5 Marks Questions

Question 15 Marks

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer

For cone Radius of the top (r) = 5 cm
Height(h) = 8 cm
$\therefore $ Volume =$\frac{1}{3}\pi {r^2}h$
=$\frac{1}{3}\pi {\left( r \right)^2}8$
=$\frac{{200}}{3}\pi c{m^3}$
For spherical lead shot
Radius (R) = 0.5 cm
$\therefore $ The volume of a spherical lead shot =$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {\left( {0.5} \right)^3} = \frac{\pi }{6}c{m^3}$
The volume of water that flows out =$\frac{1}{4}$volume of the cone
=$\frac{1}{4}\left( {\frac{{200\pi }}{3}} \right)\,c{m^3} = \frac{{50\pi }}{3}\,c{m^3}$
Let the number of lead shot dropped in the vessel be 'n'.
Then, Volume of a lead shot = $\frac{{n\pi }}{6} = \frac{{50\pi }}{3}$
According to the questions, $\frac{{n\pi }}{6} = \frac{{50\pi }}{3}$
$\Rightarrow $ $n = \frac{{50\pi }}{3}\frac{6}{\pi } \Rightarrow n = 100$
Hence, the number of lead shot dropped in the vessel is 100.

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Question 25 Marks

A solid is consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm. It is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer

According to the question,A solid is consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm. It is placed upright in a right circular cylinder full of water such that it touches the bottom.
Given, height of cone, h = 120 cm, radius of cone r = 60 cm
Radius of hemisphere = 60 cm.
Volume of cone $= \frac { 1 } { 3 } \pi r ^ { 2 } h$
$= \frac { 1 } { 3 } \times 3.14 \times 60 \times 60 \times 120$
$= 3.14 \times 60 \times 60 \times 40$
$= 452160 \mathrm { cm } ^ { 3 }$
Volume of hemisphere=$\frac{2}{3}πr^3=\frac{2}{3} ×\frac{22}{7}×(60)^3$=452160
Total volume = Volume of cone + Volume of hemisphere
= 452160 + 452160
$= 904320 \mathrm { cm } ^ { 3 }$
Height of cylinder = 180 cm,
radius = 60 cm
Volume of water in the cylinder = Volume of cylinder
$= \pi r ^ { 2 } h$
$= 3.14 \times 60 \times 60 \times 180$
$= 2034720 \mathrm { cm } ^ { 3 }$
Water left in the cylinder = Volume of water - Volume of (cone + hemisphere)
= 2034720 - 904320
$= 1130400 \mathrm { cm } ^ { 3 }$

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Question 35 Marks

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamun, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Answer

Volume of a gulab jamun
$ = \frac{2}{3}\pi {(1.4)^3} + \pi {(1.4)^2}(2.2) + \frac{2}{3}\pi {(1.4)^3}$
$ = \frac{4}{3}\pi {(1.4)^3} + \pi {(1.4)^2}(2.2)$
$ = \pi {(1.4)^2}\left[ {\frac{{4 \times 1.4}}{3} + 2.2} \right]$
$ = \pi (1.96)\left[ {\frac{{5.6 + 6.6}}{3}} \right]$
$ = \frac{{\pi (1.96)(12.2)}}{3}c{m^3}$
$\therefore $ Volume of 45 gulab jamuns
$ = 45 \times \frac{{\pi (1.96)(12.2)}}{3}$
$ = 15\pi (1.96 \times 12.2)$
$ = 15 \times \frac{{22}}{7} \times 1.96 \times 12.2$
$ = 15 \times 22 \times 2.8 \times 12.2$
$= 1127.28 cm^3$
$\therefore $ Volume of syrup
$ = 1127.28 \times \frac{{30}}{{100}}$
$= 338.184 cm^3$
$= 338 cm^3 (approx.)$

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Question 45 Marks

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same).

Answer

For upper conical portion
Radius of the base $( r )=1.5 cm$
Height $\left(h_1\right)=2 cm$
$\therefore$ Volume $=\frac{1}{3} \pi r ^2 h_1=\frac{1}{3} \pi(1.5)^2(2)=1.5 \pi cm^3$

For lower conical portion
Volume $=1.5 \pi cm^3$.
For central cylindrical portion
Radius of the base $(r)=1.5 cm$
Height $\left(h_2\right)=12-(2+2)=12-4=8 cm$
$\therefore$ Volume $=\pi r ^2 h_2 \quad=\frac{1}{3} \pi(1.5)^2(8)=18 \pi cm^3$.
Therefore, volume of the model $=1.5 \pi+1.5 \pi+18 \pi=21 \pi=21 \times \frac{22}{7}=66 cm^3$.
Hence, the volume of the air contained in the model that Rechel made is $66 cm^3$.

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Question 55 Marks

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

Answer

For Hemisphere,
Radius(r) = 1 cm
$\therefore $ Volume $= \frac { 2 } { 3 } \pi r ^ { 3 }$
$= \frac { 2 } { 3 } \pi ( 1 ) ^ { 3 }$
$= \frac { 2 } { 3 } \pi \mathrm { cm } ^ { 3 }$
For cone,
Radius of the base(r) = 1 cm
Height (h) = 1 cm
$\therefore $ Volume$= \frac { 1 } { 3 } \pi r ^ { 2 } h$
$= \frac { 1 } { 3 } \pi ( 1 ) ^ { 2 } ( 1 ) = \frac { 1 } { 3 } \pi \operatorname { cm } ^ { 3 }$
Therefore, volume of the solid
=volume of the hemisphere + volume of cone
$= \frac { 2 } { 3 } \pi + \frac { 1 } { 3 } \pi = \pi \mathrm { cm } ^ { 3 }$

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Question 65 Marks

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $cm^2$.

Answer

Diameter of the solid cylinder = 1.4 cm
$\therefore$ Radius of the solid cylinder = 0.7 cm
$\therefore$ Radius of the base of the conical cavity = 0.7 cm
Height of the solid cylinder = 2.4 cm
$\therefore$ Height of the conical cavity = 2.4 cm
$\therefore$ Slant height of the conical cavity = $\sqrt{(0.7)^2\;+\;(2.4)^2\;}\;=\sqrt{0.49\;+5.76}=\;\sqrt{6.25}\;=\;2.5$ cm
$\therefore$ TSA of remaining solid
$=2 \pi(0.7)(2.4)+\pi(0.7)^2+\pi(0.7)(2.5)$
$=3.36 \pi+0.49 \pi+1.75 \pi$
$=5.6 \pi$
$=5.6 \times \frac{22}{7}$
$=17.6 cm^2=18 cm^2\left(\text { to the nearest } cm^2\right)$

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Question 75 Marks

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volume of the cylinder and toy. (Take $\pi$ = 3.14)

Answer

According to the question, A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm.
Let BPC is a hemisphere and ABC is a cone.
Radius of hemisphere = Radius of cone
$= \frac { 4 } { 2 } = 2 \mathrm { cm }$
h = Height of cone = 2 cm
Volume of toy $= \frac { 2 } { 3 } \pi r ^ { 3 } + \frac { 1 } { 3 } \pi r ^ { 2 } h$
$= \frac { 1 } { 3 } \pi r ^ { 2 } ( 2 r + h ) = \frac { 1 } { 3 } \times 3.14 \times 2 \times 2 ( 2 \times 2 + 2 )$
$= \frac { 1 } { 3 } \times 3.14 \times 4 \times 6$
$= 25.12 \mathrm { cm } ^ { 3 }$ ...(i)
Let right circular cylinder EFGH circumscribe the given solid toy.
Radius of cylinder = 2 cm,
Height of cylinder = 4 cm
Volume of right circular cylinder $= \pi r ^ { 2 } h$
$= 3.14 \times ( 2 ) ^ { 2 } \times 4 \mathrm { cm } ^ { 3 }$ ...(iii)
$= 50.24 \mathrm { cm } ^ { 3 }$
$\therefore$ Difference of two volume = Volume of cylinder - Volume of toy
$= 50.24 - 25.12$
$=25.12 cm^3$

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Question 85 Marks

A wooden toy rocket is in the shape of a cone mounted on a cylinder as shown in given below figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi $ =3.14)

Answer

TSA of the article = 2$\pi$rh + 2(2$\pi$$r^2$)
= 2$\pi$ (3.5)(10) + 2[2$\pi$ $(3.5)^2]$
= 70$\pi$ + 49$\pi$
= 119 $\pi$
= 119 $\times$ $\frac {22}7$
$= 374 cm^2$

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Question 95 Marks

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take $\pi = \frac {22}7$).

Answer

Surface area to colour = surface area of hemisphere + curved surface area of cone
Diameter of hemisphere = 3.5 cm
So radius of hemispherical portion of the lattu = r = $\frac { 3.5 } { 2 } \mathrm { cm }$ = 1.75
r = Radius of the concial portion = $\frac{3.5}2$ = 1.75
Height of the conical portion = height of top - radius of hemisphere = $$ 5 - 1.75 = 3.25 cm
Let I be the slant height of the conical part. Then,
$l^2=h^2+r^2$
$l^2=(3.25)^2+(1.75)^2$
$\Rightarrow l^2\;=10.5625+3.0625$
$\Rightarrow l^2=13.625$
$\Rightarrow l=\sqrt{13.625}$
$\Rightarrow l=3.69$
Let S be the total surface area of the top. Then,
$S = 2 \pi r ^ { 2 } + \pi r l$
$\Rightarrow \quad S = \pi r ( 2 r + l )$
$\Rightarrow S=\frac{22}7\times1.75(2\times1.75+3.7)$
$\;\;\;\;=\;5.5(3.5+3.7)$
$=5.5(7.2)$
$=39.6\;cm^2$

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Maths 5 Marks Questions

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