A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter making an angle θ with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is

Question

A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter making an angle θ with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is (A) 2 π r T+W (B) 2 π r T cos θ-W (C) 2 π r T cos θ+W (D) W-2 π r T cos θ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b3859027094807a2fb7508b25fbf0f07-.png" /> Weight of metal disc = total upward force upthrust force + force due to surface tension = weight of displaced water +T cos θ(2 π r) =W+2 π r T cos θ

Q. A thin metal disc of radius $r$ floats on water surface and bends the surface downwards along the perimeter making an angle $θ$ with vertical edge of the disc. If the disc displaces a weight of water $W$ and surface tension of water is $T$ , then the weight of metal disc is

$2 π r T + W$

$2 π r T cos θ - W$

$2 π r T cos θ + W$

$W - 2 π r T cos θ$

Weight of metal disc $=$ total upward force upthrust force + force due to surface tension
= weight of displaced water $+ T cos θ (2 π r)$
$= W + 2 π r T cos θ$

Q. A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter making an angle θ with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is

2πrT+W

2πrTcosθ−W

2πrTcosθ+W

W−2πrTcosθ