Question

The distance covered between 6.00A.M and 3.15P.M by the tip of the 12cm long hour hand in a clock is

Answer 1

Answer:

$\sf{The \ distance \ covered \ by \ the \ tip \ of \ hour}$

$\sf{hand \ is \ \dfrac{183\pi}{5} \ cm}$

Given:

• Hour hand is 6 cm long.

To find:

• The distance covered between 6.00 am and 3.15 pm by the tip of the 12 cm long hour hand.

Solution:

$\sf{Difference \ between \ 6.00 \ am \ and \ 3.15 \ pm}$

$\sf{is \ of \ 9.15 \ hours}$

$\sf{1 \ hour=60 \ mins}$

$\sf{\therefore{9.15 \ hours=9.15\times60}}$

$\sf{i.e. \ 9.15 \ hours=549 \ minutes}$

$\sf{We \ know, \ 1 \ minute=(\frac{1}{2})^\circ}$

$\sf{\therefore{549 \ minutes=(\frac{549}{2})^\circ}}$

$\sf{\theta=(\theta\times\frac{\pi}{180})^{c}}$

$\sf{\therefore{(\frac{549}{2})^\circ=\frac{549}{2}\times\frac{\pi}{180}}}$

$\sf{=(\dfrac{61\pi}{20})^{c}}$

$\boxed{\sf{Length \ of \ arc(s)=\theta^{c}\times \ r}}$

$\sf{Here, \ radius \ is \ length \ of \ arc, \ i.e. \ 12 \ cm}$

$\sf{\therefore{Length \ of \ arc=\dfrac{61\pi}{20}\times12}}$

$\sf{\therefore{Length \ of \ arc=\dfrac{183\pi}{5} \ cm}}$

$\sf{Distance \ covered \ by \ tip \ of \ hour \ hand}$

$\sf{=Length \ of \ the \ arc}$

$\sf{\therefore{Distance \ cover=\dfrac{183\pi}{5} \ cm}}$

$\sf\purple{\tt{\therefore{The \ distance \ covered \ by \ the \ tip \ of \ hour}}}$

$\sf\purple{\tt{hand \ is \ \dfrac{183\pi}{5} \ cm}}$