Question

A turbine has an initial angular velocity of 1800 rpm. Its velocity is doubled twice with uniform acceleration in 12 seconds. The revolutions during this interval is​

Answer 1

$\; \; \; \; \; \;{\large{\bold{\sf{\underbrace{\underline{Let's \; understand \; the \; concept \; 1^{st}}}}}}}$

${\bullet}$ This question says that a rectangular cardboard have 4 cm and 2 cm as it's length and breadth respectively. A circle greatest area is cut from so find area of remain portion.

${\large{\bold{\sf{\underline{Given \; that}}}}}$

${\bullet}$ Length of rectangular cardboard = 4 cm

${\bullet}$ Breadth of rectangular cardboard = 2 cm

${\bullet}$ A circle of greatest area is cut from cardboard.

${\large{\bold{\sf{\underline{To \; find}}}}}$

${\bullet}$ Area of remain portion

${\large{\bold{\sf{\underline{Solution}}}}}$

${\bullet}$ Area of remain portion = 4.86 cm²

${\large{\bold{\sf{\underline{Using \; concept}}}}}$

${\bullet}$ Formula to find radius.

${\bullet}$ Area of circle formula.

${\bullet}$ Area of Rectangle formula.

${\large{\bold{\sf{\underline{Using \; formula}}}}}$

${\bullet}$ Radius = Diameter/2

${\bullet}$ Area of circle = πr²

${\bullet}$ Area of Rectangle = Length × Breadth

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

~ According to the question,

• As we already know that a circle of greatest area is cut from the cardboard so we have to find the area of remain portion.

• So, it's cleared that the maximum complete circle can be cut from diameter of "2 cm."

~ So, now, let's convert diameter into radius

$\; \; \; \;{\bf{\longmapsto Radius \: = \dfrac{Diameter}{2}}}$

$\; \; \; \;{\bf{\longmapsto Radius \: = \dfrac{2}{2}}}$

$\; \; \; \;{\bf{\longmapsto Radius \: = 1 \: cm}}$

${\pink{\frak{Henceforth, \: 1 \: is \: radius \: of \: circle}}}$

~ Now let's find the area of circle

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: \pi r^{2}}}$

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: \dfrac{22}{7} \times 1^{2}}}$

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: \dfrac{22}{7} \times 1 \times 1}}$

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: \dfrac{22}{7} \times 1}}$

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: \dfrac{22}{7}}}$

$\; \; \; \;{\bf{\longmapsto Area \: of \: circle = \: 3.14 \: cm^{2}}}$

${\pink{\frak{Henceforth, \: 3.14 \: cm^{2} \: is \: area \: of \: circle}}}$

~ Now let's find area of rectangular carboard

$\; \; \; \;{\bf{\longmapsto Area \: = Length \times Breadth}}$

$\; \; \; \;{\bf{\longmapsto Area \: = 4 \times 2}}$

$\; \; \; \;{\bf{\longmapsto Area \: = 8\: cm^{2}}}$

${\pink{\frak{Henceforth, \: area \: of \: rectangular \: cardboard \: is \: 8 \: cm^{2}}}}$

~ Now at last let us find the area of remain portion

$\; \; \; \;{\bf{\longmapsto Area \: = \: Area \: of \: rectangular \: cardboard - \: Area \: of \: circle}}$

$\; \; \; \;{\bf{\longmapsto Area \: = \: 8 - 3.14}}$

$\; \; \; \;{\bf{\longmapsto Area \: = 4.86 \: cm^{2}}}$

${\pink{\frak{Henceforth, \: 4.86 \: cm^{2} \: is \: the \: area \: of \: remain \: portion}}}$