Download App

Surface Areas And Volumes — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsSurface Areas And Volumes4 Marks
Question
How many spherical lead shots each of diameter $4.2\ cm$ can be obtained from a solid rectangular lead piece with dimensions $66cm \times 42cm \times 21cm.$
✓

Answer

Given that, lots of spherical lead of shots made from a solid rectangular lead piece.
$\therefore$ Number of spherical lead shots
$=\frac{\text{volume of solid rectangular lead piece}}{\text{volume of a spherical lead shot}}\ ...(\text{i})$
Also, given that diameter of a spherical lead shot i.e. sphere $= 4.2\ cm$
$\therefore$ Radius of a spherical lead shot,
$\text{r}=\frac{4.2}{2}=2.1\text{cm}[\therefore\text{radius}=\frac{1}{2}\text{diameter}]$
So, volume of a spherical lead shot i.e.
$\text{sphere}=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\times\frac{22}{7}\times(2.1)^3$
$=\frac{4}{3}\times\frac{22}{7}\times2.1\times2.1\times2.1$
$=\frac{4\times22\times21\times21\times21}{3\times7\times1000}$
Now, length of rectangular lead piece, $l = 66\ cm$
Breadth of rectangular lead piece, $b = 42\ cm$
Hieght of rectangular lead piece, $h = 21\ cm$
$\therefore$ Volume of a solid rectangular lead piece i.e.
cuboid =$ l \times b \times h = 66 \times 42 \times 21$
From Eq. $(i)$, Number of spherical lead, shots
$\frac{66\times42\times21}{4\times22\times21\times21\times21}\times3\times7\times1000$
$=\frac{3\times22\times21\times2\times21\times21\times1000}{4\times22\times21\times21\times21}$
$=3\times2\times250$
$=6\times250=1500$
Hence, the required number of spherical lead shots is $1500.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks Questions" from Surface Areas And Volumes→Surface Areas And Volumes — all question types→Maths STD 10 question papers→Gujarat Board STD 10 question papers→Gujarat Board question paper generator→

Similar questions

Solve the following quadratic equations by factorization:
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1},$ $\text{x}\neq-1,\frac{1}{3}$
→
If the sum of first m terms of an AP is $(2m^2 + 3m)$ then what is its second term?
→
Find the sum of the following arithmetic series:
$34 + 32 + 30 + .... + 10.$
→
In a class test, the sum of Kamal's marks in Mathematics and English is $40$. Had he got $3$ marks more in Mathematics and $4$ marks less in English, the product of the marks would have been $360$. Find his marks in two subjects separately.
→
Find the value of k for which the following equations have real and equal roots:
$ x^2-2(k+1) x+k^2=0 $
→
If $\alpha,\ \beta,\ \gamma$ are the zeroes of the polynomial $p(x)=6 x^3+3 x^2-5 x+1 $, find the value of $\Big(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\Big).$
→
Solve the following quadratic equation:
$\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=1\frac{1}{9},$ $\text{x}\neq\frac{3}{2},\ 5$
→
In an isosceles $\triangle\text{ABC,}$ the base $AB$ is produced both the ways to $P$ and $Q$ such that $AP × BQ = AC^2$. Prove that $\triangle\text{APC}\sim \triangle\text{BCQ}.$
→
A toy is in the form of a cylinder with hemispherical ends. If e whole length of the toy is $90\ cm$ and its demeter is $42\ cm$, find the cost of painting the toy at the rate of $70$ paise per sq cm.
→
Solve the following systems of equations graphically:
$x + y = 6$
$x - y = 2$
→