MHT CET · Physics · Mechanical Properties of Fluids
A thin metal disc of radius ' \(r\) ' floats on water surface and bends the surface downwards along the perimeter making an angle ' \(\theta\) ' with the vertical edge of the disc. If the weight of water displaced by the disc is \(\mathrm{W}\), the weight of the metal disc is [ \(\mathrm{T}=\) surface tension of water \(]\)
- A \(2 \pi \mathrm{r} \cos \theta+\mathrm{W}\)
- B \(\mathrm{W}-2 \pi \mathrm{rT} \cos \theta\)
- C \(2 \pi r \mathrm{~T}+\mathrm{W}\)
- D \(2 \pi \mathrm{rT} \cos \theta-\mathrm{W}\)
Answer & Solution
Correct Answer
(A) \(2 \pi \mathrm{r} \cos \theta+\mathrm{W}\)
Step-by-step Solution
Detailed explanation
The weight of the disc is balanced by the force due to the surface tension and the upthrust of water.
The component of surface tension in vertically upward direction is \(\mathrm{T} \cos \theta\) and the force acting due to it \(2 \pi \mathrm{r} \mathrm{T} \cos \theta\).
The upthrust is equal to the weight of the water displaced (W).
\(\therefore\) Weight of the disc \(=2 \pi \mathrm{rT} \cos \theta+\mathrm{W}\)
The component of surface tension in vertically upward direction is \(\mathrm{T} \cos \theta\) and the force acting due to it \(2 \pi \mathrm{r} \mathrm{T} \cos \theta\).
The upthrust is equal to the weight of the water displaced (W).
\(\therefore\) Weight of the disc \(=2 \pi \mathrm{rT} \cos \theta+\mathrm{W}\)
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