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Standard IX

Physics

Second Law of Motion

A fisherman h...

Question

A fisherman hooks with a string a log of wood of weight $12 N$ and volume $1000 c m^{3}$ . He pulls the half log out of the water. The tension in string at this instant is $(g = 10 m / s^{2}, p_{w a t e r} = 1000 k g / m^{3}$

12 N

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5 N

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7 N

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10 N

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Solution

The correct option is C $7 N$
Given,
Weight of the log of the wood $= 12 N$

Volume of the log of the wood, $V = 1000 c m^{3}$

Tension in string will be equal to = weight - buoyant force

$T = m g - R$

Here, half of the log pulls out of the water.

$B = \frac{V}{2} p g$

Therefore,

$T = m g - \frac{V}{2} p g$

$T = 12 - \frac{(1000 \times 10^{- 6})}{2} \times 1000 \times 10$

$= 7 N$

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Second Law of Motion

Standard IX Physics

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A fisherman hooks with a string a log of wood of weight 12 N and volume 1000 cm3. He pulls the half log out of the water. The tension in string at this instant is (g=10 m/s2, pwater=1000 kg/m3

The correct option is C 7 NGiven, Weight of the log of the wood =12 N Volume of the log of the wood, V=1000 cm3 Tension in string will be equal to = weight - buoyant force T=mg−R Here, half of the log pulls out of the water. B=V2pg Therefore, T=mg−V2pg T=12−(1000×10−6)2×1000×10 =7 N

A fisherman hooks with a string a log of wood of weight $12 N$ and volume $1000 c m^{3}$ . He pulls the half log out of the water. The tension in string at this instant is $(g = 10 m / s^{2}, p_{w a t e r} = 1000 k g / m^{3}$