Question

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╰──➤ANSWER THE FOLLOWING QUESTIONS-
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Answer 1

Answer:

1)Volume of cylinder =πr²hcm

Outer radius = 16/2=8 cm

Inner radius = 8−2=6 cm

Volume =π(r′²-r²)h, where r' = outer radius and r = inner radius

=π×100×(8²-6²)=8792cm³

Hence, 8792cm³ of iron has been used in making the tube.

2)Let the side of solid cube = a

TSA of cube = 6a²

When the cube is cut into two cuboid with equal volume, the cube will be cut into two equal halves through its side.

the cuboid thus formed will have dimension,

length (l)= a

breadth (b)= a

height (h)= a/2

TSA = 2 (lb + bh + lh)

= 2 (a² + a²/2 + a²/2)

= 2 (2a²)

= 4a²

Ratio = (TSA of cuboid) : (TSA of cube) = 4a² : 6a² = 2:3

The Ratio is 2:3.

3) As we know that the circumference of a circle is given as-

Circumference =πr

Whereas, r is the radius of circle

Diameter of circular sheet =28cm

∴ Radius of circular sheet =2/28=14cm

Therefore,

Circumference of circular sheet =14π

When a semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

Slant height of cup (l)= Radius of circular sheet =14cm

Circumference of the base of cone=circumference of circular sheet =14π

Let r be the radius of the base of cone

∴2πr=14

⇒r=7cm

Let h be the height of cup.

Therefore,

l² =r² +h²

(14)²=(7)²+h²

⇒h= √196−49 =√147=7√2cm

Now,

Capacity of cup = Volume of cone

As we know that, volume of cone is given as-

V=1/3 πr²h

Therefore,

Capacity of cup =1/3 × 22/7 ×(7)²×7√3=622.4cm³

Thus the capacity of the cup is 622.4cm³

4)Given :- Weight of the heavier sphere = 5920 g and weight of the lighter sphere = 740 g, and diameter of lighter sphere = 5 cm or radius = 2.5 cm

Let the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'. Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 2.5 cm.

Weight of an object = density × volume of that object

So,

(Weight of the heavier sphere/weight of the lighter sphere) = (Density × V1/density × V2)

As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio of their volumes.

(weight of the heavier sphere/weight of the lighter sphere) = (V1/V2)

⇒ (5920/740) = (4/3πr₁³)/(4/3πr³)

⇒ (5920/740) = (4/3 × 22/7× r₁³)/(4/3 × 22/7 × 2.5 × 2.5 × 2.5)

⇒ 8 = r₁³ × 1/2.5 × 1/2.5 × 1/2.5

⇒ 8 = r₁³ × 1/15.625

⇒ r₁³ = 8 × 15.625

⇒ r₁³ = 125

⇒ r₁ = 5 cm

So, the radius of the heavier sphere is 5 cm.

Hope it helps u!

Answer 2

Answer:

79.Volume of cylinder =$π{r}^{2}hc{m}^{2}$

Outer radius = $16/2=8 cm$

Inner radius = 8−2=6 cm

Volume =$π({r'}^{2} - {r}^{2})h$, where r' = outer radius and r = inner radius

=$π×100×({8}^{2} - {6}^{2})=8792{cm}^{3}$

Hence, $\large\underline\bold{8792c{m}^{3} }$of iron has been used in making the tube.

__________________________

80.Let the side of the cube be a.

Total surface area of a cube=$6{a}^{2}$

Length of the each resulting cuboid is half of the side of the cube =$a/2.$

Height and breadth of the cuboid remain same as the side of the cube a.

Total surface area of a cuboid =$2(l×b+b×h+l×h)$=$2( a/2×a+a×a+a/2×a)=4{a}^{2}.$

Ratio of the total surface area of the given cube and that of one of the cuboids =$6{a}^{2} :4{a}^{2}$

=$\large\underline\bold{3:2}$

__________________________

81. As we know that the circumference of a circle is given as-

Circumference =πr

Whereas, r is the radius of circle

Diameter of circular sheet =28cm

∴ Radius of circular sheet =$28/2$

=14 cm

Therefore,

Circumference of circular sheet =14π

[When a semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.]

Slant height of cup (l)= Radius of circular sheet =14cm

Circumference of the base of cone = circumference of circular sheet =14π

Let r be the radius of the base of cone

∴2πr=14

⇒r=7cm

Let h be the height of cup.

Therefore,

${l}^{2}={r}^{2}+{h}^{2}$

${(14)}^{2} ={(7)}^{2}+ {h}^{2}$

$⇒h= \sqrt{196−49}=\sqrt{147}$ $=7\sqrt{2} cm$

Now,

Capacity of cup = Volume of cone

As we know that, volume of cone is given as-

$V= 1/3 π{r}^{2} h$

Therefore,

Capacity of cup = $1/3 × 22/7×{(7)}^{2}×7\sqrt{3}$ =$622.4c{m}^{3}$

Thus the capacity of the cup is $\large\underline\bold{622.4c{m}^{3}.}$

__________________________

82.Mass is directly proportional to volume for same metal (Density).

Let Mass of Solid 1 be M1, Volume be V1 ,

Mass of Solid 2 be M2 and Volume be V2

Now

$\frac{M1}{M2}$ = $\frac{V1}{V2}$

Volume of sphere is directly proportional to ${R}^{3}$

So,

$\frac{M1}{M2}$=$\frac{V1}{V2}$= ${R1}^{3}/{R2}^{3}$

⇒ $5920/740$= ${R1}^{3}/{R2}^{3}$

⇒ ${R1}^{3}/{R2}^{3}$=8

⇒${R1}^{3}/{R2}^{3}$ =$\sqrt[3]{8}$

⇒ $R1/R2=2$

So, $R1 =R2×2$

So, $R1 =2.5×2=\large\underline\bold{5cm}.$

Hope this Helps Mi amigo!

¡Adiós!

P.S.Hope you have that the Amazing Day You deserve !