INTRODUCTION

Screw gauges are precision measuring devices that are widely used in engineering, manufacturing, laboratories, and research to precisely measure linear distances, wire radius, and glass plate thickness, among other things.

To get correct readings, common errors like zero error, least count error, and parallax error must be addressed.While concentrating on the measurement, keep the screw gauge’s errors to a least.In this part, we will look at screw gauge formula and the important practice questions for NEET and JEE.

Identify the Pitch:

The pitch of the screw is the distance it moves axially in one complete rotation. This is usually provided by the manufacturer or can be measured using a standard ruler.

What is the Least Count?

The least count of a measuring instrument is the smallest measurement increment that it can accurately measure. For a screw gauge, this is determined by the pitch of the screw and the number of divisions on its circular scale.

Least Count of a Screw Gauge

Zero Error in Screw Gauge

On the off chance that the zero of the circular scale (Head scale) precisely matches (exactly coincides) with the zero of the pitch scale (main scale), then, at that point, the mechanical assembly is said to have no zero error.

Positive Error in Screw Gauge

In case zero of the Circular/Head scale is under the reference line then the zero error should be positive.

Exactly when the zero error is positive, the contrasting zero correction is negative and this value will be deducted from the measured value.

Negative Error in Screw Gauge

If zero of the Head/Circular scale is above the reference line then the error should be negative. Exactly when the error is negative, the looking at zero correction in the device is positive and this value will be added to the measured value.

Diameter and Radius of a wire

Numerical on least count:

Question -1

The pitch of a screw gauge is 0.75 mm and the circular scale has  100 divisions. A student records the circular scale reading as 40 divisions and the main scale reading as 3 mm. Find the total reading.

Solution:

Least count = 0.75 mm/100 = 0.0075 mm
P.S.R + (H.S.R x L.C)
= 3+(40 x 0.0075) mm
= 3+0.3
= 3.3 mm

Question -2

A screw gauge with a least count of 0.01 mm is used to measure the thickness of a metal sheet.The main  scale reading is 1 mm and the circular scale  reading is 20 divisions. The zero error is +0,02 mm.Find the corrected thickness of the metal.

Solution:

P.S.R+(H.S.R×L.C)±Z.C

=1+(20×0.01)−0.02

=1.20−0.02=1.18mm

Question -3

A screw gauge with a least count of 0.01 mm is used to measure the diameter of a thin wire. The main reading is  4 mm and the circular reading is 32 divisions. If the zero error is -0.03 mm. Find the actual diameter of the wire.

Solution:

Diameter of the wire =P.S.R + (H.S.R x L.C)土 Z.C
= 4 + (32 x 0.01) +0.03
= 4.32+0.03
= 4 35 mm

Question -4

A student measures the diameter of a spherical ball using a screw gauge and obtains the following readings.

Main scale reading = 5 mm

Circular scale reading = 60 divisions

Least count 0.01 .mm

Find the diameter of the ball

Solution:

M.S.R + (C.S.R x L.C)

= 5+(60 x 0.01) mm

= 5+0.6 

= 5.6 mm

Question -5

From the figure find the object thickness

Question and Solution:

Question -6

What is the least count of a screw gauge if the pitch is 0.5 mm and the circular scale has 50 divisions?

Solution:

Least count = Pitch /Total number of divisions in the circular scale

 = 0.5 mm/50

 = 0.01 mm 

Question -7

A screw gauge has a least count of 0.005 mm and 100 divisions on the circular scale.What is the pitch of the screw gauge.

Solution:

Least count      = Pitch/ Total number of divisions in the circular scale

 Pitch = Least count x total number of divisions in the circular scale

 = 0.005 mm x 100

  = 0.5 mm

Question -8

Screw gauge A  has a pitch of 0.5 mm and 50 divisions, while screw gauge B has a pitch of 1 mm and 100 divisions. Which screw gauge  has a smaller least count?

Solution:

For screw gauge A 

Least count = Pitch/ Total number of divisions in the circular scale

 = 0.5 mm/50

 = 0.01 mm

For screw gauge B 

Least count = Pitch /Total number of divisions in the circular scale

= 1 mm/100

 = 0.01 mm

Both screw gauges have the same least count.

Question -9

A screw gauge  has a pitch of 0.5 mm and a least count of 0.005 mm. How many divisions are there on the circular scale?

Solution:

Least count = Pitch / Total number of divisions in the circular scale

 Total number of divisions in the circular scale  = Pitch /Least count

 = 0.5 mm/0.005 mm

 = 100

Question -10

A screw gauge has a zero error of  +0.005. The main scale reading is 3.5 cm and the circular scale reading is 35 divisions.If the least count is 0.001 cm. Find the corrected diameter.

Solution:

Pitch ScaleReading (Main Scale Reading)  P.S.R  =  3.5 cm

Coinciding Head Scale Division  H.S.D =  35 th Division

Least Count of the screw gauge    L.C     =  0.001 cm  

Head Scale Reading (Circular Scale Reading)    H.S.R =  H.S.D x L.C 

=  35 x 0.001 = 0.035

Zero error of the screw gauge   Z.E    =  +0.005

Corresponding  Zero Correction   Z.C    =  -0.005

Hint: Zero correction is the negative of zero error

Corrected Diameter = Observed Diameter ± Zero correction

 = (P.S.R + H.S.R)  ± Zero correction

Hint:If the zero correction is positive,you add it to the observed reading.     

 If the zero correction is negative,you subtract it from the observed reading.

Corrected Diameter = (3.5 cm +0.035 cm) -0.005

= 3.535 – 0.005

= 3.530 cm

Answer: Corrected Diameter is 3.530 cm

Question -11

A screw gauge has a zero error of  -0.004. The main scale reading is 1.5 cm and the circular scale reading is 20 divisions.If the least count is 0.001 cm. Find the corrected diameter.

Solution:

Pitch Scale Reading (Main Scale Reading) P.S.R  =  1.5 cm

Coinciding Head Scale Division    H.S.D =  20 th Division

Least Count of the screw gauge   L.C     =  0.001 cm  

Head Scale Reading (Circular Scale Reading)  H.S.R =  H.S.D x L.C 

 =  20 x 0.001 = 0.020 cm

Zero error of the screw gauge        Z.E     =  -0.004

Corresponding  Zero Correction   Z.C    =  +0.004

Hint: Zero correction is the negative of zero error

Corrected Diameter = Observed Diameter ± Zero correction

 = (P.S.R + H.S.R)  ± Zero correction

Hint:If the zero correction is positive,you add it to the observed reading.     

If the zero correction is negative,you subtract it from the observed reading.

Corrected Diameter =(1.5 cm +0.020 cm) +0.004

= 1.520 + 0.004

 = 1.524 cm

Answer: Corrected Diameter is 1.524 cm

Question -12

A screw gauge has a zero error of  +0.003. The main scale reading is 1.0 cm and the circular scale reading is 10 divisions.If the least count is 0.001 cm. Find the corrected diameter.

Solution:

Pitch ScaleReading (Main Scale Reading)  P.S.R  =  1.0 cm

Coinciding Head Scale Division   H.S.D  =  10 th Division

Least Count of the screw gauge  L.C       =  0.001 cm  

Head Scale Reading (Circular Scale Reading)    H.S.R  =  H.S.D x L.C 

=  10 x 0.001 = 0.010

Zero error of the screw gauge    Z.E      =  +0.003

Corresponding  Zero Correction    Z.C      =  -0.003

Hint: Zero correction is the negative of zero error

Corrected Diameter = Observed Diameter ± Zero correction

= (P.S.R + H.S.R)  ± Zero correction

Hint:If the zero correction is positive,you add it to the observed reading.     

If the zero correction is negative,you subtract it from the observed reading.

Corrected Diameter  = (1.0 cm +0.010 cm) -0.003

= 1.010 – 0.003

= 1.007 cm

Question and Answer from 13 to 23

Zero error (Z.E)    (cm)	Least coun(L.C)  (cm)	P.S.R (cm)	H.S.D (cm)	H.S.R	Zero Correction  (Z.C)	Observed reading    (cm)	Corrected Diameter    (cm)
+0.002	0.001	2.5	45	0.045	-0.002	2.545	2.543
-0.003	0.001	1.2	25	0.025	+0.003	1.225	1.228
+0.004	0.001	3.0	30	0.030	-0.004	3.030	3.026
-0.002	0.001	2.0	15	0.015	+0.002	2.015	2.017
+0.005	0.001	4.5	50	0.050	-0.005	4.550	4.545
-0.004	0.001	1.8	40	0.040	+0.004	1.840	1.844
+0.003	0.001	2.2	35	0.035	-0.003	2.235	2.232
-0.005	0.001	3.5	20	0.020	+0.005	3.520	3.525
+0.002	0.001	1.0	10	0.010	-0.002	1.010	1.008
-0.003	0.001	2.8	25	0.025	+0.003	2.825	2.828

Conclusion:

You have tackled some essential screw gauge practice questions, crucial for acing the NEET physics section. Mastering this instrument requires a solid grasp of its pitch, least count, zero errors (both positive and negative error) and the ability to accurately read the main scale and circular scale.

Remember consistent practice is key to perfecting your technique and avoiding common pitfalls. By diligently working through problems like these,you’ll build the confidence and precision needed to accurate measure and interpret readings from the screw gauge, ultimately boosting your scores in NEET physics. Keep practicing, and good luck.