Taylor's tool life equation is given by $V T^{n} = C$ where V is in m/min and T is in min. In a turning operation, two tools X and Y are used. For tool X, n = 03 and C = 60 and for tool Y, n = 0.6 and C = 90. Both the tools will have the same tool life for the cutting speed (in m/min, round off to one decimal place) of

40

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Solution

The correct option is A 40
PIC IS NOT CLEAR.

Let x is the cutting speed for sametool life in

$Then, x T_{x}^{0.3} = 60 o r T_{x} = {(\frac{60}{x})}^{1 / 0.3}$

$x T_{y}^{0.3} = 90 o r T_{y} = {(\frac{90}{x})}^{1 / 0.6}$

$N o w, T_{x} = T_{y}$

${(\frac{60}{x})}^{1 / 0.3} = {(\frac{90}{x})}^{1 / 0.6}$

$o r {(\frac{60}{x})}^{0.6} = {(\frac{90}{x})}^{0.3}$

$\frac{x^{0.3}}{x^{0.6}} = \frac{90^{0.3}}{60^{0.6}}$

$o r x^{- 0.3} = 0.33066$

$o r x = {(0.33066)}^{V - 0.3} = 40 m / min$

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The correct option is A 40PIC IS NOT CLEAR. Let x is the cutting speed for sametool life in Then,xT0.3x=60orTx=(60x)1/0.3 xT0.3y=90orTy=(90x)1/0.6 Now,Tx=Ty (60x)1/0.3=(90x)1/0.6 or(60x)0.6=(90x)0.3 x0.3x0.6=900.3600.6 orx−0.3=0.33066 orx=(0.33066)V−0.3=40m/min