CBSE BOARD X, asked by goraisuma449, 3 months ago

A mechanic can open a nut by applying a force of 160N while using a lever handle of length 50cm. What is the length of the handle required if he wants to apply the force of only 80N.​

Answers

Answered by IamSameerhii
0

\huge \underline{ \underline{ {   \cal \: {★Q}} \large \red{ \tt \: U} \green{ \tt \: E} \blue{ \tt \: S} \orange{  \tt \: T} \purple{ \tt \: I} \red{ \tt \: O} \blue{ \tt \: N}}} :  -\\  \\

  • A mechanic can open a nut by applying a force of 160N while using a lever handle of length 50cm. What is the length of the handle required if he wants to apply the force of only 80N.

 \underline{ \huge{  \sf \:   \ ☯\:\: T}   \large\tt{{{ \underline{ \blue{HINGS}}}}}} \\  \\  \\   \underline{ \huge{  \sf \:   \  G}\large\tt{{{ \underline{ \blue{IVEN}   : -  }}}}} \\  \\

  •   \boldsymbol   { \underline{★Force \:  \:  required  \:  \: to  \: \:  open  \: \:  a \:  \:  nut  \:  \:     \orange➙  \:  \red{\tt {\: 160  \: N}}}} \\  \\

  •   \boldsymbol   { \underline{★Length \:  \:  of  \:  \: the  \: \:  levers  \: \:  handle \:  \:      \orange➙  \:  \red{\tt {\: 50   \: \bf{cm }}}}} \\  \\

  •   \boldsymbol   { \underline{★Force \:  \:  required  \:  \: to  \: \:  open  \: \:  the \:  \: nut \:  \: of \:  \: another \:  \: handle \:       \orange➙  \:  \red{\tt {\: 80  \: N}}}} \\  \\  \\

 \underline{ \huge{  \sf \:   \ ☯\:\: T}   \large\tt{{{ \underline{ \pink{O}}}}}} \\  \\  \\   \underline{ \huge{  \sf \:   \  F}\large\tt{{{ \underline{ \pink{IND}   : -  }}}}} \\  \\

  •   \boldsymbol   { \underline{★Length \:  \:  of  \:  \: the  \: \:  lever  \: \: of \: \:another\: \: handle \:  \:      \orange➙  \:  \red{\tt {\:??}}}} \\  \\

 \underline{ \huge{  \sf \:   \ ☯ \:\:F}   \large\tt{{{ \underline{ \purple{ORMULA}}}}}} \\  \\  \\  \\ \underline{ \huge{  \sf \:   \  U}\large\tt{{{ \underline{ \purple{SED}   : -  }}}}} \\  \\  \\

\boxed{⚫{\bf{\red{\underline{\: \:Note\: :- \: (i) \: means \: case \: 1, \: (ii) \:means \:case\: 2\:. }}}}} \\  \\

  • \sf{★Effort\: arm\: (ii) =\dfrac {Effort\: applied\: (i) \times \:Handle's\: length\: (i)}{ Effort\: applied\: (ii)}} \\   \\

\huge \underline{ \underline{ {   \cal \: {★S}} \large \red{ \tt \: O} \green{ \tt \: L} \blue{ \tt \: U} \orange{  \tt \: T} \purple{ \tt \: I} \red{ \tt \: O} \blue{ \tt \: N}}} : - \\  \\

\large\frak{\underline{\dag\:The \:work\: done \:to \:open \:the\:nut \:in \:both \:the\: cases\: is\:same.}} \\  \\

\large\sf{\underline{\therefore\: \:work\: done \:in \:case \:(i):\implies \:work\: done \:in \:case \:(ii)}} \\  \\

\huge { \underline{ {   \sf \: {•\:\;T}} \large \blue{ \bf{ HEN}, }}} \\\\

\large\sf{\underline{★Effort\: \times \:effort\: arm\: in\:(i)\: \implies \:effort \: \times \:effort\: arm\: in \:case\: (ii)}} \\  \\

\sf{\underline{★160\:N \times \:50\: cm\: in\:(i)\: \implies \:80 \:N \times \:Effort\: arm\: in \:case\: (ii)}} \\  \\

\sf{\therefore\:\:Effort\: arm\: (ii) =\dfrac {160 \: \times\: 50\: }{ 80}} \\   \\

\frak{\underline{\dag\:Multiply \:160\: and\: 50 \:to \:get \:8000}} \\  \\

\sf:{\implies{\dfrac{8000}{80}}} \\ \\

\frak{\underline{\dag\:Expand\: \dfrac{8000}{80}}} \\  \\

\sf:{\implies{\dfrac{\cancel{8000}\:^{100}}{\cancel{80}\:^{1}}}} \\ \\

\sf:{\implies{\dfrac{8000}{80} \approx 100\:cm}} \\  \\  \\  \\

\huge \underline{ \underline{ {   \cal \: {★A}} \large \red{ \tt \: N} \green{ \tt \: S} \blue{ \tt \: W} \orange{  \tt \: E} \purple{ \tt \: R} }} :  - \\  \\

 \boxed{\sf{★\:\:\: \bigg\lgroup { \red{\dfrac{8000}{80} \approx 100\:\:cm }  \bigg\rgroup★}}}\\\\

\huge \underline{ \underline{ {   \cal \: {♪}} \large \red{ \tt \: ♪} \green{ \tt \: •} \blue{ \tt \: •} \orange{  \tt \: •} \purple{ \tt \: ✓} \red{ \tt \: ✓} \blue{ \tt \: ✓}}}

Similar questions