A conical vessel of slant height $10$ $m$ and radius $5 m$ is being filled with water at a uniform rate of $1.5$ $m^{3} / m i n$ . How long will it take to fill the vessel?

41

min

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59

min

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72

min

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151

min

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Solution

The correct option is A $151$ min
Given,
$R a d i u s (r) = 5 m$
$S l a n t h e i g h t (l) = 10 m$
Therefore,
$H e i g h t (h) = \sqrt{l^{2} - r^{2}} \Rightarrow \sqrt{100 - 25} = 5 \sqrt{3} m$
Therefore,
$V o l u m e o f c o n e = \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 5 \sqrt{3} = 226.81 m^{3}$

Time taken to fill the vessel $= \frac{226.81 m^{3}}{1.5 m^{3} p e r m i n u t e} = 151.21 m i n u t e s \approx 151 m i n u t e s$

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A conical vessel of slant height 10 m and radius 5 m is being filled with water at a uniform rate of 1.5 m3/min. How long will it take to fill the vessel?

The correct option is A 151 minGiven,Radius(r)=5 mSlant height(l)=10 mTherefore, Height(h)=√l2−r2⇒√100−25=5√3 mTherefore, Volume of cone=13πr2h=13×227×5×5×5√3=226.81 m3Time taken to fill the vessel =226.81 m31.5 m3 per minute=151.21 minutes≈151 minutes

A conical vessel of slant height $10$ $m$ and radius $5 m$ is being filled with water at a uniform rate of $1.5$ $m^{3} / m i n$ . How long will it take to fill the vessel?