2 Marks Questions - MATHS STD 9 Questions - Vidyadip

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21 questions · timed · auto-graded

Question 12 Marks

Twenty$-$seven solid iron spheres, each of radius r and surface area $S$ are melted to form a sphere with surface area $S\ '$. Find the

radius $r\ '$ of the new sphere, and
ratio of $S$ and $S\ '.$

Answer

Volume of $27$ solid sphere, each of radius, $r = 27 \times \frac{4}{3} \pi r^3 $
$= 36 \pi r^3 $
According to the question,
Volume of sphere of radius $r\ ' =$ Volume of $27$ solid spheres
$\Rightarrow \frac{4}{3} \pi (r')^3 = 36 \pi r^3$
$\Rightarrow (r\ ')^3 = 27r^3 = (3r)^3$
$\Rightarrow r\ ' = 3r$
We have,
$S' = 4 \pi r\ '^2 = 4 \pi(3r)^2 $
$= 36 \pi r^2$
$\therefore$ $\frac{S}{S\ '}$ = $\frac{4 \pi r^2}{36 \pi r^2}$ = $\frac{1}{9}$
$\Rightarrow S : S\ ' = 1 : 9.$

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Question 22 Marks

Find the volume of a sphere whose surface area is $154 \ cm^2.$

Answer

Let the radius of the sphere be $r \ cm.$
Surface area $= 154 \ cm^2$
$\Rightarrow\ 4\pi r^2 = 154$
$\Rightarrow\ 4\times{\frac{22}{7}}\times r^2=154$
$\Rightarrow\ r^2={154\times\frac{7}{4} \times22}$
$\Rightarrow {r^2={\frac{49}{4}} }$
$\Rightarrow r=\sqrt{\frac{49}{4}}$
$\Rightarrow r={\frac{7}{2}}\ cm$
$\therefore$Volume of the sphere $={\frac{4}{3}}\pi r^3$
$={\frac{4}{3}}\times{\frac{22}{7}}\times({\frac{7}{2}})^2$
$={\frac{539}{3}}\ cm^3$
$=179{\frac{2}{3}}\ cm^3$

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Question 32 Marks

A hemispherical tank is made up of an iron sheet $1 \ cm$ thick. If the inner radius is $1\ m,$ then find the volume of the iron used to make the tank.

Answer

Inner radius of hemispherical tank $(r) = 1 m = 100 \ cm$
Thickness of sheet $= 1 \ cm$
$\therefore$ Outer radius of hemispherical tank $(R) = 100 + 1 = 101 \ cm$
Volume of iron of hemisphere $= \frac{2}{3} \pi [R^3 - r^3] \ cm^2$
$= \frac{2}{3}\times \frac{22}{7}\times \left[ {{\left( 101 \right)}^{3}}-{{\left( 100 \right)}^{3}} \right] cm^2$
= $\frac{44}{21} [1030301 - 1000000] \ cm^2$
$= 0.06348\ m^2$

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Question 42 Marks

The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the metal weighs 8.9 g per $cm^3$?

Answer

Diameter of metallic ball = 4.2 cm
$\therefore$ Radius of metallic ball $\left( r \right)$=$\frac{4.2}{2}$ = 2.1 cm
Volume of metallic ball =$\frac{4}{3}\pi {{r}^{3}}$=$\frac{4}{3}\times \frac{22}{7}\times 2.1\times 2.1\times 2.1$
=$\frac{4}{3}\times \frac{22}{7}\times \frac{21}{10}\times \frac{21}{10}\times \frac{21}{10} =38.808 c{{m}^{3}}$
Density of metal = 8.9 g per $c{{m}^{3}}$
DensityxVolume=Mass
$\because$ Mass of 1 $c{{m}^{3}}$ = 8.9 g
$\therefore$ Mass of 38.808 $c{{m}^{3}}4 =\text{ }8.9\times 38.808$ = 345.3912 g = 345.39 g (approx.)

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Question 52 Marks

A right triangle $\text{ABC}$ with slides $5 \ cm, 12 \ cm$ and $13 \ cm$ is revolved about the side $12 \ cm.$ Find the volume of the solid so obtained.

Answer

The solid obtained will be a right circular cone whose radius of the base is $5 \ cm.$ and height is $12 \ cm$
$\therefore r = 5 \ cm, h = 12 \ cm$
$\therefore$ Volume $={\frac{1}{3}}\pi r^2h$
${\frac{1}{3}}\times\pi \times (5)^2\times12\ cm^3$
$= 100\pi \ cm^3$
The volume of the solid so obtained is $100\pi \ cm^3$

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Question 62 Marks

A conical pit of top diameter $3.5 \ cm$ is $12\ m$ deep. What is its capacity in kilolitres?

Answer

For conical pit : Diameter $= 3.5 \ cm$.
$\therefore$ Radius $(r) = \frac{3.5}2\ m = 1.75\ m$
Depth $(h) = 12\ m$
$\therefore$ Capacity of the conical pit $={\frac13} \pi r^2h$
${\frac 1 3}\times{\frac {22}7}\times(1.75)^2\times12\ m^3$
$= 38.5\ m^3 = 38.5 \times 1000\ l$
$= 38.5\ kl.$

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Question 72 Marks

If the volume of a right circular cone of height $9 \ cm$ is $48\pi \ cm^3,$ find the diameter of its base.

Answer

Let the radius of the base of the right circular cone be $r \ cm.$
$h = 9 \ cm,$ volume $= 48\pi \ cm^3$
$\Rightarrow {{\frac13}\pi r^2h}= 48\pi$
$\Rightarrow {\frac13}r^2h=48$
$\Rightarrow {\frac13}\times r^2\times9=48$
$\Rightarrow r^2 = \frac{48\times3}9$
$\Rightarrow r^2 = 16$
$\Rightarrow r = \sqrt{16} = 4 \ cm$
$\Rightarrow 2r = 2(4) = 8 \ cm.$
$\therefore$ the diameter of the base of the right circular cone is $8 \ cm.$

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Question 82 Marks

The height of a cone is $15 \ cm.$ If its volume is $1570 \ cm^3.$ Find the radius of the base.

Answer

Let the radius of the base of the cone be $r \ cm.$
$h = 15 \ cm,$ Volume $= 1570 \ cm^3$
$\Rightarrow {\frac13}\pi r^2h$
$= 1570$
$\Rightarrow \frac13 \times 3.14 \times r^2 \times 15 $
$= 1570$
$\Rightarrow r^2={\frac{1570\times3}3.14\times15}$
$\Rightarrow r^2=100$
$\Rightarrow r=\sqrt{100}$
$\Rightarrow r = 10 \ cm.$
$\therefore$ the radius of the base of the cone is $10 \ cm.$

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Question 92 Marks

Find the capacity in litres of a conical vessel with height $12 \ cm,$ slant height $13 \ cm.$

Answer

$12 \ cm, l = 13 \ cm.$
$r^2 + h^2 = l^2$
$\Rightarrow r^2 + (12)^2 = (13)^2$
$\Rightarrow r^2 + 144 = 169$
$\Rightarrow r^2 = 169^– 144$
$\Rightarrow r^2 = 25$
$\Rightarrow r = \sqrt{25}$
$\Rightarrow r = 5 \ cm$
$\therefore$ Capacity $={\frac13}\pi r^2h$
$={\frac13}\times{\frac{22}7}\times(5)^2\times12$
$={\frac{2200}7}\ cm^3$
$={\frac{2200}{7000}}l$
$={\frac{11}{35}}l$

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Question 102 Marks

Find the capacity in litres of a conical vessel with radius $7 \ cm,$ slant height $25 \ cm.$

Answer

$r = 7 \ cm, l = 25 \ cm.$
$r^2 + h^2 = l^2$
$\Rightarrow (7)^2 + h^2 = (25)^2$
$\Rightarrow h^2 = (25)^2– (7)^2$
$\Rightarrow h^2 = 625 ^– 49$
$\Rightarrow h^2 = 576$
$\Rightarrow h = \sqrt{576}$
$\Rightarrow h = 24 \ cm$
$\therefore$ Capacity $={\frac13}\pi r^2h$
$={\frac13}\times{\frac{22}7}\times(7)^2\times24$
$= 1232 \ cm^3 $
$= 1.232\ l$

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Question 112 Marks

A right circular cylinder just encloses a sphere of radius r. Find

surface area of the sphere.
curved surface of the cylinder.
ratio of the area obtained in (i) and (ii).

Answer

Surface area of the sphere =$4\pi r^2$
For cylinder
Radius of the base = r
Height = 2r
$\therefore$ Curved surface area of the cylinder = $2\pi rh=2\pi (r)(2r)=4\pi r^2$
Ratio of the areas obtained in (i) and (ii)
$={{surface\ area\ of\ the\ sphere}\over{curved\ surface\ area\ of\ the\ cylinder}}$
$={{4\pi r^2}\over{4\pi r^2}}={1\over1}=1:1$

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Question 122 Marks

Find the radius of sphere whose surface area is $154 \ cm^2.$

Answer

Let the radius of the sphere be $r \ cm.$
Surface area $= 154 \ cm^2$
$\Rightarrow 4\pi r^2 = 154$
$\Rightarrow 4\times{\frac{22}7}\times r^2=154$
$\Rightarrow r^2={\frac{154\times7}4\times22}={\frac{49}4}$
$\Rightarrow r = {\sqrt{\frac{49}4}}={\frac72}$
$\Rightarrow r = 3.5 \ cm$
$\therefore$ the radius of the sphere is $3.5 \ cm.$

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Question 132 Marks

A hemispherical bowl made of brass has inner diameter $10.5 \ cm$. Find the cost of tin$-$plating it on the inside at the rate of $₹16$ per $100\ cm^2.$

Answer

Inner diameter of bowl $= 10.5 \ cm$
$\therefore $ Inner radius of bowl $\left( r \right)=\frac{10.5}{2} = 5.25 \ cm$
Now, Inner surface area of bowl $= 2\pi {{r}^{2}}$
$=2\times \frac{22}{7}\times 5.25\times 5.25$
$=2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}$
=$\frac{693}{4}c{{m}^{2}}$
$\because $ Cost of tin$-$plating per $100\text{ }c{{ m}^{2}} = ₹ 16$
$\therefore $ Cost of tin$-$plating per $1\text{ }c{{ m}^{2}}=\frac{16}{100}$
$\therefore $ Cost of tin$-$plating per $\frac{693}{4}c{{ m}^{2}}$
$=\frac{16}{100}\times \frac{693}{4} $
$= ₹27.72$

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Question 142 Marks

The radius of a spherical balloon increases from $7 \ cm$ to $14 \ cm$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Case $I : r = 7 \ cm$
Surface area $= {4\pi r^2}$
$= 4 \times \frac{22}7 \times (7)^2 $
$= 616 \ cm^2$
Case $II : r = 14 \ cm$
Surface area $= {4\pi r^2}$
$= 4 \times\ \frac {22}7 \times (14)^2 $
$= 2464 \ cm^2$
$\therefore$ Ratio of surface area of the balloon $= 616 : 2464 $
$=1:4$

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Question 152 Marks

The slant height and base diameter of a conical tomb are $25\ m$ and $14\ m$ respectively. Find the cost of white washing its curved surface at the rate of $₹ 210$ per $100\ m^2.$

Answer

Slant height$ (l) = 25\ m$
Base diameter $(d) = 14\ m$
$\therefore$ Base radius $(r) = \frac{14}2\ m = 7\ m$
$\therefore$ Curved surface area of the tomb $= \pi rl$
$= \frac{22}7 \times 7 \times 25 = 550\ m^2$
$\therefore$ Cost of white$-$washing the curved surface of the tomb at the rate of $Rs. 210$ per $100\ m^2$
$= \frac{210}{100} \times 550 $
$= Rs. 1155.$

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Question 162 Marks

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer

Slant height (l) = 21 m
Diameter of base = 24 m
∴ Radius of base (r) = $24\over2$m = 12 m
∴ Total curved surface area of the cone =$\pi r(l+r)$
= $22\over7$× 12 × (21 + 12)
=${22\over7}\times12\times33={8712\over7}$
=$1244{4\over7}\ m^2$

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Question 172 Marks

The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of $7\ m.$ Find the area available to the motorcyclist for riding.

Answer

We are given,
The diameter of the sphere $= 7 \ m$
Therefore, the radius is $3.5\ m$
So, the riding space available for the motorcyclist is the surface area of the ‘sphere’ which is given by
$4\pi r^2 $
$= 4 \times \frac {22}7 \times 3.5 \times 3.5\ m^2$
$= 154\ m^2$

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Question 182 Marks

The height of a cone is $16 \ cm$ and its base radius is $12 \ cm.$ Find the curved surface area and the total surface area of the cone $($Use $\pi = 3.14).$

Answer

We are given that,Here, $h = 16 \ cm$ and $r = 12 \ cm.$
So, from $l^2 = h^2 + r^2$,we have
$l = \sqrt {16^2 + 12^2} \ cm = 20 \ cm$
So, curved surface area $= \pi rl$
$= 3.14 \times 12 \times 20 \ cm^2 = 753.6 \ cm^2$
Further, total surface area $= \pi rl + \pi r^2 $
$= (753.6 + 3.14 \times 12 \times 12) \ cm^2$
$= (753.6 + 452.16) \ cm^2$
$= 1205.76 \ cm^2$

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Question 192 Marks

Hameed has built a cubical water tank with lid for his house, with each other edge $1.5\ m$ long. He gets the outer surface of the tank excluding the base, covered with square tiles of side $25 \ cm.$ Find how much he would spend for the tiles if the cost of tiles is $₹ 360$ per dozen.

Answer

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.
Edge of the cubic tank, $a = 1.5\ m = 150 \ cm$
So, surface area of the tank $= 5 \times 150 \times 150 \ cm^2.$
Area of each square title = $\frac{surface \ area \ of \ the \ tank}{area \ of \ each \ title}$ = $\frac{5 \times 150 \times 150}{25 \times 25} = 180$
Cost of $1$ dozen tiles, $i.e.,$ cost of $12$ tiles $= Rs. 360$
Therefore, cost of one tile $= Rs \frac{360}{12} = Rs. 30$
So the cost of 180 tiles $= 180 \times Rs.30 = Rs. 5400$

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Question 202 Marks

A shot$-$put is a metallic sphere of radius $4.9 \ cm$. If the density of the metal is $7.8\ g$ per $\ cm^3,$ find the mass of the shot-put.

Answer

Since the shot$-$putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere is given by $= \frac 43 \pi r^3$
$= \frac 43 \times \frac {22}7 \times 4.9 \times 4.9 \times 4.9 \ cm^3$
$= 493 \ cm^3 ($nearly$)$
Further, mass of $1 \ cm^3$ of metal is $7.8\ g.$
Therefore, mass of the shot$-$putt will be $= 7.8 \times 493\ g$
$= 3845.44 \ g $
$= 3.85 \ kg \ ($nearly$)$

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Question 212 Marks

The pillars of a temple are cylindrically shaped if each pillar has a circular base of radius 20cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?

Answer

Radius of base of cylinder = 20cm
Height of pillar = 10m = 1000cm
Volume of each cylinder = $\pi {{r}^{2}}h$
$=\frac{22}{7}\times 20\times 20\times 1000\text{ }c{{m}^{3}}$
$=\frac{8800000}{7}\text{ }c{{m}^{3}}$
$=\frac{8.8}{7}{{m}^{3}}\left[ \therefore 1000000c{{m}^{3}}=1{{m}^{3}} \right]$
$\therefore$ Volume of 14 pillars = volume of each cylinder $\times$ 14
$=\frac{8.8}{7}\times 14c{{m}^{3}}=17.6{{m}^{3}}$
So 14 pillars would need $17.6{{m}^{3}}$ of concrete mixture

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2 Marks Questions - MATHS STD 9 Questions - Vidyadip