Question

A water sprinkler completes one rotation in 5 sec at a speed at 44 m/s. Calculate the area drenched by the sprinkler in 5 sec.
A) 2200 m2
B) 8800 m2
C) 3650 m2
D) 3850 m2​

Answer 1

Given :-

Time taken to complete one rotation by the water sprinkler at a speed of 44 m/s = 5 s

To Find :-

The area drenched by the sprinkler in 5 sec.

Analysis :-

Firstly find the radius of the sprinkler by substituting the given values accordingly using the formula of circumference of a circle.

Then using the formula of area of circle you can easily find the area drenched by the sprinkler in 5 sec.

Solution :-

We know that,

• c = Circumference
• r = Radius
• a = Area

Using the formula,

$\underline{\boxed{\sf Circumference \ of \ a \ circle=2 \pi r }}$

Substituting their values,

⇒ 2πr = 44 × 5

⇒ 2πr = 220

⇒ r = 220/2π

⇒ r = 110/π

⇒ r = 110 × 7/22

⇒ r = 5 × 7

⇒ r = 35 m

Therefore, the radius of the land is 35 m.

Using the formula,

$\underline{\boxed{\sf Area \ of \ the \ circle=\pi r^2}}$

Substituting their values,

⇒ a = 22/7 × 35²

⇒ a = 22/7 × 1225

⇒ a = 22 × 175

⇒ a = 3850 m²

Therefore, the area drenched by the sprinkler in 5 sec is (d) 3850 m².

Answer 2

$\Large{\bold{\color{cyan}{\underline{GiVeN,}}}}$

• A water sprinkler completes one rotation in 5 second at a speed of 44 m/s.

$\longmapsto\:\:\bf{Velocity\:(v)\:=\:44\:m/s}$

$\longmapsto\:\:\bf{Time\:period\:(t)\:=\:5\:s}$

$\Large{\bold{\pink{\underline{To\:FiNd,}}}}$

• The area drenched by the sprinkler in 5 seconds.

$\\ \Large{\bold{\purple{\underline{CaLcUlAtIoN,}}}}$

$\bf\orange{We\:know\:that,}$

$\Large\red\bigstar\:\:{\underline{\green{\boxed{\bf{\blue{v\:=\:\omega\:r}}}}}}$

$\bf\red{Where,}$

• v is the velocity of the object.

• ω is the angular velocity.

• r is the radius of circular path.

[NOTE ➛ ω = 2πf, where f is the angular frequency.]

$:\implies\:\:\bf{v\:=\:(2\:\pi\:f)\:r}$

[NOTE ➛ f = ¹/ₜ , where t is the time period.]

$:\implies\:\:\bf{v\:=\:\Big(2\pi\times{\dfrac{1}{t}}\Big)\:r}$

$:\implies\:\:\bf{r\:=\:\dfrac{v}{2\pi\times{\frac{1}{t}}}\:}$

$:\implies\:\:\bf{r\:=\:\dfrac{v}{2\pi}\times{t}\:}$

$:\implies\:\:\bf{r\:=\:\dfrac{44}{2\times{\frac{22}{7}}}\times{5}\:}$

$:\implies\:\:\bf{r\:=\:\dfrac{2}{2}\times{7}\times{5}\:}$

$:\implies\:\:\bf\green{r\:=\:35\:m}$

$\Large\bf\pink{Now,}$

$\red\checkmark\:\:\bf\blue{Area\:drenched\:=\:\pi\:r^2\:}$

$:\implies\:\:\bf{Area\:drenched\:=\:\dfrac{22}{7}\times{(35)^2}\:}$

$:\implies\:\:\bf{Area\:drenched\:=\:\dfrac{22}{7}\times{35}\times{35}}$

$:\implies\:\:\bf{Area\:drenched\:=\:22\times{5}\times{35}}$

$:\implies\:\:\bf\purple{Area\:drenched\:=\:3850\:m^2}$

━─━─━─━─━─━─━─━─━─━─━─━─━─━─━

$\Large\bf\orange{Therefore,}$

D) The area drenched by the sprinkler in 5 seconds is 3850 m².