Question

4. According to the five-kingdom classification, the correct sequence is (from simplest to complex)
a. Protista, Fungi, Monera, Plantae, Animalia b. Monera, Protista, Fungi, Plantae, Animalia
c. Monera, Fungi, Protista, Plantae, Animalia d. Monera, Protista, Fungi, Animalia, Plantae​

Answer 1

$\underline{ \huge\bf\red{QuestiØn} :}$

4. According to the five-kingdom classification, the correct sequence is (from simplest to complex)

a. ☺ Protista ➡ Fungi ➡ Monera ➡ Plantae ➡ Animalia

b. ☺ Monera ➡ Protista ➡ Fungi ➡ Plantae ➡ Animalia ✅

c. ☺ Monera ➡ Fungi ➡ Protista ➡ Plantae ➡ Animalia

d. ☺ Monera ➡ Protista ➡ Fungi ➡ Animalia ➡ Plantae

$\underline{ \: \huge\bf\green{SolutiØn} \: :}$

$\rm\red{(b)}\blue{\underline{\bf\green{ \quad Monera \red{\Rightarrow} Protista \red{\Rightarrow} Fungi \red{\Rightarrow} Plantae \red{\Rightarrow} Animalia \quad }}}$ is the correct sequence.

$\red{\rule{5.8cm}{0.02cm}}$

$\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

Answer 2

QuestiØn:

4. According to the five-kingdom classification, the correct sequence is (from simplest to complex)

a. ☺ Protista ➡ Fungi ➡ Monera ➡ Plantae ➡ Animalia

b. ☺ Monera ➡ Protista ➡ Fungi ➡ Plantae ➡ Animalia ✅

c. ☺ Monera ➡ Fungi ➡ Protista ➡ Plantae ➡ Animalia

d. ☺ Monera ➡ Protista ➡ Fungi ➡ Animalia ➡ Plantae

$\underline{ \: \huge\bf\green{SolutiØn} \: :}$

$\rm\red{(b)}\blue{\underline{\bf\green{ \quad Monera \red{\Rightarrow} Protista \red{\Rightarrow} Fungi \red{\Rightarrow} Plantae \red{\Rightarrow} Animalia \quad }}}$ is the correct sequence.

$\red{\rule{5.8cm}{0.02cm}}$

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$\begin{gathered}\boxed{\bigstar{\sf \ Cylinder :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cylinder= \pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ cylinder= 2\pi r h\\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ cylinder= 2\pi r (h+r)\end{gathered}$

$\begin{gathered}\boxed{\bigstar{\sf \ Cone :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cone= \dfrac{1}{3}\pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Cone = \pi r l \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Cone = \pi r (l+r) \\ \\ \\ \sf {\textcircled{\footnotesize4}} Slant \ Height \ of \ cone (l)= \sqrt{r^2+h^2}\end{gathered}$

$\begin{gathered}\boxed{\bigstar{\sf \ Hemisphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Hemisphere= \dfrac{2}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Hemisphere = 2 \pi r^2 \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Hemisphere = 3 \pi r^2\end{gathered}$

$\begin{gathered}\boxed{\bigstar{\sf \ Sphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Sphere= \dfrac{4}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Surface\ Area \ of \ Sphere = 4 \pi r^2\end{gathered}$