Water filled in an empty tank of area 10A through a tap of cross sectional area A. The speed of water flowing out of tap is given by V(m/s)=101-sinπ30t where ‘t’ is in seconds. The height of water level from the bottom of the tank at t=15 sec will be

We know Qt=AVtdV=Avtdt⇒V=A∫015vtdt⇒V=10A∫0151-sinπ30tdt⇒V=10At+cosπ30tπ3015=10A15-30π⇒h=V10A=15-30π

Water filled in an empty tank of area 10A through a tap of cross sectional area A. The speed of water flowing out of tap is given by V(m/s)=101-sinπ30t where ‘t’ is in seconds. The height of water level from the bottom of the tank at t=15 sec will be

We know Qt=AVtdV=Avtdt⇒V=A∫015vtdt⇒V=10A∫0151-sinπ30tdt⇒V=10At+cosπ30tπ3015=10A15-30π⇒h=V10A=15-30π

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Water filled in an empty tank of area 10A through a tap of cross sectional area A. The speed of water flowing out of tap is given by $V (m / s) = 10 (1 - \sin \frac{π}{30} t)$ where ‘t’ is in seconds. The height of water level from the bottom of the tank at t=15 sec will be

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10m

$\frac{5}{4} m$

$15 + \frac{30}{π} m$

$15 - \frac{30}{π} m$

answer is D.

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Detailed Solution

We know $Q_{t} = {AV}_{t}$

$dV = {Av}_{t} dt$

$\Rightarrow V = A \int_{0}^{15} v_{t} dt$

$\Rightarrow V = 10 A \int_{0}^{15} (1 - \sin \frac{π}{30} t) dt \Rightarrow V = 10 A {[t + \frac{\cos \frac{π}{30} t}{\frac{π}{30}}]}^{15} = 10 A [15 - \frac{30}{π}]$

$\Rightarrow h = \frac{V}{10 A} = [15 - \frac{30}{π}]$

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Water filled in an empty tank of area 10A through a tap of cross sectional area A. The speed of water flowing out of tap is given by V(m/s)=101-sinπ30t where ‘t’ is in seconds. The height of water level from the bottom of the tank at t=15 sec will be

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Water filled in an empty tank of area 10A through a tap of cross sectional area A. The speed of water flowing out of tap is given by $V (m / s) = 10 (1 - \sin \frac{π}{30} t)$ where ‘t’ is in seconds. The height of water level from the bottom of the tank at t=15 sec will be