A beaker containing water is placed on the platform of a spring balance. The balance reads `1.5 kg.` A stone of mass `0.5 kg` and density `10^(4) kg//m^(3)` is immersed in water without touching the walls of the beaker. What will be the balance reading now?

Options

A. 2kg
B. 2.5 kg
C. 1 kg
D. 3 kg

Answer

A

Explanation

As water exerts upthrust on the stone, the stone also exerts the same force on water, according to Newton’s ` 3 ^(rd)` law.
` rArr F = V_("stone" rho _(omega )) g = (m_("stone"))/(rho_("stone")) rho _(omega) g = 5 N `
Also, the weight is exerted on the beaker. Therefore, the reading will be 1.5 + 0.5 = 2kg