SSC JE Civil Engineering 2018 Set 6 Solved Paper

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Question : 137 of 200

Marks: +1, -0

Which of the following expression represent the simplified form of Colebrook equation use to calculate the friction factor, if variable have their standard meanings?

$1/√\f= 1.14+ 2\log (k_s/D+9.35/ {Re√ \f })$
$1/√\f= 1.14-2\log (k_s/D+9.35/ {Re√ \f })$
$1/√\f= 1.14+ 2\log (k_s/ {Re}+9.35/ {D√ \f })$
$1/√\f= 1.14+ 2\log (k_s/ {Re}+18.7/ {D√ \f })$

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