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(b) 17cm

Explanation:

We know that length of diagonal of the cuboid= √ (l²+b²+h²)

=√ (12²+9²+8²)

=√ (144+81+64)

=17

It is given that

Radius of lead ball = 1cm

Radius of sphere = 8cm

We know that

Number of lead balls = volume of sphere/ volume of one lead ball

So we get

Number of lead balls = (4/3 πR³)/ (4/3 πr³)

By substituting the values

Number of lead balls = (4/3 × (22/7) × 8³)/ (4/3 × (22/7) × 1³)

On further calculation

Number of lead balls = (4/3 × (22/7) × 512)/ (4/3 × (22/7) × 1)

We get

Number of lead balls = 512

Therefore, 512 lead balls can be made from the sphere.

The surface solid will be a sphere of radius r whose total surface area is 4πr².

Given,

A metallic block of dimension 11dm x 1m x 5dm

The diameter of each bullet = 5 cm

We know that,

Volume of the sphere = 4/3 πr³

Since, 1 dm = 10^-1m = 0.1 m

The volume of the rectangular block = 1.1 x 1 x 0.5 = 0.55 m³

Radius of the bullet = 5/2 = 2.5 cm

Let the number of bullets made from the rectangular block be n.

Then from the question,

The volume of the rectangular block = sum of the volumes of the n spherical bullets

0.55 = n x 4/3 π(2.5)³

Solving for n, we have

n = 8400

Therefore, 8400 can be cast from the rectangular block of metal.

Volume of the spherical ball of radius 8 cm = 4/3π x 8³c³ 

Also, volume of each smaller spherical ball of radius 1cm = 4/3π x 1³cm³.

Let n be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of n smaller balls.

Hence, 4/3π x n = 4/3π x 8³

=> n = 8³ = 512

Hence, the required number of balls = 512.

Solution:

False

Curved surface area of a hemisphere = 2 πr²

Here, two identical solid hemispheres of equal radius are stuck together. So, base of both hemispheres is common.

∴ Total surface area of the combination

= 2 πr² + 2 πr² = 4πr²

The correct option is (a) 4πr².

Explanation:

Because curved surface area of a hemisphere is 2 w² and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.

Hence, the curved surface area of new solid = 2 πr² + 2 πr² = 4πr².

Let radius of the third ball be r. 

Volume of the larger ball = \(\frac{4}{3}πR^3\) 

= \(\frac{4}{3}π{3}^3\)

Volume of larger ball = sum of volume of smaller balls

⇒ \(\frac{4}{3}π{3}^3\)= \(\frac{4}{3}π(1.5)^3 +{\frac{4}{3}π(2)^3} + {\frac{4}{3}πr^3}\)

⇒ r³ = 27 – 3.375 – 8 

⇒ r = 2.5 cm

Let R and r be the radii of the top and base of the bucket respectively,

Let h be its height.

Then, we have R = 20 cm, r = 10 cm, h = 30 cm

Capacity of the bucket = Volume of the frustum of the cone

= 1/3 π(R² + r² + R r )h

= 1/3 π(20² + 10² + 20 x 10 ) x 30

= 3.14 x 10 (400 + 100 + 200)

= 21980 cm³ = 21.98 litres

Now,

Surface area of the bucket = CSA of the bucket + Surface area of the bottom

= π l (R + r) + πr²

We know that,

l = √h² + (R – r)²

= √[30² + (20 – 10)²] = √(900 + 100)

= √1000 = 31.62 cm

So,

The Surface area of the bucket = (3.14) x 31.62 x (20 + 10) + (3.14) x 10²

= 2978.60 + 314

= 3292.60 cm²

Next, given that the cost of 1 litre milk = Rs 25

Thus, the cost of 21.98 litres of milk = Rs (25 x 21.98) = Rs 549.50

Given, 

h = 24 cm, r’ = 15 cm and r’’ = 5 cm 

Let slant height be l 

Slant height, l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(15 - 5)^2 + 24^2}\)

⇒ l = 26 cm 

Total surface area of the frustum = π(r’ + r’’)l+ πr’’² 

= π(15 + 5)(26) + π (5)² 

= 1711.3 cm² 

Cost of metal sheet used at the rate of Rs. 10 per 100 cm² 

= \(\frac{1711.3}{100}\times10\)

= Rs 171

Given,

Slant height of frustum of cone (l) = 4 cm

Let ratio of the top and bottom circles be r₁ and r₂

And given perimeters of its circular ends as 18 cm and 6 cm

⟹ 2πr₁ = 18 cm; 2πr₂ = 6 cm

⟹ πr₁= 9 cm and πr₂ = 3 cm

We know that,

Curved surface area of frustum of a cone = π(r₁ + r₂)l

= π(r₁ + r₂)l

= (πr₁+πr₂)l = (9 + 3) × 4 = (12) × 4 = 48 cm²

Therefore, the curved surface area of the frustum = 48 cm².

Given,

Height of the conical bucket = 45 cm

Radii of the 2 circular ends of the conical bucket are 28 cm and 7 cm

So, r₁ = 28 cm r₂ = 7 cm

Volume of the conical bucket = 1/3 π(r₁² + r₂² + r₁ r₂ )h

= 1/3 π(28² + 7² + 28 × 7)45 = 15435π

Therefore, the volume/ capacity of the bucket is 48510 cm³.

Given: 

for the bucket, 

Height = 45 cm 

r’ = 5 cm 

r’’ = 15 cm 

Curved surface area = π(r’ + r’’)l + π r’’² 

= π(5 + 15)(26) + π (15)² 

= 745 π cm² 

Curved surface area of bucket = 745π cm²

From the figure, diameter of the cone 5.2 m

Thus its radius ’r’ = 5.2/2 = 2.6 m

∴ Height of the cone = h = 6.8 m 

Volume of the cone = 1/3 πr²h

= 1/3 x 22/7 x 2.6 x 2.6 x 6.8

= 1011.296/21

= 48.156 m³

Quantity of water that flows per minute

= 1.8 m³

∴ Total time required = Total volume/1.8

= 48.156/1.8 = 26.753

27 minutes

We know that volume of the cylinder = πr²h

Here r = 5.6m h = 1.25m

V = 22/7 × 5.6 × 5.6 × 1.25

V = 123.2 m³

Also we know that curved surface area of cylinder = 2πrh

Curved surface area = 2 × 22/7 × 5.6 × 1.25

Curved surface area = 44 m²

We know that total surface area of cylinder = 2πr(r + h)

Total surface area = 2 × 22/7 × 5.6 (5.6 + 1.25)

Total surface area = 241.12 m²

We know that volume of the cylinder = πr²h

Here r = 7cm h = 50cm

V = 22/7 × 7 × 7 × 50

V = 22 × 7 × 50

V = 7700 cm³

Also we know that curved surface area of cylinder = 2πrh

Curved surface area = 2 × 22/7 × 7 × 50

Curved surface area = 2200 cm²

We know that total surface area of cylinder = 2πr(r + h)

Total surface area = 2 × 22/7 × 7 (7 + 50)

Total surface area = 2580 cm²

Volume = 753.60 cm³

Radius (r) = 12 cm

Height (h) = ?

Volume of cone = \(\frac{1}{3} \pi r^2 h\)

753.60 cm³ = \(\frac{1}{3} \times \frac{22}{7} \times 12 cm \times 12 cm \times h\)

\(\frac{753.60 cm^3 \times 3 \times 7}{22 \times 12 \times 12 cm^2}\) = h

\(\frac{15825.6 \,cm}{3168} = h\)

h = 5 cm

Hence, the height of the cone is 5 cm.

(c) 1200 π cm, 426.24 π² gm

One round of wire covers 4 mm = \(\frac4{10}\) cm in thickness of the surface of the cylinder 

Length of the cylinder = 24 cm

∴ Number of rounds to cover 24 cm = \(\frac{24}{\frac4{10}}=\frac{24\times10}{4}=60\) 

Diameter of the cylinder = 20 cm 

⇒ Radius of cylinder = 10 cm 

Length of the wire in completing one round 

= 2πr = 2π x 10 cm = 20 π cm.

∴ Length of the wire in covering the whole surface 

= Length of the wire in completing 60 rounds 

= (20 π × 60) cm = 1200 π cm. 

Radius of copper wire = 2 mm = \(\frac2{10}\) cm 

∴ Volume of wire = \(\big(π\times\frac2{10}\times\frac2{10}\times1200π\big)\)cm³ = 48 π² cm³

So, weight of wire = (48π² × 8.88) gm = 426.24 π² gm.

Given r = 7m and h=10 cm

But we know that volume of the cylinder = πr²h

V = 22/7 × 0.1 × 0.1 × 7

V = 0.22 cm³

Given weight of the wood 225kg per cubic meter

Weight of the pole = 0.22 × 225 = 49.5 kg

Area of the curved surface = Area of the rectangle = (11 × 8) cm² = 88 cm². 

⇒ 2πrh = 88 cm ⇒ 2 × \(\frac{22}{7}\times{r}\times8 = 88\) ⇒ r = \(\frac{88\times7}{44\times8}=\frac74\) cm. 

∴ Volume of the cylinder = πr²h = \(\frac{22}{7}\times\frac74\times\frac74\times8\) = 77 cm³ .

It is given that

Radius of cylinder = 10.5cm

Height of cylinder = 60cm

We know that

Volume of cylinder = πr²h

By substituting the values

Volume of cylinder = (22/7) × (10.5)² × 60

So we get

Volume of cylinder = 20790 cm³

So the weight of the cylinder if the material weighs 5 g per cm³ = 20790 × 5 = 103950g

We know that 1000g = 1kg

Weight of the cylinder = 103950/1000 = 103.95kg

Therefore, the weight of solid cylinder is 103.95kg.

We know that

Curved surface area = 1/3 × Total surface area

By substituting the values

Curved surface area = 1/3 × 462

So we get

Curved surface area = 154 cm²

So the total surface area – curved surface area = 462 – 154 = 308cm²

We know that

2 πr² = 308

By substituting the values

2 × (22/7) × r² = 308

On further calculation

r² = (308 × 7)/44

So we get

r² = 49

By taking square root

r = √49 = 7cm

We know that

Curved surface area = 2 πrh

By substituting the values

154 = 2 × (22/7) × 7 × h

So we get

h = 154/44

By division

h = 3.5 cm

So we get r = 7cm and h = 3.5 cm

We know that

Volume of the cylinder = πr²h

By substituting the values

Volume of cylinder = (22/7) × (7)² × 3.5

So we get

Volume of cylinder = 539 cm³

Therefore, volume of the cylinder is 539 cm³.

It is given that

Curved surface area = 1210 cm²

Diameter of the cylinder = 20cm

Radius of the cylinder = 20/2 = 10cm

We know that

Curved surface area of the cylinder = 2 πrh

By substituting the values

1210 = 2 × (22/7) × 10 × h

So we get

h = (1210 × 7) / (2 × 22 × 10) = 19.25 cm

We know that

Volume of cylinder = πr²h

By substituting the values

Volume of cylinder = (22/7) × (10)² × 19.25

So we get

Volume of cylinder = 6050 cm³

Therefore, the height of cylinder is 19.25 cm and the volume is 6050 cm³.

We know that

Curved surface area = 2/3 × Total surface area

By substituting the values

Curved surface area = 2/3 × 231

So we get

Curved surface area = 154 cm²

So the total surface area – curved surface area = 231 – 154 = 77cm²

We know that

2 πr² = 77

By substituting the values

2 × (22/7) × r² = 77

On further calculation

r² = (77 × 7)/44

So we get

r² = 49/4

By taking square root

r = √49/4 = 7/2cm

We know that

Curved surface area = 2 πrh

By substituting the values

154 = 2 × (22/7) × (7/2) × h

So we get

h = 154/22

By division

h = 7 cm

So we get r = 7/2 cm and h = 7 cm

We know that

Volume of the cylinder = πr²h

By substituting the values

Volume of cylinder = (22/7) × (7/2)² × 7

So we get

Volume of cylinder = 269.5 cm³

Therefore, volume of the cylinder is 269.5 cm³.

Consider r as the radius of the cone

It is given that

Slant height of the cone = 14cm

Curved surface area of the cone = 308cm²

It can be written as

πrl = 308

By substituting the values

(22/7) × r × 14 = 308

On further calculation

22 × r × 2 = 308

So we get

r = 308/ (22 × 2)

r = 7cm

We know that

Total surface area of a cone = πr (l + r)

By substituting the values

Total surface area of a cone = (22/7) × 7 × (14 + 7)

On further calculation

Total surface area of a cone = 22 ×21 = 462 cm²

Therefore, the base radius of the cone is 7cm and the total surface area is 462 cm².

Let r and h be the radius and height of the cylinder respectively. Then, after 1% increase, 

New radius = 1.1 r and new height = 1.1 h 

Original volume = πr²h , 

New volume = π(1.01r)² (1.01h) =1.030301 πr²h

∴ % increase = \(\bigg(\frac{1.030301πr^2h-πr^2h}{πr^h}\times100\bigg)\)% = \(\bigg(\frac{0.030301πr^2h}{πr^h}\times100\bigg)\)%

= (0.030301 × 100)% = 3.03%

Volume of solid cylinder V= πr²h

If initial radius and height are 100cm each then increased radius and height will be 101cm each.

Initial volume V₁=π×100×100²

^{Increased volume V₂}= π×101×101²

Percentage increase 

=[ (V2-V1)/V1]×100

=(101^3-100^3)/100^3×100

= 3.0301%

It is given that

Lateral surface area of a cylinder = 94.2 cm²

Height = 5cm

(i) We know that

Lateral surface area of cylinder = 2 πrh

By substituting the values

94.2 = 2 × 3.14 × r × 5

On further calculation

r = 94.2/ (2 × 3.14 × 5)

So we get

r = 3cm

(ii) We know that

Volume of a cylinder = πr²h

By substituting the values

Volume of a cylinder = (3.14) × (3)² × 5

So we get

Volume of a cylinder = 141.3 cm³

In the question it is given that 

Length of the cylinder = 14 m 

Which means, 

That the height of the cylinder = 14m 

Lateral surface area of the cylinder = \(2\pi rh\)

220 = \(2\times\frac{22}{7}\times r\times14\)

r = \(\frac{10}{4}\)

r = 2.5 m 

Hence, 

We can find the volume as, 

Volume of the cylinder = \(\pi r^2h\)

= \(\frac{22}{7}\times (2.5)^2\times14\)

= 22 × 2 × 2.5 × 2.5 

= 275 m³

Given that, 

Diameter = 14 cm 

So, Radius = 7 cm 

We know that, 

Curved surface area of cylinder = \(2\pi rh\)

220 cm² = \(2\pi rh\)

h = \(\frac{220\times7}{2\times22\times7}\)

h = 5 cm 

Therefore, 

Volume of cylinder = \(2\pi rh\)

= \(\frac{22}{7}\times7\times7\times5\)

= 770 cm³ 

Hence, option A is correct

Let the side of cube be ‘a’ 

Hence total surface area of cube = 6a² 

Cost of painting the cube = 6a² × 10 

264.6 = 60 a²

a² = \(\frac{264.6}{60}\)

a² = 4.41 

a = 2.1 

Hence, 

Volume of the cube = a³ 

= (2.1)³ 

= 9.261 cm³ 

Hence, option B is correct

Given details are,

Population of village = 4000

Dimensions of water tank = 20m × 15m × 6m

Water required per head per day = 150 litres

Total requirement of water per day = 150 × 4000 = 600000 litres

Volume of water tank = l × b × h

= 20 × 15 × 6

= 1800m³

= 1800000 litres

We know that,

Number of days water last in the tank = volume of tank / total requirement

= 1800000/600000

= 3 days

∴ Water in the tank last for 3 days.

Volume of cylindrical rod = Volume of the slab of ice 

⇒ π x (4)² x l = 8 x 11 x 2 (l = length of slab)

⇒ l = \(\frac{8\times11\times2\times7}{4\times4\times22}\) inches = 3.5 inches.

We know that, 

Curved surface area of the cylinder = \(2\pi rh\)

264 = \(2\pi rh\)

r = \(\frac{264\times7}{2\times22\times14}\)

r = 3 cm

We know that, 

Volume of cylinder = \(\pi r^2h\)

= \(\frac{22}{7}\times3\times3\times14\)

= 396 cm³ 

Hence, option B is correct

Length of cuboid water tank, l = 20 m 

breadth, b = 15 m 

height, h = 6 m 

Volume, V = ? 

Volume of water tank, V = l × b × h 

= 20 × 15 × 6 

= 1800 m³ . 

1m³ = 1000 litres 

∴ 1800m³ = 1800 x 1000 = 1800000 litres. 

Water required daily per man= 150 litres 

∴ Quantity of water required for 4000 men? 

= 4000 × 150 = 600000 litres. 

Quantity of water for 1 day = 600000 

Number of days for consuming 1800000 litres … ? … 

= \(\frac{1800000}{600000}\)

= 3 Days.

Let the height and radius of the cylinder be h and r respectively. Then,

\(\frac{\text{Total surface area}}{\text{lateral surface area}}\) = \(\frac{2πrh+2πr^2}{2πrh}\) = \(\frac{2πr(r+h)}{2πrh}\) = \(\frac{r+h}{h}\)

Given, \(\frac{r+h}{h}\) = \(\frac53\) ⇒ \(\frac{12+h}{h}\) = \(\frac53\) ⇒ 36 + 3h = 5h ⇒ 2h = 36 ⇒ h = 18 cm.

Correct option is (B) 2π xy cm²

r = x cm & h = y cm

\(\therefore\) Curved surface area of cylinder \(=2\pi rh\)

= \(2\pi\,xy\,cm^2\)

Correct option is  B) 2π xy cm² 

Volume of cuboidal tank, V = 50000 lit. 

1 c.c = 1000 litres 

∴ Volume of tank = \(\frac{50000}{1000}\) = 50 c.c 

length of tank, l = 2.5 m 

height, h = 10m 

breadth,b = ? 

Volume of cuboid, V = l × b × h 

50 = 2.5 x 10 x b

50 = 25 x b

b = \(\frac{50}{2.5}\)

∴ b = 2m.

Surface area of the sphere = 4πr²

⇒ 4πr² = 616 

⇒ 4 x \(\frac{22}7\) x r² = 616

⇒ r² = \(\frac{{616}\times{7}}{{4}\times{22}}\)

⇒ r² = 49 

⇒ r = \(\sqrt{4}9\) 

⇒ r = 7 cm 

Diameter of base = 7 × 2 

= 14 cm

Correct option is (C) Decreased by 50%

Let r and h be parameters of original cylinder & R and H be parameters of another cylinder.

Let R = 2r

Also \(2\pi RH=2\pi rh\)    (Given)

\(\Rightarrow\) \(H=\frac{2\pi rh}{2\pi R}=\frac{rh}{2r}=\frac h2\)   \((\because\) R = 2r)

\(\therefore\) Height is decreased by 50%.

C) Decreased by 50%

We know that, 

Volume of cuboid = Length × Breadth × Height 

576 = 3x × 4x × 6x 

576 = 72x³

x = \(\sqrt[3]{\frac{576}{72}}\)

= 2 

Therefore, 

Total surface area of cuboid = 2 (lb + bh + hl) 

= 2 (3x × 4x + 4x × 6x + 6x × 3x) 

= 2 (48 + 96 + 72) 

= 432 cm²

Volume of the cylinder, V = πr²h = 616 

Diameter of the tank = 5.6 m 

Thus its radius, r =d/2 =5.6/2 = 2.8 m

Height = h (say)

∴ πr²h = 616

22/7 × 2.8 × 2.8 × h = 616

h = \(\frac {616\times7}{22\times2.8\times2.8}\) = 25

∴ Height = 25 m

Volume of the cylinder V = πr² h = 308 cm 

Height of the cylinder h = 8 cm 

∴ 308 = 22/7 . r² x 8

r² = 308 × 7/22 x 1/8

r² = 12.25

∴ r = √12.25 = 3.5cm

L.S.A. = 2πrh

= 2 × 22/7 × 3.5 × 8 = 176cm²

T.S.A. = 2πr (r + h) 

2 × 22/7 × 3.5 (3.5 + 8)

= 2 × 22 × 0.5 × 11.5 = 253 cm²

Dimensions of the metal cuboid 

= 22 cm × 15 cm × 7.5 cm 

Height of the cylinder, h = 14 cm 

Cuboid made as cylinder 

∴ Volume of cuboid 

= Volume of cylinder lbh = 2πr²h

⇒ 22 × 15 × 7.5 = 22/7 × r² × 14

⇒ r² = 7.5 × 7.5 

r = 7.5 cm

Correct option is (A) 3 cm

Height of cuboid \(h=\frac V{lb}\)

\(=\frac{60}{5\times4}\) = 3 cm

Correct option is  A) 3 cm

Given that, 

Length = 18 cm 

Breadth = 10 cm 

Height = 8 cm 

We know that, 

Total surface area of cuboid = 2 (lb + bh + hl) 

= 2 (18 × 10 + 18 × 8 + 10 × 8) 

= 2 (180 + 144 + 80) 

= 808 cm²

Given,

Diameter of the coin = 1.75 cm

So, its radius = 1.74/2 = 0.875 cm

Thickness or the height = 2 mm = 0.2 cm

We know that,

Volume of the cylinder (V₁) = πr²h

= π 0.875² × 0.2

And, the volume of the cuboid (V₂) = 11 × 10 × 7 cm³

Let the number of coins needed to be melted be n.

So, we have

V₂ = V₁ × n

11 × 10 × 7 = π 0.875² × 0.2 x n

11 × 10 × 7 = 22/7 x 0.875² × 0.2 x n

On solving we get, n = 1600

Therefore, the number of coins required are 1600.

Volume of cuboid = l × b × h = 11 × 10 × 7 

Diameter of coin = 1.75 cm 

Radius = \(\frac{1.75}2\) cm = 0.875 cm 

Thickness of coin = 2 mm = 0.2 cm 

Volume of coin (cylinder) = πr²h

= (\(\frac{22}7\)) × 0.875 × 0.875 × 0.2 cm³ 

Let number of coins be n 

⇒ n × volume of coins = volume of cuboid 

⇒ n × (22/7) × 0.875 × 0.875 × 0.2 = 11 × 10 × 7

⇒ n = \(\frac{{11} \times{10} \times{7} \times{7}}{{22} \times{0.875} \times{0.875} \times{0.2}}\)

⇒ n = 1600

odd

We know that, the cubes of all odd natural numbers are odd.

We have,

= √(1.96)

= √(196/100)

= √((14 × 14)/(10 × 10))

= √(14² / 10²)

= 14/10

= 1.4