Question

For tool A, Taylor's tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, n=0.3 and K=60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is (A) 26.7 (B) 42.5 (C) 80.7 (D) 142.9

Answer 1

answer : option (A) 26.7

explanation : use formula, $VT^n=K$

for case 1 : $V_1T_1^{n_1}=K_1$

where,$n_1=0.45,K_1=90$

$V_1T_1^{0.45}=90\implies T_1=\left(\frac{90}{V_1}\right)^{\frac{1}{0.45}}$

for case 2 : $V_2T_2^{n_2}=K_2$

where, $n_1=0.30,K_1=60$

$V_2T_2^{0.3}=60\implies T_2=\left(\frac{60}{V_2}\right)^{\frac{1}{0.30}}$

at limiting condition,

$V_1=V_2=V$ and $T_1=T_2$

$\left(\frac{90}{V}\right)^{\frac{1}{0.45}}=\left(\frac{60}{V}\right)^{\frac{1}{0.30}}$

hence, $V=26.7$