this value into the formula,we get:
$g' = g(1 - \frac{R_e/2}{R_e}) = g(1 - \frac{1}{2}) = \frac{1}{2}g$.
The weight of the body at depth $d$ is $W' = mg'$.
Substituting $g' = \frac{1}{2}g$,we get $W' = m(\frac{1}{2}g) = \frac{1}{2}mg = \frac{1}{2}W$.
Since $W = 250 \; N$,the new weight is $W' = \frac{1}{2} \times 250 = 125 \; N$.

(A) The centripetal force is given by $F_{c} = m\omega^{2}r$,where $r$ is the distance from the axis of rotation.
At the poles,the radius of rotation $r$ is effectively zero,but the question refers to the effective force experienced due to Earth's rotation.
However,considering the standard physics curriculum regarding the effective acceleration due to gravity $g' = g - \omega^{2}R \cos^{2}\phi$,the centripetal component is maximum at the equator where $\phi = 0^{\circ}$ and $\cos\phi = 1$.
Thus,the centripetal force is maximum at the equator.

(N/A) Consider the Earth to be a sphere of radius $R_E$ and mass $M_E$ with uniform density $\rho$.
Let a point mass $m$ be situated at a point $P$ at a distance $r$ from the centre,where $r = R_E - d$.
According to the shell theorem,for a point inside the Earth,the gravitational force is exerted only by the mass of the sphere of radius $r$,denoted by $M_r$. The shells outside this radius exert no net gravitational force on the mass $m$.
The mass of the inner sphere is $M_r = \rho \cdot \frac{4}{3} \pi r^3$.
Since the density is uniform,$\rho = \frac{M_E}{\frac{4}{3} \pi R_E^3}$.
Substituting $\rho$ into the expression for $M_r$,we get $M_r = M_E \left( \frac{r^3}{R_E^3} \right)$.
The gravitational force on mass $m$ at distance $r$ is $F = \frac{G M_r m}{r^2}$.
Substituting $M_r$,we get $F = \frac{G m}{r^2} \cdot M_E \frac{r^3}{R_E^3} = \frac{G M_E m r}{R_E^3}$.
The acceleration due to gravity $g(r)$ is given by $g(r) = \frac{F}{m} = \frac{G M_E r}{R_E^3}$.
Substituting $r = R_E - d$,we get $g(d) = \frac{G M_E (R_E - d)}{R_E^3} = \frac{G M_E}{R_E^2} \left( 1 - \frac{d}{R_E} \right)$.
Since $g = \frac{G M_E}{R_E^2}$ is the acceleration due to gravity at the surface,the expression becomes $g(d) = g \left( 1 - \frac{d}{R_E} \right)$.

(N/A) Imagine the Earth to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface.
$A$ point outside the Earth is outside all the shells. Thus,all the shells exert a gravitational force at the point outside just as if their masses were concentrated at their common centre.
For a point inside the Earth,the situation is different. Consider the Earth to be made up of concentric shells. $A$ point mass $m$ is situated at a distance $r$ $(r < R_E)$ from the centre. The point $P$ lies outside the sphere of radius $r$. For the shells of radius greater than $r$,the point $P$ lies inside. Hence,no net gravitational force is exerted by these outer shells on mass $m$ kept at $P$.
If the mass of the particle is $m$ at $P$,and the mass of the inner sphere of radius $r$ is $M_r$,then the force on the mass $m$ at $P$ has a magnitude:
$F = \frac{G m M_r}{r^2}$
Assuming the entire Earth has a uniform density $\rho$,its total mass is $M_E = (\frac{4}{3} \pi R_E^3) \rho$.
Therefore,$\rho = \frac{M_E}{\frac{4}{3} \pi R_E^3}$.
The mass of the inner sphere is $M_r = (\frac{4}{3} \pi r^3) \rho = (\frac{4}{3} \pi r^3) \frac{M_E}{\frac{4}{3} \pi R_E^3} = M_E \frac{r^3}{R_E^3}$.
Substituting $M_r$ into the force equation:
$F = \frac{G m}{r^2} (M_E \frac{r^3}{R_E^3}) = \frac{G M_E m r}{R_E^3}$.
Since $F = mg_r$,the acceleration due to gravity at distance $r$ is $g_r = \frac{G M_E r}{R_E^3}$.
On the surface of the Earth,$r = R_E$. Substituting this into the equation:
$g = \frac{G M_E R_E}{R_E^3} = \frac{G M_E}{R_E^2}$.

(N/A) Acceleration due to gravity is defined as the uniform acceleration produced in a body falling freely under the sole influence of the earth's gravitational force. It is denoted by the symbol $g$.
The magnitude of acceleration due to gravity on the surface of the earth is approximately $9.8 \ m/s^2$.

(A) According to the shell theorem for a spherical mass distribution,the gravitational force inside a uniform spherical shell is zero.
Since the Earth can be considered as a collection of concentric spherical shells,a particle at the centre of the Earth experiences an equal gravitational pull from all directions.
Therefore,the net gravitational force on a particle at the centre of the Earth is $0$.

(A) The gravitational force $F$ on a mass $m$ at a distance $r$ from the center of the Earth (where $r < R_e$) is given by $F = \frac{G M_r m}{r^2}$.
Assuming the Earth has a uniform density $\rho$,the mass $M_r$ contained within a sphere of radius $r$ is $M_r = \rho \cdot \frac{4}{3} \pi r^3$.
Substituting $M_r$ into the force equation: $F = \frac{G m}{r^2} \cdot \rho \cdot \frac{4}{3} \pi r^3$.
Simplifying this,we get $F = (\frac{4}{3} \pi G \rho m) r$.
Since $G, \pi, \rho, m$ are constants,it follows that $F \propto r$.

(21 KM) The Earth is not a perfect sphere; it is an oblate spheroid,meaning it bulges at the equator and is flattened at the poles.
The equatorial radius of the Earth is approximately $6378 \; km$.
The polar radius of the Earth is approximately $6357 \; km$.
The difference between the equatorial radius and the polar radius is $6378 \; km - 6357 \; km = 21 \; km$.

(N/A) $G$ is the Universal Gravitational Constant,while $g$ is the acceleration due to gravity.
$1$. $G$ is a constant value throughout the universe $(G = 6.67 \times 10^{-11} \ N \ m^2/kg^2)$,whereas $g$ varies from place to place on the Earth and other celestial bodies.
$2$. $G$ is a scalar quantity,while $g$ is a vector quantity.
$3$. The $SI$ unit of $G$ is $N \ m^2/kg^2$,whereas the $SI$ unit of $g$ is $m/s^2$.

(N/A) Consider a point mass $m$ at a height $h$ above the surface of the Earth as shown in the figure. This body is placed at point $P$ at a distance $r = R_{E} + h$ from the center of the Earth.
The magnitude of the gravitational force on the body is:
$F(h) = \frac{GM_{E}m}{(R_{E} + h)^{2}}$
Since $F = mg(h)$,the acceleration due to gravity at height $h$ is:
$g(h) = \frac{F(h)}{m} = \frac{GM_{E}}{(R_{E} + h)^{2}} \quad ......(1)$
Acceleration due to gravity on the surface of the Earth $(h = 0)$:
$g = \frac{GM_{E}}{R_{E}^{2}} \quad ......(2)$
Taking the ratio of equation $(1)$ and $(2)$:
$\frac{g(h)}{g} = \frac{GM_{E}}{(R_{E} + h)^{2}} \times \frac{R_{E}^{2}}{GM_{E}} = \frac{R_{E}^{2}}{(R_{E} + h)^{2}}$
$\frac{g(h)}{g} = \frac{R_{E}^{2}}{R_{E}^{2}(1 + \frac{h}{R_{E}})^{2}} = \left(1 + \frac{h}{R_{E}}\right)^{-2}$
$g(h) = g \left(1 + \frac{h}{R_{E}}\right)^{-2} \quad ......(3)$
For small heights $(h \ll R_{E})$,we can use the binomial expansion $(1 + x)^{n} \approx 1 + nx$:
$g(h) \approx g \left(1 - \frac{2h}{R_{E}}\right) \quad ......(4)$
Equation $(3)$ is valid for any height,while equation $(4)$ is an approximation valid only when $h \ll R_{E}$.

(N/A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula $g_{h} = g_{e} \left(1 - \frac{2h}{R_{e}}\right)$,where $h = 10 \, km = 10^{4} \, m$,$R_{e} = 6.4 \times 10^{6} \, m$,and $g_{e} = 9.8 \, m/s^2$.
Substituting the values:
$g_{h} = 9.8 \left[1 - \frac{2 \times 10^{4}}{6.4 \times 10^{6}}\right]$
$g_{h} = 9.8 [1 - 0.003125]$
$g_{h} = 9.8 [0.996875]$
$g_{h} = 9.769375 \, m/s^2$
Rounding to appropriate significant figures,we get $g_{h} \approx 9.77 \, m/s^2$.

(N/A) Consider a point mass $m$ at a depth $d$ below the surface of the Earth. Let $R_{E}$ be the radius of the Earth. Its distance from the center of the Earth is $r = (R_{E} - d)$.
The gravitational force on mass $m$ due to the outer shell of thickness $d$ is zero.
The gravitational force on the body is exerted only by the inner sphere of radius $(R_{E} - d)$.
Assuming the Earth has a uniform density $\rho$,the mass of the inner sphere $M_{s}$ is given by:
$M_{s} = \frac{4}{3} \pi (R_{E} - d)^{3} \rho$ $......(1)$
The total mass of the Earth $M_{E}$ is:
$M_{E} = \frac{4}{3} \pi R_{E}^{3} \rho$ $......(2)$
Dividing equation $(1)$ by $(2)$:
$\frac{M_{s}}{M_{E}} = \frac{(R_{E} - d)^{3}}{R_{E}^{3}}$
$M_{s} = M_{E} \frac{(R_{E} - d)^{3}}{R_{E}^{3}}$ $......(3)$
The gravitational force $F(d)$ exerted on the point mass $m$ at depth $d$ is:
$F(d) = \frac{G M_{s} m}{(R_{E} - d)^{2}}$ $......(4)$
Substituting the value of $M_{s}$ from equation $(3)$ into equation $(4)$:
$F(d) = \frac{G m}{(R_{E} - d)^{2}} \left[ M_{E} \frac{(R_{E} - d)^{3}}{R_{E}^{3}} \right]$
$F(d) = \frac{G M_{E} m (R_{E} - d)}{R_{E}^{3}}$
Since $F(d) = m g_{d}$,where $g_{d}$ is the acceleration due to gravity at depth $d$:
$m g_{d} = \frac{G M_{E} m (R_{E} - d)}{R_{E}^{3}}$
$g_{d} = \frac{G M_{E}}{R_{E}^{2}} \left( 1 - \frac{d}{R_{E}} \right)$
Since $g = \frac{G M_{E}}{R_{E}^{2}}$ is the acceleration due to gravity at the surface,we get:
$g_{d} = g \left( 1 - \frac{d}{R_{E}} \right)$

(N/A) $1$. Inside the Earth $(r < R_E)$: The acceleration due to gravity at a distance $r$ from the center is given by $g(r) = \frac{4}{3} \pi G \rho r$,where $\rho$ is the density of the Earth. Since $\frac{4}{3} \pi G \rho$ is constant,we have $g(r) \propto r$. This means $g$ increases linearly as we move from the center to the surface.
$2$. Outside the Earth $(r > R_E)$: The acceleration due to gravity at a distance $r$ from the center is given by $g(r) = \frac{GM}{r^2}$. Thus,$g(r) \propto \frac{1}{r^2}$. This means $g$ decreases as the inverse square of the distance from the center.
$3$. At the surface $(r = R_E)$: The value of $g$ is maximum,given by $g = \frac{GM}{R_E^2}$.
The graph shows $g(r)$ on the y-axis and $r$ on the x-axis,illustrating the linear increase inside the Earth and the inverse-square decrease outside the Earth.

(N/A) The angle made by the line joining a given place on the Earth's surface to the center of the Earth with the equatorial plane is called the latitude $(\lambda)$ of that place.
For the equator,latitude $\lambda = 0^{\circ}$ and for the poles,latitude $\lambda = 90^{\circ}$.
As shown in the figure,the latitude of the place $P$ on the Earth's surface is $\lambda = \angle POE$. Consider a particle of mass $m$ at this position. Two forces act on it:
$(1)$ The Earth's gravitational force $mg$ acting towards the center $O$ along the line $PO$.
$(2)$ Due to the Earth's rotation,the particle experiences a centrifugal force $m r \omega^2$ acting outwards along the radius of the circular path $PM$ (where $r$ is the radius of the circle of latitude).
From the geometry of the figure,$r = R_e \cos \lambda$,where $R_e$ is the radius of the Earth.
The component of the centrifugal force $m r \omega^2$ acting along the line $PO$ (towards the center) is $m r \omega^2 \cos \lambda$.
Substituting $r = R_e \cos \lambda$,the component is $m (R_e \cos \lambda) \omega^2 \cos \lambda = m R_e \omega^2 \cos^2 \lambda$.
The effective force $F$ on the particle towards the center is $F = mg - m R_e \omega^2 \cos^2 \lambda$.
If $g'$ is the effective gravitational acceleration,then $mg' = F = mg - m R_e \omega^2 \cos^2 \lambda$.
Therefore,the expression for effective gravitational acceleration is $g' = g - R_e \omega^2 \cos^2 \lambda$.

(N/A) The gravitational acceleration $g$ 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:
$g_h$ is the acceleration due to gravity at height $h$.
$g$ is the acceleration due to gravity at the Earth's surface $(g \approx 9.8 \ m/s^2)$.
$R_e$ is the radius of the Earth.
$h$ is the height above the Earth's surface.

(N/A) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by the formula:
$g_h = g \left( 1 + \frac{h}{R_e} \right)^{-2}$
Using the binomial expansion for $h << R_e$,we can approximate this as:
$g_h \approx g \left( 1 - \frac{2h}{R_e} \right)$
Where $g$ is the acceleration due to gravity at the surface of the Earth,$h$ is the height,and $R_e$ is the radius of the Earth.