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Download JNTU Anantapur (JNTUA) B-Tech 2019 Final Year First Semester R19 Regulation previous year R15 VLSI Design Questions for Reference and Download the important Question bank of R19

VLSI Design

Download JNTU Anantapur (JNTUA) B-Tech 2019 Final Year First Semester R19 Rrgulation - VLSI Design - (19A04702T) syllabus Latest updated from the below link

Download JNTU Anantapur (JNTUA) B-Tech 2019 Final Year First Semester R19 Rrgulation - VLSI Design - (19A04702T) -Introdcution to Course objective , CO PO PSO mapping of the syllabus and NBA Formt. Download the Slide for the Bloom's Taxonomy level.

Download JNTU Anantapur (JNTUA) B-Tech 2019 Final Year First Semester R19 Regulation previous year R15 VLSI Design Questions for Reference and Download the important Question bank of R19

Content	Link
Syllabus	Click Here for Syllabus
Introduction	Introdcution, Vision- Msiion-CO-PO-PSO- PEO
VLSI LAB	Experiment 1
Unit 1	Questions
University Question papers	University Qustions
Unit 1 Question and Answer	Unit 1 Question and Answer
Unit 1 PPT	Unit 1 PPT

Journal Paper Writing Tips to Students

 

Students can get more information about how to write a paper for an academic journal.  Here you can find the Youtube video link. Click the below link to access the video 

Click Here for Part 1

Click here for Part 2

TCS NQT Test preparation - Work and Time Module

 Important Formule 

1. Work from Days:

   If A can do a piece of work in n days, then A's 1 day's work =	1	.
   	n

2. Days from Work:

   If A's 1 day's work =	1	,	then A can finish the work in n days.
   	n

3. Ratio:

   If A is thrice as good a workman as B, then:

   Ratio of work done by A and B = 3 : 1.

   Ratio of times taken by A and B to finish a work = 1 : 3.

4. Work Done by A and B
   

   A and B can do a piece of work in ‘a’ days and ‘b’ days respectively.

   When working together they will take $\frac{ab}{a+b}$ days to finish the work

   In one day, they will finish $( \frac{a+b}{ab})^{th}$  part of work.

5. Efficiency is inversely proportional to the

   Time taken when the amount of work done is constant.

   Efficiency is inversely proportional to the

   Time taken when the amount of work done is constant.

   Efficiency α = $\frac{1}{Time  Taken}$

   Rule 6: Work of one day = $\frac{Total  work}{Total  number  of  working  days}$

   Total work = one day work × Total number of working days

   Remaining work =  1 – work done

   Work done by A = A’s one day work × Total number of working days of A

   Rule 7:If A can finish $\frac{m}{n}$ part of the work in D days.

   Then total time taken to finish the work by A = $\frac{D}{\frac{m}{n}}$ = $\frac{n}{m}$ × D days

   Rule 8:

   If A can do a work in ‘x’ days

   B can do the same work in ‘y’ days

   When they started working together, B left the work ‘m’ days before completion then total time taken to complete the work = (y+m)x/(x+y)

   Rule 9: A and B finish work in a days.

   They work together for ‘b days and then A or B left the work.

   B or A finished the rest of the work in ‘d’ days.

   Total time taken by A or B alone to complete the work = $\frac{ad}{a – b}$ or $\frac{bd}{a-b}$

   Work and Time Formulas & Three Universal Rules

   • If M1 persons can do W1 work in D1 days and M2 persons can do W2 works in D2 days then the formula $\frac{M_{1}D_{1}}{W_{1}}$ =   $\frac{M_{2}D_{2}}{W_{2}}$ .It can be written as M1D1W2 = M2D2W1.
     
   • If the persons work T1 and T2 hours per day respectively $\frac{M_{1}D_{1}T_{1}}{W_{1}}$ =  $\frac{M_{2}D_{2}T_{2}}{W_{2}}$
   • It can be written as, M1D1 T1 W2 = M2D2 T2W1
   • If the persons has efficiency of E1 and E2 respectively then $\frac{M_{1}D_{1}T_{1}E_{1}}{W_{1}}$ = $\frac{M_{2}D_{2}T_{2}E_{2}}{W_{2}}$ Therefore, M1D1T1E1W2 = M2D2T2E2W1.

   In all the above formula,
   

   M = Number of workers

   D = Number of days
   T = Time required
   W = Units of work
   E = Efficiency of worker

Example Q&A

1.    Question: A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs.3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?

Solution:

Let the total amount of work to be done be W units.

Productivity of A, Pa = W/6 units per day.

Productivity of B, Pb = W/8 units per day.

3 days x [Pa + Pb + Pc] = W => Pc = W/24 units per day

Ratio of wages of A: B: C = Ratios of their productivities = (W/6): (W/8): (W/24) = 4: 3: 1.

Amount to be paid to C = Rs.3200 x (1/8) = Rs.400

2.    Question: It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. We want to use three pumps: A, B and another pump C to fill the tank in 2 hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank?

Solution:

Let the total capacity of the tank be C liters.

Fill rate of pump A, Fa = C/6 liters per hr

Fill rate of pump B, Fb = C/8 liters per hr

2 hrs x [Fa + Fb + Fc] = C => Fc = 5C/24 liters per hr

Let‘t’ be the time taken by only pump C to fill the tank.

‘t’ hrs x 5C/24 = C => t = 24/5 = 4.8 hrs