A gardener wants to wet a garden without moving from his place with a water whose velocity is $20 m s^{- 1}$ the maximum area that he can wet $(g = 10 m / s^{2})$ (in meter $^{2}$ )

1600 π

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40 π

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400 π

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200 π

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Solution

The correct option is A $1600 π$
Given,
$u = 20 m / s$
$g = 10 m / s^{2}$
The maximum range, $R_{m a x} = \frac{u^{2}}{g}$
$R_{m a x} = \frac{20 \times 20}{10} = 40 m$
Maximum area, $A = π R_{m a x}^{2}$
$A = π \times 40 \times 40 = 1600 π$
The correct option is A.

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A gardener wants to wet a garden without moving from his place with a water whose velocity is 20ms−1 the maximum area that he can wet (g=10m/s2) (in meter2)

The correct option is A 1600 πGiven,u=20m/sg=10m/s2The maximum range, Rmax=u2gRmax=20×2010=40mMaximum area, A=πR2maxA=π×40×40=1600πThe correct option is A.

A gardener wants to wet a garden without moving from his place with a water whose velocity is $20 m s^{- 1}$ the maximum area that he can wet $(g = 10 m / s^{2})$ (in meter $^{2}$ )

The correct option is A $1600 π$
Given,
$u = 20 m / s$
$g = 10 m / s^{2}$
The maximum range, $R_{m a x} = \frac{u^{2}}{g}$
$R_{m a x} = \frac{20 \times 20}{10} = 40 m$
Maximum area, $A = π R_{m a x}^{2}$
$A = π \times 40 \times 40 = 1600 π$
The correct option is A.