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 10⁴ kg/m³ is immersed in water without touching the walls of the beaker. What will be the balance reading now?

As the water exerts upthrust on the stone, the stone also exerts the same force on water, according to Newton's third law. 

Upthrust force acting on the stone is

$\begin{array}{c}U=V_{\text{stone }}{\rho }_{\omega }g\\ =\left(\frac{m_{\text{stone }}}{{\rho }_{\text{stone }}}\right){\rho }_{\omega }g=\frac{0.5}{{10}^4}\times 1000\times 10=0.5\mathrm{N}\end{array}$

Also, the weight is exerted on the beaker. Therefore the reading will be 1.5 + 0.5 = 2kg.