Question

Q. 15. A vessel full of water is in the form of an inverted
cone of height 8 cm and the radius of its top,
which is open, is 5 cm. 100 spherical lead balls are
dropped into vessel. One-fourth of the water flows
out of the vessel. Find the radius of a spherical
ball​

Answer 1

$\huge\pink{\tt{QUESTION:-}}$

A vessel full of water is in the form of an inverted cone of height 8 cm and the radius of its top, which is open, is 5 cm. 100 spherical lead balls are dropped into vessel. One-fourth of the water flowsout of the vessel. Find the radius of a spherical ball.

$\huge\green{\tt{SOLUTION:-}}$

$\rightarrow$ Volume of cone,

$\rightarrow$ ⅓πr²h

$\rightarrow$ .°. ⅓×π×(5)²×8

$\implies$ .°. 200/3π -----(1)

NOW,

$\rightarrow$ Volume of liquid displaced = volume of balls dropped

$\rightarrow$ .°. ¼(200/3π)=100×¾πr³ (from eq.(1) )

$\rightarrow$ 50 = 100 ×4r³

$\rightarrow$ 50/400 =r³

$\rightarrow$ .°. r³= ⅛

$\rightarrow$ .°. r³ = (⅛)³

$\rightarrow$ .°. r = ½cm

$\implies$ .°. $\large\red{\tt{r = 0.5cm}}$

$\huge\pink{\tt{ANSWER:}}$ the radius of spherical ball is 0.5cm.

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

${\huge\tt\bf{Given}}\begin{cases}\sf{Vessel's\:shape\:is\:in\:the\:form\:of\:inverted\:cone}\\\sf{Height\:of\:the\:cone=8\:cm}\\ \sf{The\:radius\:of\:its\:op en\:top=5\:cm}\\\sf{100\:spherical\:balls\:are\:dropped\:in}\\\sf{One-forth\:of\:water\:flows\: out}\end{cases}$

$\huge\bf\tt{To\:Find:-}$

• The radius of the spherical balls.

$\huge\bf\tt{Solution:-}$

Let's first find out the volume of the cone.

We know that, Volume of the water or the Volume of a cone:-

$\huge{\boxed{\tt{\dfrac{1}{3}\times\pi{r}^{2}h}}}$

Putting in the values,

$\leadsto\tt{\dfrac{1}{3}\times\pi\times5\times5\times8}$

$\huge\leadsto{\boxed{\tt{\dfrac{200}{3}\pi\:{cm}^{3}}}}$

Now, by Archimedes principle we know that,

The volume of the balls dropped in the vessel = The volume of the water displaced.

Now, we know that the formula to find out the volume of a sphere is:-

$\huge{\boxed{\tt{\dfrac{4}{3}\times\pi{r}^{3}}}}$

Now, using this we can write,

$\leadsto\tt{100\times\dfrac{4}{\cancel{3}}\times\cancel{\pi}{r}^{3}=\dfrac{1}{4}\times\dfrac{200}{\cancel{3}}\cancel{\pi}}$

$\leadsto\tt{400\times\times{r}^{3}=50}$

$\leadsto\tt{{r}^{3}=\dfrac{50}{400}}$

$\leadsto\tt{{r}^{3}=\dfrac{1}{8}}$

$\leadsto\tt{r=\sqrt[3]{\dfrac{1}{8}}}$

$\huge\leadsto{\boxed{\tt{\red{r=\dfrac{1}{2}\:cm=0.5\:cm}}}}$

$\rule{200}4$