Question

Gravity meter measures in units of mGal where 1 Gal (or 1 galileo) is the C.G.S. unit of acceleration due to gravity. A gravity meter shows a reading of $+ 300\,{\text{mGal}}$ at a place where the normal ‘g’ is $9.78\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$. The correct ‘g’ at that place where the reading was taken is - - - ${\text{m}} \cdot {{\text{s}}^{ - 2}}$.
A. $9.777$
B. $9.783$
C. $9.750$
D. $9.810$

Answer 1

First convert the given value of the acceleration due to gravity in galileo from milli galileo and then convert it in the SI system of units. The final correct value of acceleration due to gravity is the sum of acceleration due to gravity measured by gravity meter and normal acceleration due to gravity.

Complete step by step answer:
We have given that the reading of gravity meter for acceleration due to gravity ${g_g}$ at a place is $+ 300\,{\text{mGal}}$.
${g_g} = + 300\,{\text{mGal}}$

The value of normal acceleration due to gravity ${g_n}$ at the same place is $9.78\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$.
${g_n} = 9.78\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$

We have given that Gal or galileo is the C.G.S. unit of acceleration due to gravity. We also know that the unit of acceleration due to gravity in the C.G.S. system is centimeter per second square.

Convert the measured value of acceleration due to gravity from milli galileo to the C.G.S. system of units.
${g_g} = \left( { + 300\,{\text{mGal}}} \right)\left( {\dfrac{{{{10}^{ - 3}}}}{{1\,{\text{m}}}}} \right)$
$\Rightarrow {g_g} = + 0.3\,{\text{Gal}}$
Hence, the acceleration due to gravity in Gal is $+ 0.3\,{\text{Gal}}$.

The above value of acceleration due to gravity can also be written as $+ 0.3\,{\text{cm/}}{{\text{s}}^2}$.
${g_g} = + 0.3\,{\text{cm/}}{{\text{s}}^2}$

Let us now convert this value of acceleration due to gravity in the SI system of units.
${g_g} = \left( { + 0.3\,\dfrac{{{\text{cm}}}}{{{{\text{s}}^2}}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)$
$\Rightarrow {g_g} = 3 \times {10^{ - 3}}\,{\text{m/}}{{\text{s}}^2}$

The correct reading of acceleration due to gravity at that place is the sum of acceleration due to gravity ${g_g}$ measured by the gravity meter and the normal acceleration due to gravity ${g_n}$.
$g = {g_g} + {g_n}$

Substitute $3 \times {10^{ - 3}}\,{\text{m/}}{{\text{s}}^2}$ for ${g_g}$ and $9.78\,{\text{m/}}{{\text{s}}^2}$ for ${g_n}$ in the above equation.
$g = \left( {3 \times {{10}^{ - 3}}\,{\text{m/}}{{\text{s}}^2}} \right) + \left( {9.78\,{\text{m/}}{{\text{s}}^2}} \right)$
$\therefore g = 9.783\,{\text{m/}}{{\text{s}}^2}$
Therefore, the correct value of acceleration due to gravity is $9.783\,{\text{m/}}{{\text{s}}^2}$.

Hence, the correct option is B.

Note: The students may get confused between the two C.G.S. units of acceleration due to gravity. But these two units galileo and centimeter per second square are equivalent, so these two units can be replaced with each other. But the students should not forget to convert the unit of acceleration due to gravity measured by the gravity meter in the SI system of units as the final answer is in the SI system of units.