A solid sphere of mass M and radius a is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance 3a from the center will be:
A. $\dfrac {2GM}{9{a}^{2}}$
B. $\dfrac {GM}{3{a}^{2}}$
C. $\dfrac {GM}{9{a}^{2}}$
D. $\dfrac {2GM}{3{a}^{2}}$

Hint: To solve this problem, use the Gauss’s law of gravitation. Firstly, find the total mass by adding mass of solid sphere and concentric spherical shell. Then, find the total radius by adding the radius of solid sphere and spherical shell. Substitute these values in the formula for Gauss’s law. Rearrange it and find the value of g i.e. gravitational field. This obtained value of g will be the gravitational field at distance 3a from the center.
Formula used:
$\oint { \overrightarrow { g } .\overrightarrow { dA } =4\pi Gm }$

Complete answer:
Gauss's law of gravitation is given by,
$\oint { \overrightarrow { g } .\overrightarrow { dA } =4\pi Gm }$
$\Rightarrow g. 4 \pi {r}^{2}= 4 \pi Gm$ …(1)
From the figure above, we can infer that the total radius is given by,
$r= a + 2a$
$\Rightarrow r=3a$ …(2)
Similarly, the total mass is given by,
$m= M+ 2M$
$\Rightarrow m=3M$ …(3)
Substituting equation. (2) and (3) in equation. (1) we ger,
$ g. 4 \pi {3a}^{2}= 4 \pi G \times 3M$
$\Rightarrow 36g \pi {a}^{2}= 12 \pi GM$
$\Rightarrow 36g {a}^{2}= 12 GM$
$\Rightarrow g= \dfrac {12GM}{36{a}^{2}}$
$\Rightarrow g= \dfrac {GM}{3{a}^{2}}$
Hence, the gravitational field at distance 3a from the center will be $\dfrac {GM}{3{a}^{2}}$.
So, the correct answer is option B i.e. $\dfrac {GM}{3{a}^{2}}$.
Note: There is an alternate method to solve this problem. The alternate method is given below:
Gauss's law of gravitation is given by,
$\oint { \overrightarrow { g } .\overrightarrow { dA } =4\pi Gm }$
$\Rightarrow g. 4 \pi {r}^{2}= 4 \pi Gm$ …(1)
For the solid sphere, mass is M. So, the gravitational field at distance 3a will be,
$ {g}_{1}. 4 \pi {3a}^{2}= 4 \pi GM$
$\Rightarrow {g}_{1} = \dfrac {GM}{9 {a}^{2}}$
Similarly, for the conteric shell, mass is 2M. So, the gravitational field at the distance 3a will be
$ {g}_{2}. 4 \pi {3a}^{2}= 4 \pi G \times 2M$
$\Rightarrow {g}_{2} = \dfrac {2GM}{9 {a}^{2}}$
So, the total gravitational field at distance 3a will be,
$g= {g}_{1}+ {g}_{2}$
Substituting the values in above equation we get,
$g= \dfrac {GM}{9 {a}^{2}} + \dfrac {2GM}{9 {a}^{2}}$
$\Rightarrow g= \dfrac {3GM}{9 {a}^{2}}$
$\Rightarrow g = \dfrac {GM}{3 {a}^{2}}$
Hence, the gravitational field at distance 3a from the center will be $\dfrac {GM}{3{a}^{2}}$.

So, the correct answer is “Option B”.

Note:
To solve these types of problems, students must remember the formula for Gauss's law of gravitation and area and volume of each shape such as sphere, cylinder, etc. Without knowing formulas for area and volume most of the equations cannot be solved. The gravitational field does not depend upon the intervening medium. The gravitational field has its own energy and momentum.