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The total ang...

Question

The total angular movement (in degrees) of a lead screw with a pitch of 5.0 mm to drive the work table by a distance of 20 mm in a NC machine is

A

14400

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B

57600

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C

72000

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D

28800

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Solution

The correct option is A 14400
Distance travelled in one revolution =5 mm
No. of revolution to travel a distance of 200 mm
$= \frac{200}{5} = 40 r e v$
1 revolution $= 360^{\circ}$
$∴ 40$ revolution $= 360 \times 40 = 14400^{\circ}$

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