Question

A thin hemispherical shell of mass M and radius R is placed as shown in the figure The magnitude of gravitational field at P due to thin hemispherical shell is ${{I}_{0}}$.​ The magnitude of gravitational field at 'Q' due to thin hemispherical shell is given by

A. $\dfrac{{{I}_{0}}}{2}$

B. ${{I}_{0}}$

C. $\dfrac{2GM}{9{{R}^{2}}}-{{I}_{0}}$

D. $\dfrac{2GM}{9R}+{{I}_{0}}$

Answer 1

Newton's law of gravitation states that every particle attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. So, whenever there is mass it has a field called a gravitational field around it which exerts an attractive force on another mass placed in it.

Complete step by step answer:

Let us consider a thin hemispherical shell of mass M and radius R

The formula for gravitational field strength is:

$E=\dfrac{GM}{{{r}^{2}}}$ in N/kg

Where, $G$ – gravitational constant, $M$- the mass which has a gravitational field around it AND $R$ – is the distance between the two masses.

Now, to find the gravitational field at point Q, consider that the spherical shell is made up of two hemispheres each of mass M. So now the gravitational field at point Q at distance 3R due to spherical shell of mass 2M is,

{{E}_{P}}=\dfrac{G(2M)}{{{(3R)}^{2}}} \\\

\Rightarrow {{E}_{P}}=\dfrac{2GM}{9{{R}^{2}}} \\\

It is given that at point P gravitational field is ${{I}_{0}}$, using symmetry the gravitational field due to the imaginary hemispherical shell at point Q would also be ${{I}_{0}}$. Thus, now if we remove lower hemispherical shell then gravitational field due to the given hemisphere at point Q will be = (Gravitational field due to whole sphere at Q) - (Gravitational field due to imaginary hemisphere at Q)

i.e., gravitational field at point Q is

${{E}_{Q}}={{E}_{P}}-{{I}_{0}}$

$\therefore {{E}_{Q}}=\dfrac{2GM}{9{{R}^{2}}}-{{I}_{0}}$

Thus, the correct option is C.

Note: Students may get confused as gravitational field at a distance 3R up is ${{I}_{0}}$ so down at same distance is ${{I}_{0}}$ an think that option B is correct but it is wrong and also if we think, mass is hemisphere so electric field below is half of that of point P and mark option A, then also it is wrong. It is that we consider it as a total sphere and by removing the lower part the total gravitational field reduces by ${{I}_{0}}$. Option D is incorrect. We can say it just by looking at it as in the denominator of the solution it is R not ${{R}^{2}}$, which is dimensionally incorrect.