# Workshop Calculation And Science 1st Year Book PDF

### Workshop Calculation

Direct proportion the more. Solve the problem by first writing the statement and proceeding to single and then to the multiple according to the type of proportion that is Involved.

#### Introduction

Proportional fundamentals, as applicable to motor vehicle calculations are discussed below.

Simple Proportion Proportion This is an equality between two ratios.

Examples.

If one vehicle fleet uses 30 liters of petrol per day how much petrol is used by 6? Vehicles operating under similar conditions.

One vehicle uses petrol = 30 liters per day. Then six vehicles will use = 6 Times as much

= 6 x 30= 180 litres/day.

If 4 vehicles of a fleet use 120 gallons of petrol per day how much petrol will be used by 12 vehicles operating under the same condition?

4 vehicles use 120 gallons per day

1 Vehicle will use

120 = 30 gallons/day

12 vehicles will use

Both examples are called simple proportions because only two quantities were used and the day is common for both ratios.

Compound and Inverse proportions

Compound proportion

Example

15 Fitter take 21 days to complete overhauling of vehicles how long 7 Fittern will take to overhaul S vehicles (Anaume time of overhauling each vehicle is constant)

In this, both direct and indirect proportions are used. 1 Fitter will overhaul 1 vehicle in days (shorter time).

Quantities (No. of days) are taken in last as that is the answer required in this case.

Ans: 7 Fitters will overhaul 8 vehicles in 20 days.

Inverse proportion

Sometimes proportions are taken inversely.

Examples

If one water pump fills the fuel tank in 12 minutes, two pumps will take half the time taken.

The time should not be doubled. If two pumps take 30 minutes to fill up a tank how long will 6 similar pumps take to fill this tank?

Ans: Time taken by 6 pumps = 30×2

-10 minutes

Proportional parts in the combustion equation

Introduction Proportion of quantities forms an important factor in the combustion process of fuel. The following happens during the combustion process. Fuel is a hydrocarbon substance.

The combustion air is supplied from the atmosphere and contains oxygen and nitrogen. Now the following chemical changes take place during the combustion of a fuel.

Carbon burns with oxygen and forms Co and Co, (Carbon monoxide and their carbon dioxide.)

Hydrogen burns with oxygen and becomes water (HO) Sulphur burns with oxygen and becomes sulfur dioxide.

Nitrogen is an inert gas and does not take part in combustion. Method of finding proportional parts in one lb of substance To be found out now Proportion of oxygen and hydrogen in one lb/kg of water.

In a gearing arrangement of a vehicle, a gear has 20 teeth meshing with a gear of 52 teeth. The din of 52 teeth gear in 200mm. Find out the diameter of the 20 teeth gear wheel.

10 If two water pumps take 45 minutes to fill up a tank how long will 4 similar pumps will take to this tank 11 In a belt-pulley drive the driving pulley in of 12 cm in diameter and rotates at 360 rpm.

Find the rpm of the driven pulley whose diameter is 20 cm diameter. 12 To overhaul a gearbox, 12 mechanics are needed to complete the work in 5 days.

If only 7 mechanics are available, how many days they will be able to complete the overhauling work?

13 Express in simple ratios the following

a 45+60

b 40 paise+Rs4.00

20mm

d 4°C+100°C

C 4 meters

14 Air contains 24% oxygen and 78% nitrogen by mass (weight). Calculate the quantity of air (mass of air) required for complete combustion of unit mass fuel (The main constituents that take part in the combustion process are carbon, hydrogen, and sulfur) Note: Given the following data (Solve the problem)

a 1 kg of carbon requires kg of oxygen.

b 1 kg of hydrogen requires 8 kg of oxygen.

cn1 kg of sulfur requires 1 kg of oxygen.

15 A fuel is a hydrocarbon substance of C,H, This shows each molecule of fuel contains 7 atoms of carbon and 14 atoms of hydrogen.

If the carbon atomic weight is 12 times greater than the hydrogen atom, find out the proportionate parts of hydrogen and carbon in one kg of fuel.

16 A vehicle worth Rs.20,000/-can be insured at a cost of Rs.150 How much wired it cost to insure a vehicle worth Rs 24000/- for one year and 3 months at the same rate (Compound proportion)

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