Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : If we move form poles to equator, the direction of acceleration due to gravity of earth always points towards the centre of earth without any variation in its magnitude.

Reason R : At equator, the direction of acceleration due to the gravity is towards the centre of earth.

In the light of above statements, chose the correct answer from the options given below

$g=\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}$  $\frac{}{{\left({}^{}{}^{}\right)}^{}}$

$\Rightarrow \mathrm{}{\int }_v^0?dV=-{\int }_x^{\mathrm{\infty }}?gdx$

$\Rightarrow \mathrm{}O-V=-\left[\int \frac{Ax}{{\left(a^2+x^2\right)}^{3/2}}\right]$

Let, $a^2+x^2=t^2$ ${}^{}{}^{}{}^{}$

$\Rightarrow \mathrm{}2xdx=2tdt$

$\Rightarrow \mathrm{}\mathrm{x}\mathrm{d}x=\mathrm{t}\mathrm{d}\mathrm{t}$

$\Rightarrow \mathrm{}V=\int \frac{Atdt}{t^3}\Rightarrow -\frac{A}{t}\Rightarrow -{\left.\frac{A}{\sqrt[]{a^2+x^2}}\right|}_x^{\mathrm{\infty }}$

$\Rightarrow \mathrm{}V=\frac{A}{\sqrt[]{a^2+x^2}}$

F = $\frac{G{m}_2}{d^2}$ - (i)

Now,  $F\text{'}=\frac{G{m}_1{m}_2}{d^2}=\frac{G2m}{3}\times \frac{4m}{3d^2}$

$k\text{'}=\frac{8}{9}G\frac{m^2}{d^2}$

$\therefore F\text{'}=\frac{8}{G}F$

$M_A={\rho }_A\times \frac{4}{3}\pi R_A^3$ ${\rho }_B=4{\rho }_A$

$M_B={\rho }_B\times \frac{4}{3}\pi R_B^3$ $R_B=\frac{R_A}{2}$

$\frac{M_A}{M_B}=2,\frac{R_A}{R_B}=2$ $\frac{V_{EA}}{V_{EB}}=\sqrt[]{\frac{2G_1{M}_A}{R_A}\times \frac{R_B}{2G_1{M}_B}}$

$v_{EA}=v_{EB}=12kmse{c}^{-1}$

$\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}\times r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

 $g^1=g\left(1-\frac{2h}{R}\right)=g\left(1-2\times \frac{32}{6400}\right)$

$g^1=\frac{99g}{100}=0.99g$

% decrease is wt = $\frac{g-g^1}{g}\times 100=1\mathrm{\%}$