Sunday 8 April 2012

Five weighing blocks and 121kgs....



Using five weighing blocks how can we weigh any weight up to 121 kgs? What shall be the five weights that have to be considered?
            Because we use a common balance with two compartments it is possible to add as well as reduce weights. If we have 1 kg block 1kg can be weighed. We don’t need a 2kg block because we can measure2kgs by using 3kgs block on one compartment and 1kg weight on the other. It is evident that for weighing 3, 4 kgs we have 1 and 3 kg blocks. Let us see what we will be our next block, blocks 5,6,7,8 are not required, As we add 9 kg block with 1&3  kgs blocks up to 13 kgs can be weighed .As we begin using 1 kg block  1kg is weighed. Here is our little calculation for next block of weight, just 1 more than the two fold.ie, 1*2=2,2+1=3.Using these two blocks up to 4kgs can be weighed. Our next block is 1 more than the double of 4kgs.ie, 9kgs.Using this we can measure up to 13kgs.Next block one more than the double of 13.ie, 13*2=26, 26+1=27.Let us put this in a series form 1, 3,9,27 .We find all of them a multiple of 3.Next number will be 81, very next will be 243.Thus if we have 1,3,9,27,81,243 designations .What is the amount up to which we can make a measurement? To find out, multiply 243 with 3 and reduce 1, ie, 243*3=729,729-1=728.Find the half of the result728/2=364.It can be found that we can measure various weights up to 364 kgs.If we consider 1,3,9,27,81 blocks, 81*3=243,243-1=242,242/2=121.
            We can conclude that a measurement up to 121 kgs can be made with the blocks of 1, 3,9,27 and 81 kgs blocks.

Sunday 4 March 2012

A puzzle about discrete distribution


One is carrying one thousand gold coins. These coins are packed in separate small bags. It is supposed to give away coins whatever the discrete amount is demanded without unpacking the bags. How is it going to be packed in ten bags?

Here is the solution

Number of coins in first bag = 1
Number of coins in second bag = 2
Number of coins in third bag = (2+2) = 4
Number of coins in fourth bag = (4+4) = 8
Number of coins in fifth bag = (8+8) = 16
Number of coins in sixth bag = (16+16) = 32
Number of coins in seventh bag = (32+32) = 64
Number of coins in eighth bag = (64+64) = 128
Number of coins in ninth bag = (128+128) = 256

Sum of the coins in all nine bags = 1+2+4+8+16+32+64+128+256 = 511

Number of coins in the tenth bag = 1000-511= 489

For example, if one demands 300 coins 256+32+8+4 = 300 i.e., 9th bag+ 6th bag +4th bag+3rd bag are given away. Likewise any number of coins can be given selecting appropriate bags from 1 to 9 and also by adding them, so that the sum is any number between 1 and 1000.




Thursday 23 February 2012

Work and time


12 people finish a work in 10 days then how long it takes for 15 people to finish the work. This problem can be solved using equations
People * days = different number of people * days required for them.
Deriving, people * days / diff number of people = days required 
From our previous example 12 * 10 = 15 * x,
12 * 10 / 15 = 8
Take the other problem if work is done by A in 12 days B in 4 days and C in 6 days, then together all of them how many days it take to finish the work?. Consider unit amount of work done by each and find the total
1/12 + 1/6 + ¼ = (1 + 2 + 3)/ 12 = 6/12 = ½
It is found that all of them together it takes 2 days to finish this work.
The other problem is solved by subtraction of unit work done by each. Take A& B together finishes a work in 4 days. This work can be done by A alone in 12days.Then how long it takes for B alone to finish this work?
            Both finish unit amount of work in ¼ days then B takes 1/4 – 1/12 = (3-1) / 12 =
2/12 = 1/6 i.e., 6days to finish this work.
And another problem, 5 people works for 7 hours to finish a work in 4 days, then a man if work for 10 hours, how long it takes to finish the work?
Again it is a game of equations
(5 * 7 * 4) / (1* 10) = 140/10 = 14
Our friend takes 14 days to finish the work!

Thursday 5 January 2012

Miniature circuit breaker


Miniature circuit breaker



            In earth leakage circuit breaker system, the parts of an installation which are sought to be protected are connected to an earth electrode through the coil of an earth leakage circuit breaker, which controls the supply to that part of the installation. In addition to the equipment earth. It provides the means of disconnecting the faulty circuit at a current level (about 30 mille amperes) which is only a fraction of the normal full load current, and is the most effective method of protection
            The trip circuit breaker may be provided with time-lag feature to allow the operation of the protective devices protecting individual circuits.
            MCB body is made of flame retardant high strength plastic with high melting point and high dielectric strength. The contacts are of silver, silver alloy, or of silver graphite with wiping action to ensure maintenance-free long life. The arc produced on separation of the contacts is rapidly moved under the influence of a magnetic field, under the arc chute stack where it is broken down in to partial arcs, resulting in rapid extinction of arc. Most MCBs are equipped with thermal and/or magnetic and current limiting protection for normal overload inverse time current characteristic protection. Thermal operation is achieved with a bimetal strip which deflects when heated by the overload current flowing through it. In doing so, it moves the trip lever releasing the latch mechanism and causing the contacts to open under spring action, when short circuit fault occurs, the rising fault current energizes the solenoid of hammer trip mechanism and the trip lever causes immediate opening of the contacts.
            “L”series circuit breakers are recommended for installations with low inductive loads like heaters,ovens,geysers,GLS lamp etc.They have pickup current five times more than the normal current.
            “G”series or motor duty MCBs are recommended for installations with high inrush current peaks and which, at the same time require closer overload protection such as motors, air conditioners, fluorescent lamps, sodium vapor lamps, machine tools etc.These MCBs have eight times the normal rated current as magnetic pickup current.
            Back up protection-If the prospective short circuit current is expected to be more than 9000Amps, MCBs will require suitable back up protection for example HRC fuse or higher capacity circuit breakers.