A screw gauge has a negative zero error. Which is correct among these?
A) Correction = - coinciding divisions of C.S $\times$ L.O
B) Correction = + coinciding divisions of C.S $\times$ L.O
C) Correction = + [ln - coinciding divisions C.SI $\times$ L.C
D) Correction = + [ln + coinciding divisions C.SI $\times$ L.C

Hint: For solving question, we will use the generalised formula of Correction = +[L.C.×(n−p)
N is the total number of circular scale divisions
Where L.C is the least and this formula is only valid in case of negative the zero error. In the solution part we were going to discuss what is negative zero error.

Complete solution:
Let us understand first what a screw gauge is. Micrometer screw gauge is an instrument used to measure up-to 1/10 of mm (or 0.01 mm= 0.001 cm) which is usually called the least count of micrometer. It also measures the thickness of small sheets of glass, plastic and the diameter of thin wires, , etc.
Least count: Minimum value up to which a screw gauge can measure and is known as its least count A screw gauge of 100 divisions will move the cap scale along the main scale by 1/100 of mm-0.01 mm. Or it is defined as the ratio between the pitch of the screw and the number of divisions on the circular scale.
$least\,count = \dfrac{{Pitch\,of\,screw\,gauge}}{{total\,number\,of\,divisions\,of\,circular\,scale}}$
\[\Rightarrow\] $\dfrac{{1mm}}{{100}} = 0.01mm$
If the zero of the circular scale above the reference line (index) line, the error is negative. So, zero correction will be positive.
Formula for correction is =+[L.C.×(n−p)]
when the zero error is positive zero error, it will be subtracted from the measured value and When when the zero error is negative zero error, it will be added to the measured reading.
Generally, the zero error will be coming in terms of circular scale division.

Thus, correction for negative zero error = + [coincide division of circular scale ×L.C].

Note: Alternate method,
Total number of divisions on the circular sale is given as n.
Let ${p^{th}}$ circular scale division coincides with the main scale division.
Negative zero error is calculated by, Correction =+[L.C.×(n−p)]
So, option C is correct.