Question

a sphere of d=6cm is dropped in right circular cylindrical vessel, partly filled with water. the d of vessel is 12 cm. if the sphere is completely submerged in water, by how much will the water rise in vessel?

Answer 1

||✪✪ QUESTION ✪✪||

a sphere of d=6cm is dropped in right circular cylindrical vessel, partly filled with water. the d of vessel is 12 cm. if the sphere is completely submerged in water, by how much will the water rise in vessel ?

|| ★★ FORMULA USED ★★ ||

• Volume of sphere = (4/3) * π * (r)³
• volume of cylinder = π * R² * h
• Radius = Diameter /2 .
• Rise in water level = Volume of sphere

|| ✰✰ ANSWER ✰✰ ||

Given that, Diameter of sphere 6cm and diameter of cylinder is 12cm.

So,

→ Radius of sphere = r = (6/2) = 3cm .

→ Radius of Cylinder = R = (12/2) = 6cm.

Now, Let , Rise in water level of cylinderical vessel is h cm.

So,

→ volume of the water raised in the vessel = volume of the sphere

→ πR²h = 4/3 × πr³

→ h = ( 4 × r³ ) / ( 3 × R² )

→ h = ( 4 × 3 × 3 × 3 ) / ( 3 × 6 × 6 )

→ h = 1 cm..

Hence level of water raised in the vessel is 1cm.

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\rm Given \begin{cases} \tt{Diameter \: of \: Sphere \: = \: 6 \: cm} \\ \tt{Diameter \: of \: Cylinder \: = \: 12 \: cm} \\ \tt{Rise \: in \: height \: (h) \: = \: ?} \end{cases}$

$\Large{\underline{\underline{\rm{Splution :}}}}$

We know that

• Radius of Sphere (Rs) = 6/2 = 3 cm
• Radius of Cylinder (Rc) = 12/2 = 6 cm

________________________________

We have formula for Volume of Sphere and volume of cylinder

$\large{\boxed{\sf{Volume \: of \: Sphere \: = \: \dfrac{4}{3} \pi r^3}}} \\ \\ \large {\boxed{\sf{Volume \: of \: Cylinder \: = \: \pi r^2 h}}}$

Now equate volume of sphere and volume of cylinder.

$\large {\boxed{\sf{\dfrac{4}{3} \pi R_s^3 \: = \: \pi R_c ^2 h}}} \\ \\ \implies {\sf{\dfrac{4}{3} \pi 3^3 \: = \: \pi 6^2 h}} \\ \\ \implies {\sf{\dfrac{4}{3} \: \times \: 27 \: = \: 36 h}} \\ \\ \implies {\sf{4 \: \times \: 9 \: = \: 36h}} \\ \\ \implies {\sf{h \: = \: \dfrac{36}{36}}} \\ \\ \implies {\sf{h \: = \: 1 \: cm}} \\ \\ {\underline{\sf{\therefore \: Rise \: in \: height \: (h) \: = \: 1 \: cm}}}$

$\mathcal{\underline{\: \: \: \: \: \green{BE} \: \: \orange{BRAINLY} \: \: \: \: \:}}$