Boolean expression for OR gate is
(A) \(\bar{A} = A\)
(B) \(C = \overline{AB}\)
(C) A.B = 0
(D) A + B = C
Correct Answer - C The Boolean algebra is based on logic.
2 Answers 1 viewsCorrect Answer - C An OR gate has two or more inputs with one output. NOT gate is the most basic gate with one input and one output. In an OR...
2 Answers 1 views`A" "+ " "OH^(-)" "rarr` Products `t=0 " " 0.002 " "0.3` `k=(2.303)/(30xx(0.002-03))log_(10).(0.3[0.002-(0.002xx1)/(100)])/(0.002[0.3-(0.002xx1)/(100)])` `k=1.12xx10^(-3)"litre" "mol"^(-1)s^(-1)`
2 Answers 1 viewsCorrect Answer - B `~(pvvq) vv(~p^^q)` `=(~p^^~q)vv(~p^^q)` `=~p^^(~p^^q)` `=~p`
2 Answers 1 views(i)`f_(1)(x) = 2 log_(10)x" is defined for "x gt 0` ` f_(2)(x) = log_(10)x^(2)" is defined for "x^(2) gt 0 or x in R - {0}` Therefore ` f_(1)(x) and...
2 Answers 1 viewsCorrect Answer - B `{:(,2AB_(2)(g),hArr,2AB(g)+,B_(2)(g)),("Initial",1,,0,0),("moles",,,,),("At equil.",2(1-x),,2x,x):}` where, x = degree of dissociation Total moles at equilibrium `=2-2x+2x+x=(2+x)` So, `p_(AB_(2))=(2(1-x)p)/((2+x)), p_(AB)=(2x p)/((2+ x))` `p_(B_(2))=(x p)/((2+x))` `K_(p)=((p_(AB))^(2)(p_(B_(2))))/((p_(AB_(2)))^(2))=(((2x p)/(2+x))^(2)[((x)/(2+x))p])/([((2(1-x))/((2+x)))p]^(2))` `=(4x^(3)p^(3))/((2+x)^(3))xx((2+ x)^(2))/(p^(4)4(1-x)^(2))=(x^(3)p)/((2+x)(1-x)^(2))` `=(x^(3)p)/(2) " "...
2 Answers 1 viewsThe self inductance of a coil is equal to the induced `emf` set up in the coil, when the current passing through it change at the unit rate The `SI`...
2 Answers 1 viewsCorrect option is: A. AND
2 Answers 1 viewsCorrect option is: D. LM+MNO+MPQ
2 Answers 1 viewsCorrect option is: B. A(B+C)=AB+AC
2 Answers 1 views