Question

$\huge \bf{Question}$
$\sf \: A \: boy \: is \: cycling \: such \: that \: \\ \sf \: the \: wheels \: of \: the \: cycle \: are \: making \: 140 \\ \sf revolution \: per \: minute. \: If \: the \: diameter \: \\ \sf \: of \: the \: wheel \: is \: 60 \: cm, \: calculate \: the \: \\ \sf \: speed \: per \: hour \: at \: which \: the \: boy \: is \\ \sf \: cycling.$
​

Answer 1

$\huge\tt{\red{\underline{\underline{\star \: Solution \: \red{\star}}}}}$

$\sf Diameter \tt\red{(d)} \sf \: of \: the \: wheel \: = 60 \: cm$

$\therefore \sf \: Radius \: \tt \red{(r)} \sf \: of \: the \: wheel$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \red{ \frac{d}{2}} = \sf \frac{60}{2} = 30 \: cm$

$\therefore \sf \: Circumference \: of \: the \: wheel$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \red{2\pi r }= 2\pi (30)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 60\pi \: cm$

$\purple{\implies} \sf Distance \: covered \: by \: the \: wheel \: in \\ \sf \: \: \: \: \: \: \: \: \: \: \: one \: complete \: revolution.$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 60\pi \: cm$

$\therefore \sf \: Distance \: covered \: by \: the \: wheel \: in \: \bf \red{140} \\ \: \: \: \: \: \: \: \: \sf \: revolutions$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 60\pi \times 140 \: cm$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 60 \times \frac{22}{7} \times 140 \: cm$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf = 60 \times 22 \times 2 = \bf \red{26400 \: cm}$

$\sf \: i.e., \: \: \: \ \: \: \: \: Distance \: covered \: by \: the \: wheel \: in \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \: minute$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \red{ = 26400 \: cm}$

$\therefore \sf \: Speed \: per \: hour \:$

$\sf = \: \: Distance \: covered \: by \: the \: wheel \: in \: \\ \sf \: \: \: \: \: \: \: \: 1 \: hour$

$\: \: \: \: \: \: \: \: \sf = \: 26400 \times 60 \: cm = 1584000 \: cm$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 15840 \: m = 15.84 \: km$

$\sf \: Hence, \: the \: speed \: at\: which \: the \: boy \: is \\ \sf \: \: cycling \: is \: \tt \red{15.84 \: km/hr.}$

Answer 2

$\huge\textrm{\underline{\underline \red{Answer:}}}$

ᴛʜᴇ sᴘᴇᴇᴅ ᴏғ ʙᴏʏ ɪs 15840ᴍ/ᴍɪɴ. ᴏʀ 15.84ᴋᴍ/ʜʀ.

$\huge\textrm{\underline {\underline \red{ExplanaTion :}}}$

ʟᴇᴛ's ᴜɴᴅᴇʀsᴛᴀɴᴅ ᴛʜᴇ ʙᴀsɪᴄ ᴄᴏɴᴄᴇᴘᴛ :-

ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ғɪɴᴅ sᴘᴇᴇᴅ ᴘᴇʀ ʜᴏᴜʀ ᴀɴᴅ ᴡᴇ ᴀʀᴇ ɢɪᴠᴇɴ ᴅɪᴀᴍᴇᴛᴇʀ,sᴏ ᴡᴇ ғɪʀsᴛ ғɪɴᴅ ᴛʜᴇ ʀᴀᴅɪᴜs ᴀɴᴅ ᴛʜᴇɴ ᴄɪʀᴄᴜᴍғᴇʀᴇɴᴄᴇ.

ʜᴇʀᴇ, ᴄɪʀᴄᴜᴍғᴇʀᴇɴᴄᴇ = ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ

ɴᴏᴡ,ʟᴇᴛ's ʙᴇɢɪɴ

$\huge\textrm{\underline{\underline \red{Solution:}}}$

• ᴅɪᴀᴍᴇᴛᴇʀ(ᴅ) ᴏғ ᴡʜᴇᴇʟ = 60ᴄᴍ

• ʀᴀᴅɪᴜs ᴏғ ᴡʜᴇᴇʟ = 30ᴄᴍ

$\footnotesize\sf{ \: \: \: \: \: \: \: \: \: \: (Radius \: = \: \pink{\frac{Diameter}{2} } )}$

ɴᴏᴡ;

• ᴄɪʀᴄᴜᴍғᴇʀᴇɴᴄᴇ = 2πʀ

➠$\normalsize\sf{\: 2 \times\ \pi \times\ 30}$

➠$\normalsize\sf{ \: 60 \pi}$

➠$\normalsize\sf{ Distance \: covered \: = \: 60 \pi}$

➠$\normalsize\sf \green{Distance \: in \: one \: revolution \: = \: 60\pi}$

$\rule{300}{2}$

•ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ 140 ʀᴇᴠᴏʟᴜᴛɪᴏɴs;

➠$\normalsize\sf{ 140 \times\ 60 \pi}$

$\footnotesize\sf{\: \: \: \: \: \: \: \: (Value \: of \: \pi \: = \:\pink{\frac{22}{7} }) }$

➠$\normalsize\sf{ 140 \times\ 60 \times\ \frac{22}{7} }$

➠$\normalsize\sf{ 20 \times\ 60 \times\ 22}$

➠$\normalsize\sf{ 26400 cm}$

➠$\normalsize\sf\green{Distance(140 \: revolutions)\: = \: 26400cm}$

$\rule{300}{2}$

• ɴᴏᴡ,sᴘᴇᴇᴅ ᴘᴇʀ ʜᴏᴜʀ ;

sᴘᴇᴇᴅ ᴘᴇʀ ʜᴏᴜʀ ᴍᴇᴀɴs ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ ᴏɴᴇ ʜᴏᴜʀ,ᴀɴᴅ ᴡᴇ ʜᴀᴠᴇ ᴅɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ɪɴ ᴏɴᴇ ᴍɪɴᴜᴛᴇ (26400ᴄᴍ)

sᴏ;

➠$\normalsize\sf{ 26400 \times\ 60}$

➠$\normalsize\sf{ 1584000cm}$

➠$\normalsize\sf \green{Speed \: per \: hour \: = \: 15.84km/hr}$

$\rule{300}{2}$