Linked Data:
Blind hole $10 m m$ diameter, $50 m m$ deep are being drilled in steel block, drilling spindle speed is $600 r p m$ , feed $0.2 m m / r e v,$ point angle of drill is $120$

Machining time (in $m i n$ ) per hole will be

0.08

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0.31

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0.44

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0.86

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Solution

The correct option is C $0.44$
Given data:
$D = 10 m m, t = 50 m m, N = 600$
$f = 0.2 m m / r e v$

For blind hole
$A P_{1} = 0.3 D$
$= 0.3 \times 10 = 3 m m$

$L = t + A P_{1} = 50 + 3 = 53$

$Time/hole = \frac{L}{f N} = \frac{53}{0.2 \times 600} = 0.442 min$

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Linked Data:
Blind hole $10 m m$ diameter, $50 m m$ deep are being drilled in steel block, drilling spindle speed is $600 r p m$ , feed $0.2 m m / r e v,$ point angle of drill is $120$

During the above operation, the drill wears out after producing $200$ holes.Taylor's tool life equation is of the form $V T^{0.3} = C,$ where $V =$ cutting speed in $m / m i n$ and $T =$ tool life in min. Taylor's constnat $C$ will be