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RRB ALP Study Notes – Volume Formulas for Maths Preparation

To Help you achieve success on the RRB ALP Recruitment and exam, we will be providing you very important RRB ALP Study Notes. These RRB ALP Study notes include the list of formulas to calculate Volumes of various Geometrical Figures like Cube, Sphere, Pyramid, etc. The purpose of the RRB ALP Study notes is to assist the aspirants with their preparation and help them clear the RRB ALP recruitment exams with excellent marks.

 RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes

RRB ALP Study Notes – Important Formulas to calculate Volume

Figure Formula
Cube -:

RRB ALP Study Notes Cube

 

 

 

Volume   = a3 

= a × a × a

Cylinder -:

RRB ALP Study Notes Cylinder

 

 

 

 

Volume = pi × r2 × h

pi = 3.14
h is the height
r is the radius

Rectangular Solid -:

RRB ALP Study Notes Rectangular Solid

 

Volume = l × w × h 

l is the length
w is the width
h is the height

Sphere -:

RRB ALP Study Notes sphere

 

 

 

Volume = (4 × pi × r3)/3 

pi = 3.14
r is the radius

Cone -:

RRB ALP Study Notes Cone

 

 

Volume = (pi × r2 × h)/3

pi = 3.14
r is the radius
h is the height 

Pyramid -:

RRB ALP Study Notes Pyramid

 

 

 

Volume = (B × h)/3

B is the area of the base
h is the height 

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RRB ALP Study notes – Sample Questions

RRB ALP Study notes – Sample Question 1

Q) A solid cylinder of brass 9 m high and 4 m in diameter is melted and recast into a cone of diameter 3 m. Find the height of the cone.

(a) 45 m

(b) 48 m

(c) 51 m

(d) 42 m

Answer -: Option (b) – 48 m

Given that, height of cylinder (h) = 9 m    and    diameter of cylinder = 4 m

∴ radius of cylinder (r) = 2 m

And, diameter of cone = 3 m

∴ radius of cone (R) = 1.5 m

Let height of Cone = H

Note: When an object is melted and given a different shape, the volumes of both the objects (of two different shapes) remain equal.

Volume of cylinder = Volume of cone

∴ πr²h = 1/3 × πR²H

⇒ (2) ² × 9 = 1/3 × (1.5) ² × H

∴ H = 36 × 3/2.25 = 48 m

 RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes

RRB ALP Study notes – Sample Question 2

Q) If the radius of a sphere is increased by 10% then the volume will be increased by

(a) 10%

(b) 25%

(c) 33.1%

(d) 40%

Answer -: Option (c) – 33.1%

RRB ALP Study Notes solution 2

  RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes RRB ALP Study Notes

RRB ALP Study notes – Sample Question 3

Q) A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. What is the height of the cone?

(a) 10 cm

(b) 25 cm

(c) 21 cm

(d) 14 cm

Answer -: Option (d) – 14

Given, internal diameter of hollow sphere = 4 cm, then internal radius (r) = 2 cm

And, external diameter of hollow sphere = 8 cm, then external radus (R) = 4 cm

Diameter of cone = 8 cm, then radius of cone (r1) = 4 cm

Let height of cone = h

Note: When an object is melted and given a different shape, the volumes of both the objects (of two different shapes) remain equal.

∴  Volume of cone = Volume of hollow sphere

RRB ALP Study notes solution 3

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