Download App

MATHS [2 Mark Questions]

Mensuration · Tamil Nadu Board STD 10 (Tamil - English Medium). 28 questions.

Questions

[2 Mark Questions]

🎯

Test yourself on this topic

28 questions · timed · auto-graded

Question 12 Marks
A hemispherical hollow bowl has material of volume cubic $\frac{436 \pi}{3}$ cubic $cm$. Its external diameter is 14 $cm$. Find its thickness
Answer
External radius of a hemisphere $(R)=7 cm$

Volume of a hemi-spherical bowl $=\frac{436 \pi}{3} cm ^3$
$\frac{2}{3} \pi\left( R ^3- r ^3\right)=\frac{436 \pi}{3}$
$\frac{2}{3} \pi\left(7^3- r ^3\right)=\frac{436 \pi}{3}$
$\frac{2}{3}\left(343- r ^3\right)=\frac{436}{3}$
$343-r^3=\frac{436}{3} \times \frac{3}{2}$
$343-r^3=218$
$343-218=r^3$
$125=r^3$
$\Rightarrow 5^3=r^3$
$r=5$

Internal radius $=5 cm$
Thickness of the hemisphere
= (7 – 5) cm
= 2 cm
View full question & answer→
Question 22 Marks
The slant height of a frustum of a cone is 4 m and the perimeter of circular ends is 18 m and 16 m. Find the cost of painting its curved surface area at ₹ 100 per sq. m
Answer
$
\begin{aligned}
& \text { Slant height of a frustum }( l )=4 m \\
& 2 \pi R =18 \\
& 2 \times \frac{22}{7} \times R =18 \\
& R =\frac{18 \times 7}{2 \times 22} \times \frac{63}{22} \\
& \text { Perimeter of the bottom }=16 m \\
& 2 \pi r =16 \\
& 2 \times \frac{22}{7} \times r =16 \\
& r =\frac{16 \times 7}{2 \times 22}=\frac{28}{11}
\end{aligned}
$
C.S.A of the frustum $=\pi I(R+r)$ sq.units
$
=\frac{22}{7} \times 4\left(\frac{63}{22}+\frac{28}{11}\right)
$

$
\begin{aligned}
& =\frac{22}{7} \times 4\left(\frac{63+56}{22}\right) \\
& =\frac{22}{7} \times 4 \times \frac{119}{22} \\
& =4 \times 17 \\
& =68 \text { sq.units }
\end{aligned}
$

Cost of painting
$
\begin{aligned}
& =\text{₹} 100 \times 68 \\
& =\text{₹} 6800
\end{aligned}
$
View full question & answer→
Question 32 Marks
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder
Answer
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be x
Volume of the solid cylinder = Volume of the hollow cylinder
$\begin{aligned} & \pi r^2 H=\pi h\left(R^2-r^2\right) \\ & x^2 \times 12=4\left(4.3^2-1.1^2\right) \\ & 12 x^2=4 \times 5.4 \times 3.2 \\ & 12 x^2=4(4.3+1.1)(4.3-1.1) \\ & x^2=\frac{4 \times 5.4 \times 3.2}{12} \\ & =\frac{4 \times 54 \times 32}{12 \times 100} \\ & =\frac{54 \times 32}{3 \times 100} \\ & =\frac{18 \times 32}{100} \\ & x=\sqrt{\frac{18 \times 32}{100}} \\ & =\sqrt{\frac{2 \times 9 \times 2 \times 16}{100}} \\ & =\frac{2 \times 3 \times 4}{10}\end{aligned}$
= 2.4 cm
Diameter of the solid cylinder
= 2 × 2.4
= 4.8 cm
View full question & answer→
Question 42 Marks
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer
Radius of a cone = Radius of a hemisphere = r unit

Height of a cone = r units
(height of the cone = radius of a hemisphere)
Maximum volume of the cone
$
\begin{aligned}
& =\frac{1}{3} \pi r^2 \text { cu.unis } \\
& =\frac{1}{3} \pi r^2 \times r \text { cu.units } \\
& =\frac{1}{3} \pi r^3 \text { cu.units }
\end{aligned}
$

Maximum volume of the cone
$
=\frac{1}{3} \pi r^3 \text { cu.units }
$
View full question & answer→
Question 52 Marks
A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer
Radius of the hemispherical tank $=1.75 m$
Volume of the tank $=\frac{2}{3} \pi r ^3$ cu.units

$\begin{aligned} & =\frac{2}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 1.75 \\ & =\frac{2}{3} \times \frac{22}{7} \times \frac{175 \times 175 \times 175}{100 \times 100 \times 100} m ^3 \\ & =\frac{2}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times \frac{7}{4} \\ & =\frac{11 \times 49}{3 \times 4 \times 4} m ^3 \\ & =\frac{11 \times 49}{48} m ^3 \\ & =\frac{11 \times 49}{48} \times 1000 \text { litres } \\ & =11229.17 \text { litres } \\ & \text { Time taken }=\frac{11229.17}{7} \\ & =1604.17 \text { seconds } \\ & =26.74 \text { minutes } \\ & =27 \text { minutes } \ldots(\text { approximately) }\end{aligned}$
View full question & answer→
Question 62 Marks
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder
Answer
Radius of a sphere $(r)=6 cm$
External radius of the cylinder $( R )=5 cm$
Height of the cylinder $(h)=32 cm$
Let the internal radius of the cylinder be " $x$ "
Volume of the hollow cylinder $=$ Volume of a sphere
$\pi h\left(R^2-r^2\right)=\frac{4}{3} \pi r^3$
$\pi \times 32(5+x)(5-x)=\frac{4}{3} \times \pi \times 6 \times 6 \times 6$
$32\left(25-x^2\right)=4 \times 2 \times 6 \times 6$
$
25-x^2=9
$
$
x^2=25-9=16
$
$
x=\sqrt{16}=4
$
Thickness of the cylinder $=5-4=1 cm$.
View full question & answer→
Question 72 Marks
The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder
Answer
Internal radius of the shell $(r)=3 cm$
External radius of the shell $(R)=5 cm$
Radius of the cylinder $(r)=7 cm$
Let the height of the cylinder be " $h$ "

Volume of the cylinder $=$ Volume of the hemispherical shell
$
\begin{aligned}
& \pi r^2 h=\frac{2}{3} \pi\left( R ^3- r ^3\right) \\
& \pi \times 7 \times 7 \times h =\frac{2}{3} \pi\left(5^3-3^3\right) \\
& 7 \times 7 \times h =\frac{2}{3}(125-27) \\
& 7 \times 7 \times h =\frac{2}{3} \times 98 \\
& h =\frac{2 \times 98}{3 \times 7 \times 7}
\end{aligned}
$
$
\begin{aligned}
& =\frac{4}{3} \\
& =1.33 cm
\end{aligned}
$
Height of the cylinder $=1.33 cm$
View full question & answer→
Question 82 Marks
Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially
Answer
Length of the cuboid tank $( I )=2 cm =200 cm$
Breadth of the cuboid tank $(b)=1.5 cm =150 cm$
Height of the tank $(h)=1 m =100 cm$

Volume of the cuboid $=1 \times b \times h$ cu.units
$
\begin{aligned}
& =200 \times 150 \times 100 cm ^3 \\
& =30,00,000 cm ^3
\end{aligned}
$

Radius of the tank $(r)=60 cm$
Height of the tank $(h)=105 cm$
Volume of the cylindrical tank $=\pi r^2 h$ cu.units
$
\begin{aligned}
& =\frac{22}{7} \times 60 \times 60 \times 105 cm ^3 \\
& =22 \times 60 \times 60 \times 15 cm ^3 \\
& =1188000 cm ^3
\end{aligned}
$
Volume of water left in the sump = Volume of the sump – Volume of the tank
$=3000000-1188000 cm^3$
$=1812000 cm^3$
View full question & answer→
Question 92 Marks
A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter
Answer
Radius of a cone $( V )=7 cm$
Height of a cone $(h)=8 cm$
External radius of the hollow sphere $(R)=5 cm$
Let the internal radius be " $x$ "

Volume of the hollow sphere $=$ Volume of the Cone
$
\begin{aligned}
& \frac{4}{3} \pi\left( R ^3- r ^3\right)=\frac{1}{3} \pi r ^2 h \\
& \frac{4}{3} \pi\left(5^3-x^3\right)=\frac{1}{3} \times \pi \times 7 \times 7 \times 8 \\
& 4\left(125- x ^3\right)=7 \times 7 \times 8 \\
& 500-4 x ^3=7 \times 7 \times 8 \\
& 4 x ^3=500-7 \times 7 \times 8 \\
& =500-392 \\
& 4 x ^3=108
\end{aligned}
$
$
\begin{aligned}
& x^3=\frac{108}{4}=27 \\
& x^3=3^3 \\
& \Rightarrow x=3
\end{aligned}
$
Internal diameter of the Hollowsphere $=2 \times 3=6 cm$.
View full question & answer→
Question 102 Marks
A conical flask is full of water. The flask has base radius r units and height h units, the water is poured into a cylindrical flask of base radius xr units. Find the height of water in the cylindrical flask.
Answer
Radius of the conical flask $=r$ units
Height of the conical flask $=h$ units
Volume of the conical flask $=\frac{1}{3} \pi r ^2 h$ cu.units
Radius of the cylindrical flask $=x r$ units
Let the height of the cylindrical flask be " $H$ " units

Volume of the cylindrical flask $=$ Volume of the Conical flask
$
\begin{aligned}
& \pi \times(x r)^2 \times H=\frac{1}{3} \pi r^2 h \\
& x^2 r^2 \times H=\frac{1}{3} \pi r^2 h \\
& H=\frac{h}{3 x^2}
\end{aligned}
$
Height of the cylindrical flask $=\frac{ h }{3 x^2}$ units
View full question & answer→
Question 112 Marks
An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder
Answer
$\begin{aligned} & \text { Sphere }- \text { Radius } r _1=12 cm \\ & \text { Cylinder }- \text { Radius } r _2=8 cm \\ & h _2=? \\ & \text { Volume of cylinder }=\text { Volume of sphere melted } \\ & \pi r _2^2 h _2=\frac{4}{3} \pi r_1^3 \\ & \frac{22}{7} \times 8 \times 8 \times h _2=\frac{4}{3} \times \frac{22}{7} \times 12 \times 12 \times 12 \\ & h _2=\frac{4}{3} \times 12 \times 12 \times 12 \times \frac{1}{8} \times \frac{1}{8} \\ & \therefore \text { Height of the cylinder made }=36 cm .\end{aligned}$
View full question & answer→
Question 122 Marks
A right circular cylinder just encloses a sphere of radius r units. Calculate the curved surface area of the cylinder
Answer
C.S.A of cylinder

$=2 \pi r \times h$
$=2 \pi r \times 2 r$
$=4 \pi r^2 \text { sq.units }$
View full question & answer→
Question 132 Marks
As shown in the figure a cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid.
Answer
Side of a cube $(a)=7 cm$
Radius of a hemisphere $( r )=\frac{7}{2} cm$
Surface area of the solid = T.S.A of the cube + C.S.A of the hemisphere - Area of the base of the hemisphere
$
\begin{aligned}
& =\left(6 \times 7^2+2 \times \frac{22}{7} \times\left(\frac{7}{2}\right)^2-\frac{22}{7} \times\left(\frac{7}{2}\right)^2\right) cm ^2 \\
& =\left(294+77-\frac{77}{2}\right) cm ^2 \\
& =\left(294+\frac{77}{2}\right) cm ^2 \\
& =332.5 cm ^2
\end{aligned}
$
View full question & answer→
Question 142 Marks
A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm
Answer
$
\begin{aligned}
& \text { Radius of a cone }=\text { Radius of a hemisphere }=\text { Radius of a cylinder } \\
& r =6 cm \\
& \text { Height of a cone }( h )=12 cm \\
& \text { Volume of the water displaced }=\text { Volume of the solid inside }=\text { Volume of the cone }+ \text { Volume } \\
& \text { of the hemisphere } \\
& =\frac{1}{3} \pi r ^2 h +\frac{2}{3} \pi r ^3 \\
& =\frac{1}{3} \pi r ^2( h +2 r ) cm ^3 \\
& =\frac{1}{3} \times \frac{22}{7} \times 6 \times 6(12+12) cm ^3 \\
& =\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 24 cm ^3
\end{aligned}
$
Volume of water displaced $=905.14 cm ^3$
View full question & answer→
Question 152 Marks
A right-angled triangle whose sides are 6 cm, 8 cm and 10 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two solids so formed.
Answer
Three sides of a triangle are $6 cm , 8 cm$ and $10 cm$.
Case (i): If the triangle is revolved about the side $6 cm$, the cone will be formed with radius 6 $cm$ and height $8 cm$.

Volume of the cone $=\frac{1}{3} \pi r ^2 h$ cu. units
$
\begin{aligned}
& =\frac{1}{3} \times \pi \times 6 \times 6 \times 8 \\
& =96 \pi cm ^3
\end{aligned}
$

Case (ii): If the triangle is revolved about the side $8 cm$, the cone will be formed with radius 8 $cm$ and height $6 cm$.

Volume of the cone $=\frac{1}{3} \times \pi \times 8 \times 8 \times 6$
$
=128 \pi cm ^3
$
Difference in volume of the two solids
$
\begin{aligned}
& =(128 \pi-96 \pi) cm ^3 \\
& =32 \pi cm ^3
\end{aligned}
$
$
\begin{aligned}
& =32 \times \frac{22}{7} cm ^3 \\
& =100.57 cm ^3
\end{aligned}
$
The difference in the volume of the two solids $=100.57 cm ^3$
View full question & answer→
Question 162 Marks
A conical container is fully filled with petrol. The radius is 10 m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu. meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.
Answer
The radius of the conical container $(r)=10 m$
Height of the container $(h)=15 m$
Volume of the container $=\frac{1}{3} \pi r ^2 h$ cu.units
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 10 \times 10 \times 15 cu . m \\
& =\frac{22 \times 10 \times 10 \times 5}{7} \text { cu.m }
\end{aligned}
$
Time taken $=\frac{\text { Volume of the container }}{\text { Volume of petrol release in one minute }}$

Time taken $=\frac{22 \times 10 \times 10 \times 5}{7 \times 25}$ minutes
$
=\frac{22 \times 10 \times 2}{7} \text { minutes }
$
$=62.86$ minutes
$=63$ minutes..$($ approximate $)$
View full question & answer→
Question 172 Marks
If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
Answer
Circumference of the wooden piece $=484 cm$
$
\begin{aligned}
& 2 \pi r=484 \\
& 2 \times \frac{22}{7} \times r=484 cm \\
& r=\frac{484 \times 7}{2 \times 22} \\
& r=77 cm
\end{aligned}
$

Height of the wooden piece $(h)=105 cm$
Volume of the conical wooden piece $=\frac{1}{3} \pi r ^2 h$ cu.units
$
\begin{aligned}
& =\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 cm ^3 \\
& =22 \times 11 \times 77 \times 35 cm ^3 \\
& =652190 cm ^3
\end{aligned}
$
Volume of the wooden piece $=652190 cm ^3$
View full question & answer→
Question 182 Marks
A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the raise of the water in the glass?
Answer
Radius of the cylindrical glass $(r)=10 cm$
Height of the water $(h)=9 cm$
Radius of the cylindrical metal $(R)=5 cm$
Height of the metal $( H )=4 cm$
Let the height of the water raised be " $h$ "
Volume of the water raised in the cylinder $=$ Volume of the cylindrical metal
$
\begin{aligned}
& \pi r^2 h=\pi r^2 H \\
& 10 \times 10 \times h=5 \times 5 \times 4 \\
& h=\frac{5 \times 5 \times 4}{10 \times 10}=1 cm
\end{aligned}
$
Raise of water in the glass $=1 cm$
View full question & answer→
Question 192 Marks
A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.
Answer
Radius of the well $\left(r_1\right)=5 m$
Depth of the well $(h)=14 m$
Width of the embankment $=5 m$
Outer radius $(R)=5+5=10 m$
Let the height of the embankment be $H$
Volume of Earth in the embankment $=$ Volume of the well
$
\begin{aligned}
& \pi H\left(R^2-r^2\right)=\pi r_1^2 h \\
& H\left(10^2-5^2\right)=5 \times 5 \times 14 \\
& H(100-25)=5 \times 5 \times 14 \\
& H=\frac{5 \times 5 \times 14}{75} \\
& =4.67 m
\end{aligned}
$
Height of the embankment $=4.67 m$
View full question & answer→
Question 202 Marks
A container open at the top is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends are 8 cm and 20 cm respectively. Find the cost of milk which can completely fill a container at the rate of ₹ 40 per litre.
Answer
Height of the frustum $(h)=16 cm$
Radius of the upper part $(R)=20 cm$
Radius of the lower part $(r)=8 cm$
Volume of the frustum
$
\begin{aligned}
& =\frac{1}{3} \pi h \left[ R ^2+ r ^2+ Rr \right] \text { cu.units } \\
& =\frac{1}{3} \times \frac{22}{7} \times 16\left[20^2+8^2+20 \times 8\right] \text { cu.units } \\
& =\frac{1}{3} \times \frac{22}{7} \times 16[400+64+160] \text { cu.units } \\
& =\frac{22 \times 16 \times 624}{3 \times 7} cm ^3 \\
& =\frac{22 \times 16 \times 208}{7} cm ^3 \\
& =\frac{73216}{7}
\end{aligned}
$
$
\begin{aligned}
& =10459.43 cm ^3 \\
& =\frac{10459.43}{1000} \text { litre } \\
& =10.45943 \text { litre } \\
& =10.459 \text { litre }
\end{aligned}
$
Cost of milk in the container $=10.459 \times 40=\text{₹} 418.36$
Cost of the milk $=\text{₹} 418.36$
View full question & answer→
Question 212 Marks
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at ₹ 0.14 per cm$^2$
Answer
Internal radius $(r)=\frac{20}{2}=10 cm$
External radius $( R )=\frac{28}{2}=14 cm$
T.S.A of a hollow hemisphere
$
\begin{aligned}
& =\pi\left(3 R ^2+ r ^2\right) \text { sq.units } \\
& =\frac{22}{7}\left[3 \times 14^2+10^2\right] cm ^2 \\
& =\frac{22}{7}[588+100] cm ^2 \\
& =\frac{22}{7} \times 688 cm ^2
\end{aligned}
$

Cost of painting the vessel
$
\begin{aligned}
& =\text{₹} 0.14 \times \frac{22}{7} \times 688 \\
& =\text{₹} \frac{14}{100} \times \frac{22}{7} \times 688
\end{aligned}
$
$
\begin{aligned}
& =\text{₹} \frac{2 \times 22 \times 688}{100} \\
& =\text{₹} \frac{30272}{100} \\
& =\text{₹} 302.72
\end{aligned}
$
Cost of painting $=\text{₹} 302.72$
View full question & answer→
Question 222 Marks
The ratio of the radii of two right circular cones of the same height is 1 : 3. Find the ratio of their curved surface area when the height cone is 3 times the radius of the smaller cone.
Answer
Let the radius of the first cone be ' $x$ ' and the Height of the cone be $3 x$
$
\begin{aligned}
& I=\sqrt{h^2+r^2} \\
& =\sqrt{(3 x)^2+x^2} \\
& =\sqrt{10 x^2}
\end{aligned}
$
C.S.A. of the first cone $=\pi r$ sq.units
$
\begin{aligned}
& =\pi \times x \sqrt{10 x^2} \\
& =\pi x^2 \sqrt{10}
\end{aligned}
$

The radius and the height of the second cone is $3 x$...(Given)
$
\begin{aligned}
& I=\sqrt{(3 x)^2+(3 x)^2} \\
& =\sqrt{9 x^2+9 x^2} \\
& =\sqrt{18 x^2}
\end{aligned}
$

C.S.A of the second one
$
\begin{aligned}
& =\pi \times 3 x \times \sqrt{18 x^2} \\
& =\pi \times 3 x^2 \times \sqrt{9 \times 2} \\
& =\pi 9 x^2 \sqrt{2}
\end{aligned}
$

Ratio of the curves surface area
$
\begin{aligned}
& =\pi x^2 \sqrt{10}: 9 \pi x^2 \sqrt{2} \\
& =\sqrt{10}: 9 \sqrt{2} \\
& =\sqrt{2} \times \sqrt{5}: 9 \sqrt{2} \\
& =\sqrt{5}: 9
\end{aligned}
$
View full question & answer→
Question 232 Marks
A girl wishes to prepare birthday caps in the form of right circular cones for her birthday party, using a sheet of paper whose area is 5720 cm$^2$, how many caps can be made with radius 5 cm and height 12 cm.
Answer
$
\begin{aligned}
& \text { Radius of a cap }(r)=5 cm \\
& \text { Height of a cap }(h)=12 cm \\
& I=\sqrt{ h ^2+ r ^2} \\
& =\sqrt{12^2+5^2} \\
& =\sqrt{144+25} \\
& =\sqrt{169} \\
& =13
\end{aligned}
$
C.S.A. of a cap $=\pi r l$ sq. units
$
\begin{aligned}
& =\frac{22}{7} \times 5 \times 13 cm ^2 \\
& =\frac{22 \times 65}{7} cm ^2
\end{aligned}
$

Total Area of a sheet of paper $=5720 cm ^2$
$\begin{aligned} & \text { Number of caps }=\frac{\text { Total area of the paper }}{\text { Area of paper for one cap }} \\ & =\frac{5720 \times 7}{22 \times 65} \\ & =28 \text { caps }\end{aligned}$
View full question & answer→
Question 242 Marks
4 persons live in a conical tent whose slant height is 19 cm. If each person requires 22 cm$^2$ of the floor area, then find the height of the tent
Answer
Slant height of a cone $(r)=19 cm$

Floor area for 4 persons $=4 \times 22 cm ^2$
$
\begin{aligned}
& \pi r^2=88 cm \\
& \frac{22}{7} \times r ^2=88 \\
& r ^2=\frac{88 \times 7}{22} \\
& r ^2=28 \\
& h =\sqrt{ l ^2- r ^2} \\
& =\sqrt{19^2-28} \\
& =\sqrt{361-28} \\
& =\sqrt{333} \\
& =18.248
\end{aligned}
$
Height of the tent = 18.25 cm
View full question & answer→
Question 252 Marks
A right-angled triangle PQR where ∠Q = 90° is rotated about QR and PQ. If QR = 16 cm and PR = 20 cm, compare the curved surface areas of the right circular cones so formed by the triangle
Answer

In the Right Triangle
$
\begin{aligned}
& Q^2=P R^2-Q^2 \\
& =20^2-16^2 \\
& =400-256 \\
& =144 \\
& Q P=\sqrt{144}=12 cm
\end{aligned}
$

When $P Q$ is rotated $r=12, I=20$
C.S.A of the cone $=\pi r l$ sq. units
$
\begin{aligned}
& =\pi \times 12 \times 20 cm ^2 \\
& =240 \pi cm ^2
\end{aligned}
$

When $Q R$ is rotated $r=16, I=20$
C.S.A of the cone $= nrl$ sq. units
$
\begin{aligned}
& =\pi \times 16 \times 20 \\
& =320 \pi cm ^2
\end{aligned}
$
C.S.A. of a cone when rotated about $Q R$ is larger.
View full question & answer→
Question 262 Marks
The external radius and the length of a hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then find its T.S.A
Answer
External radius of the wooden $\log (R)=16 cm$
Thickness $=4 cm$
Internal radius $(r)=16-4=12 cm$

Length of the wooden $\log (h)=13 cm$
T.S.A of the hollow cylinder $=2 \pi(R+r)(R-r+h)$ sq.cm
$
\begin{aligned}
& =2 \times \frac{22}{7} \times(16+12)(16-12+13) \text { sq.cm } \\
& =2 \times \frac{22}{7} \times 28 \times 17 sq . cm \\
& =2 \times 22 \times 4 \times 17 sq \cdot cm . \\
& =2992 sq . cm .
\end{aligned}
$
T.S.A of the hollow wooden $=2992 sq . cm$
View full question & answer→
Question 272 Marks
The radius and height of a cylinder are in the ratio 5:7 and its curved surface area is 5500 sq.cm. Find its radius and height.
Answer
Let the radius be $5 x$ and the height be $7 x$
C.S.A of a cylinder $=5500 sq . cm$.
$
\begin{aligned}
& 2 \pi r h=5500 \\
& 2 \times \frac{22}{7} \times 5 x \times 7 x=5500 \\
& 2 \times 22 \times 5 \times x^2=5500 \\
& x^2=\frac{5500}{2 \times 22 \times 5} \\
& x^2=25 cm \\
& x=5 cm
\end{aligned}
$
Radius of the cylinder $=5 \times 5=25 cm$
Height of the cylinder $=7 \times 5=35 cm$
View full question & answer→
Question 282 Marks
The frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is ₹ 2.
Answer
Slant height of the frustum (I)
$
\begin{aligned}
& =\sqrt{ h ^2+( R - r )^2} \\
& =\sqrt{8^2+(12-6)^2} \\
& =\sqrt{64+(6)^2} \\
& =\sqrt{64+36} \\
& =\sqrt{100}
\end{aligned}
$

Slant height $=10 m$

Total Area to be painted $=$ C.S.A of the Frustum + top area
$
\begin{aligned}
& =\pi l(R+r)+\pi r^2 \text { sq.units } \\
& =\pi\left[l(R+r)+r^2\right] \\
& =\frac{22}{7}\left[10(12+6)+6^2\right]
\end{aligned}
$
$
\begin{aligned}
& =\frac{22}{7}[10 \times 18+36] cm ^2 \\
& =\frac{22}{7}[180+36] cm ^2 \\
& =\frac{22 \times 216}{7} \\
& =\frac{4752}{7} cm ^2 \\
& =678.86 cm ^2
\end{aligned}
$
Cost of painting $=\text{₹} 678.86 \times 2=\text{₹} 1357.72$
View full question & answer→
Generate Paper FreeAll Mensuration topics
[2 Mark Questions] - MATHS STD 10 Questions - Vidyadip