Answer / digital_yogesh

TDS =UP TO 500 AND MAXIMUM 2000
TOTAL HARDNESS = 300 AND MAXIMUM 600
pH = 6.5 TO 8.5

Answer / ram

6.8-7.8

Answer / er. inder amar singh

The TDS reomended for human beaing is less than 100. and the
Ph should be 7.

Answer / vimal

TDS is less than 100,harness 150mg/l & PH is 6.5 to 7.5

Question 112 - In quantum computing, let the amplitude A = a | 0 > + b | 1 >, | a | | a | + | b | | b | = 1. Find the values of b if A = 0.8 | 0 > + b | 1 >.

1 Answers  

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i am giving BARC Trainee exam in chemical. Can anyone tell the syllabus and question paper pattern for OCES for BARC trainee?

1 Answers  

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QUANTUM BIOLOGY - EXAMPLE 33.7 : (a) In a DNA of a living cell, the quantum information available in the bases guanine (G) and thymine (T) are | G > = | 110 > and | T > = | 010 > respectively. Calculate | G > - | T >. (b) In a living biological cell, the step time for random walk of an electron is t. The localization time of an electron is T. If i is the geometric average of T and t, find log T as a function of t and i.

1 Answers  

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Question 46 - In a steady state one dimensional conduction with no heat generation, the differential equation is d / dx (k dT / dx) = 0. Prove that T(x) = ax + b, where k, a and b are constants. (b) At x = 0, T = c and at x = L, T = d. Prove that T(x) = (d - c) x / L + c for boundary conditions.

1 Answers  

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If sum one has HAL placement paper plz mail me ,I m Chemical Engg.my id is er.kotnala@gmail.com

0 Answers   HAL,

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What is a surfactant?

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What do you understand by wet bulb globe temperature?

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what is the symbol of sodiumj

2 Answers   HCL,

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Question 111 - There are 6 spin orbitals in p subshell in a ground state carbon atom. Only 2 electrons fill the p subshell. Number of different ways for n electrons to occupy the k spin orbitals are k! / [ (n!) (k-n)! ]. Find the number of different configurations of electrons to occupy the p subshell in a carbon atom.

1 Answers  

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Where barrel pump use?

0 Answers   Deepak Fertilisers,

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CHEMICAL FLUID MECHANIC - EXAMPLE 3.1 : Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli's equation where P + Rgy + 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid's velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.

1 Answers  

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Question 32 – Let 1 ^ 1 = 1, 2 ^ 2 = 4, 3 ^ 3 = 27. By using the Excel computer programming – either by Solver or Goal Seek, find the value of v for the Van der Waals equation (P – a / v ^2) (v - b) = RT where a = 18.82, b = 0.1193, P = 2, R = 0.082, T = 5000 for benzene. Describe briefly how to use Solver and Goal Seek in Excel program of computer to find the solution quickly.

1 Answers  

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