Physics, asked by vinayaktj007, 1 month ago

Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm.​

Answers

Answered by aadrikarastogi2
0

Explanation:

The area A and circumference C of the cross section of the bar are

A=\frac { \pi }{ 4 } { d }^{ 2 }

4

π

d

2

and C=\piπ d

Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal

semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,

V=2\left( { V }_{ Side } \right) +2\left( { V }_{ end } \right)(V

Side

)+2(V

end

)=2\left( AL \right) +2\left( \pi RA \right) \\(AL)+2(πRA)

=2\left( L+\pi R \right)(L+πR) A

or V=2\left[ 30mm+\pi \left( 10mm \right) \right] \left[ \frac { \pi }{ 4 } { \left( 6mm \right) }^{ 2 } \right][30mm+π(10mm)][

4

π

(6mm)

2

]

=3470m{ m }^{ 3 }m

3

or V=3470m{ m }^{ 3 }m

3

For the area A,

A=2\left( { A }_{ Side } \right) +2\left( { A }_{ end } \right)(A

Side

)+2(A

end

) =2\left( CL \right) +2\left( \pi RC \right) \\(CL)+2(πRC)

=2\left( L+\pi R \right)(L+πR) C

or A=2\left[ 30mm+\pi \left( 10mm \right) \right] \left[ \pi { \left( 6mm \right) } \right][30mm+π(10mm)][π(6mm)]

=2320m{ m }^{ 2 }m

2

or A=2320m{ m }^{ 2 }m

2

Hope u understand this concept thnk u...

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