A cylindrical tub container a radius …
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Sia ? 6 years, 5 months ago
According to question it is given that
Radius of cylindrical tub = 12 cm
Depth of cylindrical tub = 20 cm
Let us suppose that (r) be the radius of spherical ball
Again it is given that level of water is raised by 6.75 cm
Now, according to the question,
Volume of spherical ball = Volume of water rise in cylindrical tub
{tex}\Rightarrow \quad \frac { 4 } { 3 } \pi r ^ { 3 } = \pi ( 12 ) ^ { 2 } \times 6.75{/tex}
{tex}\Rightarrow \frac { 4 } { 3 } r ^ { 2 } = 12 \times 12 \times 6. 75{/tex}
{tex}\Rightarrow \quad r ^ { 3 } = \frac { 12 \times 12 \times 6.75 \times 3 } { 4 }{/tex}
{tex}\Rightarrow \quad r ^ { 3 } = 729{/tex}
{tex}\Rightarrow \quad r = \sqrt [ 3 ] { 729 } = 9 \mathrm { cm }{/tex}
Therefore, Radius of the ball = 9 cm
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