# GATE Papers >> Mechanical >> 2017 >> Question No 23

Question No. 23

In an arc welding process, welding speed is doubled. Assuming all other process parameters to be constant, the cross sectional area of the weld bead will

##### Answer : (D) reduce by 50%

Solution of Question No 23 of GATE 2017 Mechanical Paper

Heat given to weld pool $={\mathrm H}_\mathrm s=\frac{\mathrm V\vert{\mathrm\eta}_\mathrm h}{\mathrm\nu\;\mathrm A}$

$\mathrm V=$ arc voltage

$\mathrm I=$ current

${\mathrm\eta}_\mathrm h=$ heat transfer efficiency

$\mathrm v=$ Velocity of welding (speed)

$\mathrm A=$ Cross sectional area of weld.

${\mathrm v}_1=\mathrm v\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\mathrm A}_1=\mathrm A$

${\mathrm v}_1=2\;\mathrm v$

Now $\mathrm{vA}=\frac{{\mathrm H}_2}{\mathrm V\vert{\mathrm\eta}_\mathrm h}=$ constant (RHS unchanged)

${\mathrm v}_1{\mathrm A}_1\;=\;{\mathrm v}_2{\mathrm A}_2$ (Material deposition rate is constant)

${\mathrm A}_2\;=\;\frac{{\mathrm A}_1}2$

Area of cross section reduced by 50%

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