Question
How many spherical bullets each of $5\ cm$ in diameter can be cast from a rectangular block of metal $11dm \times 1m \times 5dm$?
✓
Answer
We are given a metallic block of dimension $= 11dm \times 1m \times 5dm.$
We know that, 1dm $= 10^{-1}m$
So, the volume of the given metallic block is $= 11 \times 10^{-1}\times 1 \times 5 \times 10^{-1} = 55 \times 10^{-2}m^3$
We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.
We know,
$\text{Volume of a sphere}=\frac{4}{3}\pi(\text{r})^3$
Here, $r = 25 \times 10^{-3}m$
Let the no. of bullets formed be n.
We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.
$\Rightarrow55\times10^{-2}=\text{n}\times\frac{4}{3}\times\frac{22}{7}\times(25\times10^{-3})^3$
$\text{n}=\frac{55\times3\times7\times10^{-2}}{4\times22\times25\times25\times25\times10^{-9}}$
$=\frac{21\times10^7}{(2\times5)^3\times25}$
$=8400$
Hence the no. of bullets that can be formed is 8400.
We know that, 1dm $= 10^{-1}m$
So, the volume of the given metallic block is $= 11 \times 10^{-1}\times 1 \times 5 \times 10^{-1} = 55 \times 10^{-2}m^3$
We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.
We know,
$\text{Volume of a sphere}=\frac{4}{3}\pi(\text{r})^3$
Here, $r = 25 \times 10^{-3}m$
Let the no. of bullets formed be n.
We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.
$\Rightarrow55\times10^{-2}=\text{n}\times\frac{4}{3}\times\frac{22}{7}\times(25\times10^{-3})^3$
$\text{n}=\frac{55\times3\times7\times10^{-2}}{4\times22\times25\times25\times25\times10^{-9}}$
$=\frac{21\times10^7}{(2\times5)^3\times25}$
$=8400$
Hence the no. of bullets that can be formed is 8400.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve for $x, x^2-(2 b-1) x+\left(b^2-b-20\right)=0$
→From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are $30^\circ$ and $45^\circ$ respectively. If bridge is at the height of $30\ m$ from the banks, find the width of the river.
→Write the value of k for which the system of equations $x + ky = 0, 2x - y = 0$ has unique solution.
→Find the points of trisection of the line segment joining the points:
(5, -6) and (-7, 5).
(5, -6) and (-7, 5).
In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If $\angle\text{PBT} = 30^\circ,$prove that
BA : AT = 2 : 1.
→BA : AT = 2 : 1.
Solve for $x$ and $y$
$\frac{5}{x-1}+\frac{1}{y-2}=2 ; \frac{6}{x-1}-\frac{3}{y-2}=1 ;[x \neq 1 y \neq 2 ]$
→$\frac{5}{x-1}+\frac{1}{y-2}=2 ; \frac{6}{x-1}-\frac{3}{y-2}=1 ;[x \neq 1 y \neq 2 ]$
If three consecutive vertices of a parallelogram are (1, -2), (3, 6) and (5, 10), find its fourth vertex.
A cone of radius 4cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of two parts.
Justify whether it is true to say that the sequence, having following $n ^{\text {th }}$ term is an A.P. $a_n=2 n-1$.
→How many numbers lie between $10$ and $300$, which when divided by $4$ leave a remainder $3$?
→