Question

A mass of $10kg$ is balanced on a sensitive physical balance. A $1000kg$ mass is placed below $10kg$ mass at a distance of $1m$. How much additional mass will be required for balancing the physical balance?
A) $66 \times {10^{ - 15}}kg$
B) $6.7 \times {10^{ - 8}}kg$
C) $66 \times {10^{ - 12}}kg$
D) $6.7 \times {10^{ - 6}}kg$

Answer 1

Hint: Newton's law of gravitation states that the force between two unknown masses is directly proportional to the force acting between them. The force is inversely proportional to the square of distance between the masses. Also these masses experience acceleration due to gravity as well.

Complete step by step solution:
Using the Newton’s law of gravitation
$F = \dfrac{{GMm}}{{{r^2}}}$
$F$ is the force between the two bodies
$G$ is the gravitational constant
$M$ is mass for first body
$m$ is mass of second body
$r$ is distance between two bodies
$F = \dfrac{{6.6 \times {{10} ^ {- 11}} \times 10 \times 1000}} {1} $
$F = 6.6 \times {10^ {- 7}} N$
We know that,
$F = mg$
$g$ is acceleration due to gravity
Therefore,
$mg = 6.6 \times {10^ {- 7}} N$
$m = 6.7 \times {10^ {- 8}} kg$
The additional mass will be required for balancing the physical balance is $6.7 \times {10^ {- 8}} kg$.

Hence the correct option is B.

Note: It depends on the masses involved and the space between them. Gravitational strength is still attractive. In the world, any object attracts some other object along a line that joins it by force. As force is the quantity of a vector, the sum of the vectors of all parts of the shell refers to the net force, which equals one force estimate from the midpoint or centre of mass of the sphere. A hollow spherical shell has a gravitational force zero on an object.