First slide

Pressure in a fluid-mechanical properties of fluids

Question

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?

Moderate

A

2kg
B

2.5kg
C

1 kg
D

3 kg

Solution

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
$\begin{matrix} U = V_{stone} ρ_{ω} g \\ = (\frac{m_{stone}}{ρ_{stone}}) ρ_{ω} g = \frac{0.5}{10^{4}} \times 1000 \times 10 = 0.5 N \end{matrix}$
Also, the weight is exerted on the beaker. Therefore the reading will be 1.5 + 0.5 = 2kg.

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