A copper wire 4 mm in diameter is evenly wound about a cylinder whose

1.	A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. Find the length and weight of the wire assuming the specific gravity to be 8.88 gm/cm3. (a) 1100 π cm, 545π2 gm (b) 1200 π cm, 441π2 gm (c) 1200 π cm, 426.24π2 gm (d) 1400 π cm, 426.24π2 gm
Answer» (c) 1200 π cm, 426.24 π² gm One round of wire covers 4 mm = \(\frac4{10}\) cm in thickness of the surface of the cylinder Length of the cylinder = 24 cm ∴ Number of rounds to cover 24 cm = \(\frac{24}{\frac4{10}}=\frac{24\times10}{4}=60\) Diameter of the cylinder = 20 cm ⇒ Radius of cylinder = 10 cm Length of the wire in completing one round = 2πr = 2π x 10 cm = 20 π cm. ∴ Length of the wire in covering the whole surface = Length of the wire in completing 60 rounds = (20 π × 60) cm = 1200 π cm. Radius of copper wire = 2 mm = \(\frac2{10}\) cm ∴ Volume of wire = \(\big(π\times\frac2{10}\times\frac2{10}\times1200π\big)\)cm³ = 48 π² cm³ So, weight of wire = (48π² × 8.88) gm = 426.24 π² gm.

A copper wire 4 mm in diameter is evenly wound about a cylinder whose length is 24 cm and diameter 20 cm so as to cover the surface. Find the length and weight of the wire assuming the specific gravity to be 8.88 gm/cm3. (a) 1100 π cm, 545π2 gm (b) 1200 π cm, 441π2 gm (c) 1200 π cm, 426.24π2 gm (d) 1400 π cm, 426.24π2 gm

Answer»

(c) 1200 π cm, 426.24 π² gm

One round of wire covers 4 mm = \(\frac4{10}\) cm in thickness of the surface of the cylinder

Length of the cylinder = 24 cm

∴ Number of rounds to cover 24 cm = \(\frac{24}{\frac4{10}}=\frac{24\times10}{4}=60\)

Diameter of the cylinder = 20 cm

⇒ Radius of cylinder = 10 cm

Length of the wire in completing one round

= 2πr = 2π x 10 cm = 20 π cm.

∴ Length of the wire in covering the whole surface

= Length of the wire in completing 60 rounds

= (20 π × 60) cm = 1200 π cm.

Radius of copper wire = 2 mm = \(\frac2{10}\) cm

∴ Volume of wire = \(\big(π\times\frac2{10}\times\frac2{10}\times1200π\big)\)cm³ = 48 π² cm³

So, weight of wire = (48π² × 8.88) gm = 426.24 π² gm.