If fish are flowing at a rate of 10 fish/ $m^{2} s$ , and the 10 $m^{2}$ net is dipped at an angle of $60^{0}$ in their path for 1 second, how many fish do we catch?

$50 \sqrt{3}$

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$50$

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$60$

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$100$

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Solution

The correct option is B
$50$

In this case, the area vector makes an angle 90- 60=30 with the fish flow vector.

So, number of fishes = 10 x 10 x sin30 = 50.

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If fish are flowing at a rate of 10 fish/m2s, and the 10 m2 net is dipped at an angle of 600 in their path for 1 second, how many fish do we catch?

The correct option is B 50 In this case, the area vector makes an angle 90- 60=30 with the fish flow vector. So, number of fishes = 10 x 10 x sin30 = 50.

If fish are flowing at a rate of 10 fish/ $m^{2} s$ , and the 10 $m^{2}$ net is dipped at an angle of $60^{0}$ in their path for 1 second, how many fish do we catch?

The correct option is B
$50$

In this case, the area vector makes an angle 90- 60=30 with the fish flow vector.

So, number of fishes = 10 x 10 x sin30 = 50.