Question

The dimensions of a tank are 70 m by 44 m by 3 m. This tank is being filled by a
pipe of radius 17.5 cm. If the pipe discharges water at the rate of 6 m per second,
how long will it take for the pipe to fill the tank?
​

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ pipe \ will \ take \ 4.44 \ hr \ to \ fill}$

$\sf{the \ tank.}$

$\sf\orange{Given:}$

$\sf{For \ cuboidal \ tank,}$

$\sf{\implies{Length(l)=70 \ m}}$

$\sf{\implies{Breadth (b)=44 \ m}}$

$\sf{\implies{Height (h)=3 \ m}}$

$\sf{For \ cylindrical \ pipe,}$

$\sf{\implies{Radius(r)=17.5 \ cm=0.175 \ m}}$

$\sf{\implies{Speed \ of \ water=6 \ m \ s^{-1}}}$

$\sf\pink{To \ find:}$

$\sf{How \ long \ will \ take \ for \ the \ pipe \ to}$

$\sf{fill \ the \ tank.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{\underline{\underline{Concept:}}}$

$\sf{To \ fill \ a \ tank \ with \ water \ the \ pipe}$

$\sf{must \ have \ same \ volume \ of \ water.}$

$\boxed{\sf{Volume \ of \ cuboid=l\times \ b\times \ h}}$

$\boxed{\sf{Volume \ of \ cylinder=\pi\times \ r^{2}\times \ H}}$

$\sf{Volume \ of \ cuboidal \ tank=}$

$\sf{Volume \ cylindrical \ pipe.}$

$\sf{\therefore{l\times \ b\times \ h=\pi\times \ r^{2}\times \ h}}$

$\sf{\therefore{70\times44\times3=\frac{22}{7}\times0.175^{2}\times \ H}}$

$\sf{\therefore{H=\frac{9240\times7}{22\times0.175^{2}}}}$

$\sf{\therefore{H=\frac{420\times7\times10^{6}}{30625}}}$

$\sf{\therefore{H=0.096\times10^{6}}}$

$\sf{\therefore{H=96000 \ m}}$

$\sf{Distance=Height}$

$\sf{\therefore{Distance=96000 \ m}}$

$\boxed{\sf{Time(t)=\frac{Distance}{Speed}}}$

$\sf{\therefore{Time(t)=\frac{96000}{6}}}$

$\sf{\therefore{Time(t)=16000 \ seconds}}$

$\sf{But, \ 1 \ second=\frac{1}{3600} \ hr}$

$\sf{\therefore{Time(t)=\frac{16000}{3600} \ hr}}$

$\sf{\therefore{Time (t)=4.44 \ hr (approx)}}$

$\sf\purple{\tt{\therefore{The \ pipe \ will \ take \ 4.44 \ hr \ to \ fill}}}$

$\sf\purple{\tt{the \ tank.}}$

Answer 2

$\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}$

$\rm{\orange{\underline{\green{The \ pipe \ will \ take \ 4.44 \ hr \ to \ fill}}}}$

$\rm{\orange{\underline{\green{the \ tank.}}}}$

_________________________________________

$\sf\large\underline\pink{Given:}$

$\rm{For \ cuboidal \ tank,}$

$\rm{\longrightarrow{Length(l)=70 \ m}}$

$\rm{\longrightarrow{Breadth (b)=44 \ m}}$

$\rm{\longrightarrow{Height (h)=3 \ m}}$

$\rm{For \ cylindrical \ pipe,}$

$\rm{\longrightarrow{Radius(r)=17.5 \ cm=0.175 \ m}}$

$\rm{\longrightarrow{Speed \ of \ water=6 \ m \ s^{-1}}}$

_________________________________________

$\sf\large\underline\blue{To \ Find}$

$\rm{Time \ taken \ to \ fill \ the \ tank}$

_________________________________________

$\purple{\underline{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}}$

$\rm{As \ we \ know}$

$\rm{To \ fill \ a \ tank \ with \ water \ the \ pipe}$

$\rm{must \ have \ same \ volume \ of \ water.}$

$\rm{\red{\boxed{\blue{Volume \ of \ cuboid=l\times \ b\times \ h}}}}$

$\rm{\red{\boxed{\blue{Volume \ of \ cylinder=\pi\times \ r^{2}\times \ H}}}}$

$\rm{Vol \ of \ cuboidal \ tank \ = \ Vol \ cylindrical \ pipe.}$

$\rm{\implies{l\times \ b\times \ h=\pi\times \ r^{2}\times \ h}}$

$\rm{\implies{70\times44\times3=\frac{22}{7}\times0.175^{2}\times \ H}}$

$\rm{\implies{H=\frac{9240\times7}{22\times0.175^{2}}}}$

$\rm{\implies{H=\frac{420\times7\times10^{6}}{30625}}}$

$\rm{\implies{H=0.096\times10^{6}}}$

$\rm{\implies{\blue{\boxed{\orange{H=96000 \ m}}}}}$

$\rm{Here \ we \ have \ to \ consider,}$

$\rm\longrightarrow{\boxed{Distance=Height}}$

$\rm{\therefore{\blue{\boxed{\orange{Distance=96000 \ m}}}}}$

$\longrightarrow\boxed{\rm{Time(t)=\frac{Distance}{Speed}}}$

$\rm{\implies{Time(t)=\frac{96000}{6}}}$

$\rm{\implies{Time(t)=16000 \ seconds}}$

$\rm{\longrightarrow{Time(t)=\frac{16000}{3600} \ hr}}$

$\rm{\implies{Time (t)=4.44 \ hr (approx \ value)}}$

$\rm{\orange{\underline{\green{The \ pipe \ will \ take \ 4.44 \ hr \ to \ fill}}}}$

$\rm{\orange{\underline{\green{the \ tank.}}}}$

___________________________________________

$\rm\orange{Hope \ it \ helps \ :))}$