Two cutting tools with tool life equations given below are being compared:
Tool 1 : $V T^{0.1}$ = 150
Too1 2 : $V T^{0.3}$ = 300
where V is cutting sped in m/minute and T is tool life in minutes. The breakeven cutting speed beyond which Tool 2 will have a higher tool life is______m/minute.
106.066

Open in App

Solution

The correct option is A 106.066
Let us assume the speed is x m/min
$x T_{1}^{0.1} = 150$

$\Rightarrow T_{1} = {(\frac{150}{x})}^{\frac{1}{0.1}} a n d x T_{2}^{0.1} = 300$

$T_{2} = {(\frac{300}{x})}^{\frac{1}{0.3}}$

$N o w T_{2} > T_{1}$

$o r {(\frac{300}{x})}^{\frac{1}{0.3}} = {(\frac{150}{x})}^{\frac{1}{0.1}}$

For solving inequality

${(\frac{300}{x})}^{\frac{1}{0.3}} = {(\frac{150}{x})}^{\frac{1}{0.1}}$

$o r {(\frac{300}{x})}^{0.1} = {(\frac{150}{x})}^{0.3}$

$o r \frac{x^{0.3}}{x^{0.1}} = \frac{150^{0.3}}{300^{0.1}}$

$o r x = 106.066 m / min$

Suggest Corrections

Similar questions

H S S : V T^{0.2} = 150

C a r b i d e : V T^{0.3} = 250

V

m / m i n

T

m i n

(n)

(k)

0.25

V T^{0.25} = 150

Linked Question:
In a machining experiment, tool life was found to vary with the cutting speed in the following manner:

What is the percentage increase in tool life when the cutting speed is havled?

Join BYJU'S Learning Program