# Civil Engineering - GATE Exam Questions - Discussion

Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 49)

49.

The 5-day BOD of a wastewater sample is obtained as 190 mg/l (with

*k*= 0.01 h^{-1}). The ultimate oxygen demand (mg/l) of the sample will beDiscussion:

5 comments Page 1 of 1.
John said:
3 years ago

Why e? Can't we use base 10?

Lt=Lo (1-10^kt).

Lt=Lo (1-10^kt).

Ravi said:
5 years ago

What is e here? Please tell me.

(1)

Shivaprasad sajjan said:
5 years ago

(BOD)5 = (BOD)u[1-e^-kt].

190 = (BOD)u[1-e^-0.01 * 120],

190 = (BOD)u * 0.70,

(BOD)u = 190/0.70 = 271.42~271.

Ans: C.

190 = (BOD)u[1-e^-0.01 * 120],

190 = (BOD)u * 0.70,

(BOD)u = 190/0.70 = 271.42~271.

Ans: C.

Roy said:
6 years ago

5-day bod is approximately 68% of the ultimate bod. So, 68%=190, so 100%= (190*100/68) =279.

So we can go with option c.

So we can go with option c.

(1)

SAMALA LAXMAN said:
1 decade ago

(BOD)5 = BODu[1- (e^-kt)].

Here give k = 0.01/hour.

t = 5days = 120 hrs.

BOD5= 190mg/lt.

Substitute these values in above formula we will get answer as 272.32.

Here give k = 0.01/hour.

t = 5days = 120 hrs.

BOD5= 190mg/lt.

Substitute these values in above formula we will get answer as 272.32.

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