During plain turning of medium carbon steel by using HSS tool, a cutting velocity of 25 m/min was observed for 1min tool life and 20 m/min for 5min tool life. The Taylor's constant for the HSS Tool is

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Solution

The correct option is B 25
$V_{1} = 25 m/min,$

$V_{2} = 20 m/min$

$T_{2} = 25 min$

$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$

$25 (1)^{n} = 20 (5)^{n}$

${(\frac{1}{5})}^{n} = \frac{20}{25}$

$n = 0.138$

$C = 25$

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During plain turning of medium carbon steel by using HSS tool, a cutting velocity of 25 m/min was observed for 1min tool life and 20 m/min for 5min tool life. The Taylor's constant for the HSS Tool is

The correct option is B 25V1=25 m/min, V2=20 m/min T2=25 min V1Tn1=V2Tn2 25(1)n=20(5)n (15)n=2025 n=0.138 C=25

The correct option is B 25
$V_{1} = 25 m/min,$

$V_{2} = 20 m/min$

$T_{2} = 25 min$

$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$

$25 (1)^{n} = 20 (5)^{n}$

${(\frac{1}{5})}^{n} = \frac{20}{25}$

$n = 0.138$

$C = 25$