Question

A hemispherical tank full of water is emptied by a pipe at the rate 3 4/7 of
litres per
second. How much time will it take to make the tank half empty, if the tank is 3m in
diameter.​

Answer 1

Answer:

$\sf \: \boxed{\sf \: Time\:taken \: to \:empty \:  half \:tank \: =  \: 16.5 \: min \: }$

Step-by-step explanation:

Given that, Diameter of hemispherical tank = 3 m

So, Radius of hemispherical tank, r = $\frac{3}{2}$ m

We know, Volume of hemispherical tank of radius r is given by

${\sf \: \sf \: Volume_{(hemispherical\:tank)} =  \dfrac{2}{3} \pi {r}^{3} \: }$

So,

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{1}{2}  \times  \dfrac{2}{3} \pi {r}^{3} \: }$

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{1}{3} \pi {r}^{3} \: }$

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{1}{3}  \times \dfrac{22}{7}  \times  {\left(\dfrac{3}{2} \right)}^{3} \: }$

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{1}{3}  \times \dfrac{22}{7}  \times  \dfrac{27}{8}  \: }$

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{22}{7}  \times  \dfrac{9}{8}  \: }$

${\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{11}{7}  \times  \dfrac{9}{4}  \: }$

$\implies\sf \: {\sf \: \sf \: Volume_{( \frac{1}{2} \:  hemispherical\:tank)} =   \dfrac{99}{28}   \:  {m}^{3}  \: }$

Now, it is given that,

$\sf \: Tank\:emptied\:by\:pipe\:in \:1 \: sec = 3 \dfrac{4}{7} \: litres$

$\sf \: Tank\:emptied\:by\:pipe\:in \:1 \: sec = \dfrac{25}{7} \: litres$

$\sf \:Tank\:emptied\:by\:pipe\:in \:1 \: sec = \dfrac{25}{7}  \times \dfrac{1}{1000} \:  {m}^{3}$

$\implies\sf \:Tank\:emptied\:by\:pipe\:in \:1 \: sec = \dfrac{25}{7000} \:  {m}^{3}$

Now,

$\sf \: Time\:taken \: to \: empty \: half \: tank \:$

$\sf \:  =  \: \dfrac{Volume_{( \frac{1}{2} \:  hemispherical\:tank)}}{Tank\:emptied \: by \: pipe\:in\:1 \: sec\:}$

$\sf \:  =  \: \dfrac{99}{28}  \div \dfrac{25}{7000}$

$\sf \:  =  \: \dfrac{99}{28} \times \dfrac{7000}{25}$

$\sf \:  =  \: \dfrac{99}{28} \times 280$

$\sf \:  =  \: 99 \times 10$

$\sf \:  =  \: 990 \: sec$

$\sf \:  =  \: \dfrac{990}{60}  \: min$

$\sf \:  =  \: 16.5 \: min$

Hence,

$\implies\sf \: \boxed{\sf \: Time\:taken \: to \:empty \:  half \:tank \: =  \: 16.5 \: min \: }$