Question

In the two set-upsX and Y, shown on the next page, the wires AB and PQR are made of the same material and have equal thickness. The length of the wire AB, (in the set-up X') is equal to the diameter (= PR) of the semi-circle, formed by the wire PQR, (in the set-up Y'). Pins, P1 and P2, are attached, to wires AB and PQR respectively, with the help of wax. Which of the two pins, PI or P2, will fall off later? Give reason for your answer,​

Answer 1

$\huge \sf \underline\red{science : }$

$\mapsto$ Given :

$\: \: \: \: \: \: \: \: \: \: \bull \large \rm \: length \: \: of \: \: wire \: \: ab \: \\ \: \: \: \: \: \large \rm \: = diameter \: \: of \: \: pr$

AnSweR :

$\:$

• P2 will fall of later...

$\:$

$\large \sf \star\underline{ \underline{ \pink{ \: explanation : }} }$

$\large \blue{ \sf \: take \: \: length \: \: of \: \: ab = 14cm}$

$\large \rm since. \:$

$\: \: \: \: \: \: \: \: \: \: \bull \large \rm \: length \: \: of \: \: wire \: \: ab \: \\ \: \: \: \: \: \large \rm \: = diameter \: \: of \: \: pr$

PQR = semi circle

$\\ \large \green{ \sf \: circumfrence = \frac{1}{2} \times 2\pi r} \\ \\ \large \underline\green{ \sf circumfrence = \pi r}$

$\large \tt \: here. \: \: r = 14cm ÷ 2= 7cm\\ \\ \large \tt \implies \: c = \frac{22}{7} \times 7cm \\ \\ \large \tt \implies \: c = \frac{22}{ \cancel7} \times \cancel{7}$

$\large \red{ \star} \: \boxed{\purple{ \therefore \: \sf \: c = 22cm }}$

$\large \mapsto \: \orange{ \sf \: 22cm > 14cm } \\ \\ \large \mapsto \: \boxed{ \underline{\purple{ \sf \: pqr > ab }}}$

So,

$\:$

• Pin 2 will fall later .

$\: \: \: \: \: \large \tt \red{this \: \: is \: \: the \: \: required}$

$\: \: \: \: \: \: \: \huge \rm \green{answer}$