Q: Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.) The le

aligned & 10 V.S.D. =9 M.S.D. \\ & 1 M.S.D. =0.1 ~cm (Given) \\ & 1 V.S.D. =0.9 M.S.D. \\ & 1 V.S.D. =0.9 0.1 ~cm \\ & V.S.D. =0.09 ~cm \\ & L.C. =1 M.S.D. -1 V.S.D. \\ & =(0.1-0.09) cm \\ & =0.01 ~cm \\ & Reading =5.08 ~cm \\ & Positive zero error =0.1 ~cm (Given) \\ & Final Corrected reading =(5.08-0.1) cm \\ & =4.98 ~cm aligned

Question 1

One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10. When both the jaws touch each other, the Vernier scale shifts to the left of zero

Accepted Answer

The least count (LC) of the Vernier calliper is given by: LC = 1 mm 10 = 0.1 mm = 0.01 cm Since the zero of the Vernier scale shifts to the left of the zero of the main scale, the instrument has a negative zero error. The zero error is calculated as: Zero error = -4 LC = -4 0.01 cm = -0.04 cm The actual length of the wire is obtained by subtracting the zero error from the measured length: Actual length = Measured length - Zero error Actual length = 1.00 cm - (-0.04 cm ) = 1.04 cm Answer: 1.04 cm

Question 2

In a vernier callipers, 20 VSD coincide with 16 MSD (each division of length 1 mm). The least count of the vernier callipers is :

Accepted Answer

Given 20 VSD = 16 MSD. 1 VSD = 16 20 MSD = 0.8 MSD. Least count is given by 1 MSD - 1 VSD. Least count = 1 MSD - 0.8 MSD = 0.2 MSD. Since 1 MSD = 1 mm, we get Least count = 0.2 mm. Converting to cm, Least count = 0.02 cm. Answer: 0.02 cm

Question 3

Which of the following measurements require ‘index correction’ ?

Accepted Answer

In an optical bench experiment, the distance between the index marks on the uprights may not be exactly equal to the actual distance between the optical centre of the lens and the tip of the pin. This difference is known as index error. To obtain the true distance, an index correction must be applied to the observed distance. Therefore, the measurement of the focal length of lenses using an optical bench requires index correction. In contrast, the resonance tube and meter bridge experiments requ

Question 4

Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.) The le

Accepted Answer

aligned & 10 V.S.D. =9 M.S.D. \ & 1 M.S.D. =0.1 ~cm (Given) \ & 1 V.S.D. =0.9 M.S.D. \ & 1 V.S.D. =0.9 0.1 ~cm \ & V.S.D. =0.09 ~cm \ & L.C. =1 M.S.D. -1 V.S.D. \ & =(0.1-0.09) cm \ & =0.01 ~cm \ & Reading =5.08 ~cm \ & Positive zero error =0.1 ~cm (Given) \ & Final Corrected reading =(5.08-0.1) cm \ & =4.98 ~cm aligned

Question 5

The pitch of an error free screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads 1 mm and 63^ rd division on the ci

Accepted Answer

Least count of screw gauge = Pitch No. of divisions on circular scale Least count = 1 100 =0.01 ~mm Final reading = MSR + CSR L.C. =1 ~mm +(63)(0.01) mm =1.63 ~mm =0.163 ~cm

Question 6

In a vernier callipers, (N+1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm , the vernier constant (in cm ) is:

Accepted Answer

V . C=M S D-V S D....(1) given : (N+1) V S D=N MSD VSD = ( N N+1 ) MSD ....(2) From (1) and (2) aligned & V C=( MSD )- N N+1 ( MSD ) \ & = MSD (1- N N+1 )= MSD N+1 \ & = 0.01 N+1 = 1 100(N+1) aligned

Experimental Physics — Previous Year Questions & Solutions