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CASE 1. A spherometer has a least count of 0.005mm and its head scale is divided into 200 equal divisions. Let the distance between consecutive threads on the spherometer screw be d1 mm.
CASE 2: A vernier caliper has each division on its main scale to be 1mm. On a vernier scale, 99mm is divided into 100 divisions. Let the vernier constant be d2 mm.
CASE 3: The pitch of a screw gauge is 1mm and there are 100 divisions on its circular scale. When no article is placed in between its jaws, the zero of the circular scale coincides with the reference line. When a steel wire is placed between the jaws, two main scale divisions are clearly visible and 67 divisions on the circular scale are observed. Let the diameter of the wire be d3mm.
The values of d1,d2 and d3 are:
A. d1=1,d2=1.99,d3=0.01
B. d1=0.01,d2=9.9,d3=1
C. d1=1,d2=0.01,d3=2.67
D. d1=0.01,d2=2.67,d3=0.01

Answer
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Hint: Recall the expression for pitch in terms of least count and number of divisions and hence solve the first two cases. Now recall the expression for the reading taken using a screw gauge. Now by substituting the given values in the question in that expression determine the answer to case 3.

Formula used:
Expression for pitch (P),
P=L.C×N
Expression for reading (d) taken using a screw gauge,
d=(M.S.R+(C.S.C×L.C))±Z.C

Complete step by step answer:
In the CASE 1 we are given the least count of spherometers as 0.005mm and also the number of divisions on the head scale is also given. If d1mm is the distance between consecutive threads on the spherometer then, we are asked to find the value of it.
We know that the distance between consecutive threads on the spherometer is called the pitch of the device. If L.Cis the least count of the device and Nis the number of divisions on the head scale then pitch d1 is given by,
d1=L.C×N
Substituting the given values,
d1=0.005mm×200
d1=1mm…………………. (1)
In CASE 2 we are asked to determine the vernier constant d2mm from the given values in the question. We know that vernier constant is actually the least count of the vernier calipers. Therefore, ifM.S.D is the main scale division and V.S.Dis the vernier scale division then, the vernier constant, that is, the least count, d2is given by,
d2=1M.S.D1V.S.D
We are given,
1M.S.D=1mm
99mm100div
1V.S.D=99100=0.99mm
Therefore,
d2=10.99=0.01mm ……………………….. (2)
For CASE 3, from the question, pitch (P) is 1mm and number of divisions (N) is 100 divisions. Therefore, the least count (L.C) is,
L.C=PN=1mm100
L.C=0.01mm
Also, the error is zero hence there is no zero correction.
Z.C=0
Also,
The main scale reading, M.S.R=2mm
The circular scale coincidence, C.S.C=67div
The reading of the diameter of a steel wire d3mm taken in a screw gauge is given by,
d3=(M.S.R+(C.S.C×L.C))±Z.C
Substituting the values,
d3=(2+(67×0.01))±0
d3=2.67mm …………………………………. (3)
From (1), (2) and (3), we get that, d1=1,d2=0.01,d3=2.67

Hence, the answer to the question is option C.

Note:
Zero correction has great significance while taking the reading on a screw gauge or any other device. In the third part of the question we are told that when no article is placed in between its jaws, the zero of the circular scale coincides with the reference line, which clearly says that the zero correction is null. If the zero of the circular had been to the left of the reference line, we should have subtracted that value from the reading and if it had been to the right then we should have added the same to the reading taken.
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