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Standard XII

Physics

Frequency

A steel ...

Question

A steel bar $200 m m$ in diameter is turned at a feed of $0.25 m m / r e v$ with a depth of cut of $4 m m$ . The rotational speed of the workpiece is $160 r p m$ . The material removal rate in $m m^{3} / s$ is

160

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167.6

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1600

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1675.5

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Solution

The correct option is D $1675.5$
$M R R = v f d$
$= \frac{π D N}{60} \times f d = \frac{π \times 200 \times 160 \times 0.25 \times 4}{60}$
$= 1675.5 m m^{3} / s$

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​​​​​A steel bar 200mm in diameter is turned at a feed of 0.25mm/rev with a depth of cut of 4mm. The rotational speed of the workpiece is 160rpm. The material removal rate in mm3/s is

The correct option is D 1675.5MRR=vfd =πDN60×fd=π×200×160×0.25×460 =1675.5mm3/s

A steel bar $200 m m$ in diameter is turned at a feed of $0.25 m m / r e v$ with a depth of cut of $4 m m$ . The rotational speed of the workpiece is $160 r p m$ . The material removal rate in $m m^{3} / s$ is

The correct option is D $1675.5$
$M R R = v f d$
$= \frac{π D N}{60} \times f d = \frac{π \times 200 \times 160 \times 0.25 \times 4}{60}$
$= 1675.5 m m^{3} / s$