How many balls of radius 1.4 , cm can be prod | vedclass

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(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.Let $R$ be the radius of the large metallic sphere $(R = 42 \, cm)$ and $

Vedclass · Accepted Answer

$27000$

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(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.Let $R$ be the radius of the large metallic sphere $(R = 42 \, cm)$ and $r$ be the radius of the small ball $(r = 1.4 \, cm)$.The number of balls $n$ that can be produced is given by the ratio of the volume of the large sphere to the volume of one small ball:$n = \frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3} = \left( \frac{R}{r} \right)^3$.Substituting the values:$n = \left( \frac{42}{1.4} \right)^3 = (30)^3$.$n = 27,000$.Therefore,$27,000$ balls can be produced.

Part $I$	Part $II$
$1.$ $CSA$ of a cylinder	$a.$ $3\pi r^2$
$2.$ $CSA$ of a sphere	$b.$ $2\pi rh$
$3.$ $CSA$ of a cone	$c.$ $4\pi r^2$
$4.$ $TSA$ of a hemisphere	$d.$ $\pi rl$

How many balls of radius $1.4 \, cm$ can be produced by melting a metallic sphere of radius $42 \, cm$?

Similar Questions

$A$ solid iron cuboidal block of dimensions $4.4 \, m \times 2.6 \, m \times 1 \, m$ is recast into a hollow cylindrical pipe of internal radius $30 \, cm$ and thickness $5 \, cm$. Find the length of the pipe in meters.

Which of the following correctly matches the information given in Part $I$ and Part $II$?

Part $I$ Part $II$

$1.$ $CSA$ of a cylinder $a.$ $3\pi r^2$

$2.$ $CSA$ of a sphere $b.$ $2\pi rh$

$3.$ $CSA$ of a cone $c.$ $4\pi r^2$

$4.$ $TSA$ of a hemisphere $d.$ $\pi rl$

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$A$ plumbline (sahul) is the combination of (see figure):

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