A mass of 10kg is balanced on a sensitive physical balance. A 1000 kg mass is placed below 10 kg mass at a distance of 1m. How much additional mass will be required for balancing the physical balance

The correct option is 6.7 x 10⁻⁸ kg.

To find the additional mass required to balance the physical balance, we need to analyze the gravitational force exerted by the large mass on the balanced mass.

Let the balanced mass be:
m = 10 kg
Let the large mass placed below it be:
M = 1000 kg
The distance between the centers of these two masses is:
r = 1 m
The universal gravitational constant is:
G = 6.67 x 10^-11 N m² kg^-2
The acceleration due to gravity is approximately:
g = 9.8 m s^-2

The gravitational force of attraction $F$ between the 10 kg mass and the 1000 kg mass is given by Newton's law of gravitation:

$F=\frac{G·M·m}{r^2}$

Substituting the values into the formula:

$F=\frac{(6.67\times {10}^{-11})·1000·10}{1^2}$

$F=6.67\times {10}^{-11}\times {10}^4$

$F=6.67\times {10}^{-7}\text{N}$

This downward gravitational pull acts as an additional force on the 10 kg mass. To balance this force, an additional mass $\mathrm{\Delta m}$ must be placed on the opposite pan of the balance. The weight of this additional mass must equal the gravitational force $F$:

$\mathrm{\Delta m}·g=F$

Solving for $\mathrm{\Delta m}$:

$\mathrm{\Delta m}=\frac{F}{g}$

$\mathrm{\Delta m}=\frac{6.67\times {10}^{-7}}{9.8}$

$\mathrm{\Delta m}\approx 0.68\times {10}^{-7}\text{kg}$

$\mathrm{\Delta m}\approx 6.8\times {10}^{-8}\text{kg}$

Rounding to the nearest option provided, the required additional mass is approximately 6.7 x 10⁻⁸ kg.