Given,

Outer dimensions of the box = 50 cm, 40 cm and 23 cm

Outer volume of the box = 50 × 40 × 23 = 46000 cu cm

Inner dimensions of the box:

Length = (50 – 2×3) = 44 cm

Breadth = (40 – 2×3) = 34 cm

Height = (23 – 30 = 20 cm

Inner volume of the box = 44 × 34 × 20 = 29920 cu cm

Volume of the wood = Outer Volume – Inner Volume = 46000 – 29920 =
16080 cu cm

Weight of the wood per cu cm = 0.5 g

Weight of the wood used for making the box = 16080 × 0.5 = 8040 g = 8.04 kg.

Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

x * 5x * 40x = 12.8

=> x^3 = 12.8/200

=> x=  0.4m = 40cm

Since V2 > V1, so the vessel can contain 100% of the beverage filled in the bow

Given : Internal diameter of hollow sphere(d)= 4 cm.
Internal radius of hollow sphere (r) = 4/2= 2 cm
external diameter of hollow sphere (D) = 8 cm.
external radius of hollow sphere( R )= 8/2= 4 cm.

Volume of the Hollow sphere = 4/3π(R³ – r³)
Volume of the Hollow sphere = 4/3π(4³ – 2³)
Volume of the Hollow sphere = 4/3π(64 – 8)
Volume of the Hollow sphere = 4/3π(56) cm³

Diameter of the cone(d1) = 8 cm
radius of the cone( r1)= 8/2 = 4 cm

Let the height of the cone be h cm.
Volume of the cone = ⅓ πr1²h
= ⅓ π × 4² × h = 16πh/3

Volume of the cone = Volume of the hollow sphere
16πh/3 = 4/3π(56)
16h = 4 ×56
h = (4 × 56)/16
h = 56/4 = 14 cm

Hence, the height of the cone is 14 cm.

Increase in volume of cylinder = volume of sphere

Let increase in volume of cylinder be = πr²h, where r is its radius and h is its height.

So, 16πh = 4/3 π3³

h = 4/3 × 27/16

= 36/16

= 9/4 = 2.25 cm

As per the question volume of cuboid=volume of sphere
volume of cuboid=lbh cubic cm
=38808cm3
volume of sphere =4/3pi r cube
4/3 pi r cube=38808cubic cm
r cube =38808×21/4×22
r=21 cm

Volume of sphere / surface are of  sphere = 27 cm
4/3πr^3  / 4πr^2 = 27
r/3 = 27
r = 81 cm

Volume of cone =1/3πr^2h

={(1/3πr^2h)/(1/3πr^2H)}

={(1/3π(2x)^2h)/(1/3π xH)}

=h/H=1/4

so h:H=1:4

Area of circle= pi r2=1.54
r=0.7
ht. of mountain= sqrt( 2.5×2.5 – .7×.7)=2.4

In Right ∆,
Hypotenuse=5cm,
Base of ∆= Radius of Cone=3cm
3rd side of ∆=Height of Cone=4cm
So,Volume of Cone=1/3πr²h=1/3×3.14×(3)²×4=37.68

∴ Curved surface area = πrl = (π × 8 × 17) cm² = 136π cm²

Coin, r=0.75cm h=0.2cm V=pi r^2h =22/7*0.75*0.75*0.2 =2.475/7cm^3

Cylinder, h=8 cm r=3cm V= 22/7*3*3*8 =1584/7 cm^3 No. of coins=1584/7*7/2.475 =640

Let height change = x cm

volume of water dropped = π × (35/2)^2 × x = 11000 cm3

x = 11000
———–π× (17.5) ^2
= 11.42 cm

Volume of the tank =  246.4 litres =  246400 cm^3

Let the  radius of the base be r cm . Then ,

(22/7 xx r^2 xx 400) = 246400

hArr    r^2 = ((246400 xx 7)/(22 xx 400))

hArr   r^2 =  196          hArr   r = 14

:.    Diameer of the base = 2r =  28 cm.

Given :

Cost of paint = ₹ 36.50/ kg

1 kg paint covers 16 ft^2.

Solution :

Side of the cubic wall = 8 ft

So, total outside surface area of the cube wall

= 6 (side)^2

= 6 × 64 sq. ft

= 384 sq. ft

Amount of paint required

= (384/16) kg

So, total cost = ₹ (384/16) × 36.50

= ₹ 24 × 36.50

= ₹ 876.00

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Area of 3 faces = lb + bh + lh
so, 120 = lb
72 = bh
60 = lh
multiply all three
l²b²h² = 518400
so, lbh =√518400 = 720 cm³

Let the water, h mtr. will rise in the tank

l×b×h=Area×speed×time

80×40×h=40/100×100×10000×1/2

h=11/60m=100/160cm=5/8cm

he height of the wall is 6 times its width and lenght of the wall is 7 times its height .if the volume of the wall be 16128 cu.m.its width is

A) 4mB) 5mC) 6mD) 7m

Answer:   A) 4m

Explanation:

Let width = x

Then, height = 6x and length = 42x

42x * 6x * x = 16128

x = 4

Let breadth = b then length = b + 10m.

Floor area of the rectangle hall = A = length × breadth = b × (b + 10)

Also area of each carpet = 6 × 4m and 34 pieces are required to cover the floor

Therefore area of hall = 34 × 6 × 4

b × (b + 10) = 34 × 24

Therefore b = 24 m and length = b + 10 = 34m.

1 litre = 1000 centimeter cube.

So, 8 litre = 8000 centimeter cube

Side = cuberoot (8000 centimeter cube) = 20 cms.

Surface area = [2(10*4 + 4*3 + 10*3)] sq.cm = 164 sq.cm