In a longitudinal section of a root, starting from the tip upward, the four zones occur in the following order

Correct option is (D) 4 – √3 Let the other root be \(\alpha.\) Then sum of roots \(=\frac{-b}a\) \(=\frac{-(-8)}1=8\) \(\therefore \alpha+4+\sqrt3=8\) \(\Rightarrow\alpha=8-(4+\sqrt3)\) = \(4-\sqrt{3}\) Hence, the other root of given quadratic equation is \(4-\sqrt{3}.\)

Correct option is (D) 2 – √3 Given quadratic equation is \(x^2-4x+1=0\) Let the other root be x. \(\therefore\) Sum of roots \(=\frac{-(-4)}1=4\) \(\Rightarrow\) \(x+2+\sqrt{3}=4\) \(\Rightarrow\) \(x=4-2-\sqrt3\) \(\Rightarrow\) x = \(2-\sqrt{3}\) Hence, the other root is \(2-\sqrt{3}.\)

Correct option is (B) 3 – √5 If \(3+\sqrt{5}\) is one root of the quadratic equation then \(3-\sqrt{5}\) is another root of that quadratic equation.

It is Anaphase.

Correct Answer - B Remainin concetration of reactant after 20 min `=(1)/(4)xx102.5"mol"L^(-1)` `"Rate"=kxx["Reactant"]=(0.693)/(t_(1//2))xx["Reactant"]` `=(0.693)/(10xx2.5=0.0693xx2.5 "mol"L^(-1) min^(-1)`

Correct Answer - `(0)` `underset(ntooo)lim[root(3)((n+1)^(2))-root(3)((n-1)^(2))]` `=underset(ntooo)limn^(2//3)[(1+(1)/(n))^(2//3)-(1-(1)/(n))^(2//3)]` `=underset(ntooo)limn^(2//3)[(1+(2)/(3).(1)/(n)+((2)/(3)((2)/(3)-1))/(2!)(1)/(n^(2))...)-(1-(2)/(3).(1)/(n)+((2)/(3)((2)/(3)-1))/(2!)(1)/(n^(2))...)]` `=underset(ntooo)limn^(2//3)[(4)/(3).(1)/(n)+(8)/(81).(1)/(n^(3))+...]` `=underset(ntooo)lim[(4)/(3).(1)/(n^(1//3))+(8)/(81).(1)/(n^(7//3))+...]=0`

Correct Answer - `[(mu_(0)I r^(2))/(2 pi d (R^(2) - r^(2)))]`