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Standard IX

Physics

Archimedes' Principle

Figure shows ...

Question

Figure shows a weighing-bridge, with a beaker P with water on one pan and a balancing weight R on the other. A solid ball Q is hanging with a thread outside water. It has a volume of 40 cm $^{3}$ and weighs 80 g. If this solid is lowered to sink fully in water, but not touching the beaker anywhere, the balancing weight R' will be

Same as R

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40 g less than R

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40 g more than R

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80 g more than R

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Solution

The correct option is C 40 g more than R
Upthrust $=$ weight of $40 c m^{3}$ of water $= 40 g$ .
$∴$ downthrust $= 40 g$ .
so, the balancing weight R' will be 40 g more than R.

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The correct option is C 40 g more than RUpthrust = weight of 40cm3 of water =40g.∴ downthrust=40g.so, the balancing weight R' will be 40 g more than R.

The correct option is C 40 g more than R
Upthrust $=$ weight of $40 c m^{3}$ of water $= 40 g$ .
$∴$ downthrust $= 40 g$ .
so, the balancing weight R' will be 40 g more than R.