Download App

Maths 1 Marks Question

Surface Areas and Volumes · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 17 questions.

Questions

1 Marks Question

🎯

Test yourself on this topic

17 questions · timed · auto-graded

Question 11 Mark
A cylinder and a cone are of same base radius and of same height. Find the ratio of the volume of cylinder to that of the cone.
Answer
Let r be the base radius and h be the height.
Then, volume of the cylinder i.e.,
$\text{V}_1=\pi\text{r}^2\text{h}$
and volume of the cone i.e.,
$\text{V}_2=\frac{1}{3}\pi\text{r}^2\text{h}$
$\therefore\ \frac{\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}=\frac{3}{1}$
Hence, the required ratio is 3 : 1.
View full question & answer→
Question 21 Mark
The slant height of the frustum of a cone is 5cm. If the difference between the radii of its two circular ends is 4cm, write the height of the frustum.
Answer
Let $r$ and $R$ be the radii of the circular ends of the frustum of the cone.
Then, $R-r=4, I=5$
We know, $I ^2=( R - r )^2+ h ^2$
$\Rightarrow 52=42+h^2 \text { or } h^2=25-16=9$
$\Rightarrow h=3 cm$
View full question & answer→
Question 31 Mark
A solid metallic cuboid of dimensions $\text{9 m} \times 8 \text{ m} \times \text{2 m}$ is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.
Answer
No. of cubes $= \frac{9 \times 8 \times 2}{2 \times 2\times 2}$
= 18
View full question & answer→
Question 41 Mark
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Answer
$\frac{2}{3} \pi\text{r}^{3} = 3\pi\text{r}^{2}\Rightarrow \text{r} = \frac{9}{2} \text{units}$
$\therefore \text{d} = 9 \text{ units}$
View full question & answer→
Question 51 Mark
The slant height of a frustum of a cone is 4cm and the perimeters (circumferences) of its circular ends are 18cm and 6cm. Find the curved surface area of the frustum. $\Big[\text{Use } \pi = \frac{22}{7}\Big]$
Answer
Slant height l = 4cm, perimeters = 18cm and 6cm
Radii can be circulated as follows:
$\text{Radius}=\frac{\text{Perimeter}}{2\pi}$
Or, $\text{R}=\frac{18}{2\pi}=\frac{9}{\pi}$
And, $\text{r}=\frac{6}{2\pi}=\frac{3}{\pi}$
Curved surface area of frustum
$=\pi(\text{R+r})\text{l}$
$=\pi\Big(\frac{9}{\pi}+\frac{3}{\pi}\Big)\times4$
$=\pi\Big(\frac{9+3}{\pi}\Big) \times4$
$=\pi\times\frac{12}{\pi}\times4$
$=48\text{cm}^2$
Hence, curved surface area of frustum $= 48cm^2.$
View full question & answer→
Question 61 Mark
What is the arithmetic mean of first n natural numbers?
Answer
First n natural numbers are, 1, 2, 3, ..........n
Sum of these numbers $=1+2+3+.....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Hence arithmetic mean $=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{{\frac{\text{n}(\text{n}+1)}{2}}{}}{\text{n}}=\frac{\text{n}+1}{2}$
View full question & answer→
Question 71 Mark
Two right circular cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1, what is the ratio of their volumes?
Answer
$\text{V}_1=\text{V}_2=\frac{1}{3}\pi(3\text{r})^2\text{h}:\frac{1}{3}\pi\text{r}^2(3\text{h})$
$=3:1$
View full question & answer→
Question 81 Mark
Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?
Answer
$\frac{\text{r}_1}{\text{r}_2}=\frac31,\frac{\text{h}_1}{\text{h}_2}=\frac13$
$\therefore$ Ratio of volumes $=\frac{\frac13\pi\text{r}_1^2\text{h}_1}{\frac13\pi\text{r}_2^2\text{h}_2}=3:1.$
View full question & answer→
Question 91 Mark
Three solid spheres of radii 3, 4 and 5cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.
Answer
Let R be the radius of single solid sphere.
Therefore,
Volume of single solid sphere = volume of all three spheres
$\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\text{r}_1^3+\frac{4}{3}\pi\text{r}_2^3+\frac{4}{3}\pi\text{r}_3^3$
$\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\pi(\text{r}_1^3+\text{r}_2^3+\text{r}_3^3)$
$\text{R}^3=(3^3+4^3+5^3)$
$\text{R}^3=27+64+125$
$\text{R}^3=216$
$\text{R}=\sqrt{216}$
$=6$
$\text{R}=6$
View full question & answer→
Question 101 Mark
Find the number of solid spheres, each of diameter 6cm, that could be moulded to form a solid metallic cylinder of height 45cm and diameter 4cm.
Answer
We have,
Radius of the sphere, $\text{R}=\frac{6}{2}=3\text{cm},$
Radius of the cylinder, $\text{r}=\frac{4}{2}=2\text{cm}$ and
Height of the cylinder, h = 45cm
Now,
The number of solid spheres $=\frac{\text{Volume or the Cylinder}}{\text{Volume of the sphere}}$
$=\frac{\pi\text{r}^2\text{h}}{\Big(\frac{4}{3}\pi\text{R}^3\Big)}$
$=\frac{3\text{r}^2\text{h}}{4\text{R}^3}$
$=\frac{3\times2\times2\times45}{4\times3\times3\times3}$
$=5$
So, the number of solid spheres so moulded is 5.
View full question & answer→
Question 111 Mark
The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7cm, respectively. Find the curved surface area of the bucket.
Answer
Let R and r be the radii of the top and base of the bucket, respectively, and let be its slant height.
Then, curved surface area of the bucket = Curved surface area of the frustum of the cone.
$=\pi\text{l}(\text{R}+\text{r})$
$=\frac{22}{7}\times45(28\times7)\text{cm}^2$
$=\Big(\frac{22}{7}\times45\times35\Big)\text{cm}^2$
$=4950\text{cm}^2$
View full question & answer→
Question 121 Mark
Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?
Answer
Let the radii of the two cylinders be r and R; and the heights be h and H.
We have,
$\frac{\text{h}}{\text{H}}=\frac{1}{2}\ ...(1)$
Now,
Volume of the first cylinder = Volume of the second sphere
$\Rightarrow\pi\text{r}^2\text{h}=\pi\text{R}^2\text{H}$
$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{\text{R}^2}{\text{r}^2}$
$\Rightarrow\frac{1}{2}=\frac{\text{R}^2}{\text{r}^2}$
$\Rightarrow\frac{\text{r}^2}{\text{R}^2}=\frac{1}{2}$
$\Rightarrow\Big(\frac{\text{r}}{\text{R}}\Big)^2=\frac{2}{1}$
$\Rightarrow\frac{\text{r}}{\text{R}}=\frac{\sqrt{2}}{1}$
$\therefore\text{r}:\text{R}\sqrt{2}:1$
So, the ratio of their radii is $\sqrt{2}:1.$
View full question & answer→
Question 131 Mark
A circus tent is cylindrical to a height of 4m and conical above it. If its diameter is 105m and its slant height is 40m, then find the total area of the canvas required.
Answer
We have,
Height of the cylindrical part, H = 4m,
Radius of the base, $\text{r}=\frac{105}{2}\text{m}$ and
Slant height of the conical part, l = 40m
Now,
The total area of canvas required = CSA of conical part + CSA of cylindrical par
$=\pi\text{r}\text{l}+2\pi\text{r}\text{H}$
$=\pi\text{r}(\text{l}+2\text{H})$
$=\frac{22}{7}\times\frac{105}{2}\times40+2\times4$
$=11\times15\times48$
$=7920\text{m}^2$
So, the area of the canvas required to make the tent is $7920 m^2.$
View full question & answer→
Question 161 Mark
If the radius of base of a right circular cone is 5cm and height is 12cm then find the slant height of the cone.
Answer
Substitute the given values into the formula :
$\begin{array}{l}l^2=r^2+h^2 \\ l^2=(5 cm)^2+(12 cm)^2\end{array}$
$l=13 cm$
The slant height of the cone is 13 cm.
View full question & answer→
Question 171 Mark
Find the length of the diagonal of a cube of side 10 cm.
Answer
Formula for the diagonal of a cube : $d=a \sqrt{3}$ = $ 10 \sqrt{3} $ cm 
View full question & answer→
Generate Paper FreeAll Surface Areas and Volumes topics
1 Marks Question - Maths STD 10 Questions - Vidyadip