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Diameter of a steel ball is measured using a Vernier callipers which has divisions of 0.1 cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as :

If the zero error is -0.03 cm, then mean corrected diameter is :

Option 1)
0.56 cm

Option 2)
0.59 cm

Option 3)
0.53 cm

Option 4)
0.52 cm

Answers (2)

best_answer

As we learnt in

To measure the diameter of small spherical cylindrical body using Vernier Callipers -

Vernier Constant

= 1 Main scale division - 1 V.S. Division

V.C= 1 M.S.D - 1 V.S.D

M.S.D= Main Scale Reading

V.S.D= Vernier Scale Reading

- wherein

Total observed reading = $N+n \times V.C$

N= N^th division

Observations:

1. Vernier constant (least count) of the Vernier Callipers:

1 M.S.D. = 1 mm

10 vernier scale divisions = 9 main scale divisions

i.e. 10 V.S.D. = 9 M.S.D.

$\therefore$ 1 V.S.D. = $=\frac{9}{10}$ M.S.D.

Vernier Constant (L.C.) = 1 M.S.D. - 1 V.S.D. = 1 M.S.D. $-\frac{9}{10}$ M.S.D.

$=\left(1-\frac{9}{10} \right )M.S.D.=\frac{1}{10}\times1M.S.D.$

$=\frac{1}{10}\times1mm=0.1mm=0.01cm$

2. Zero error: (i) ........ cm (ii) ............cm (iii) ..........cm

Mean Zero Error (e) = ............ cm

Mean Zero Correction (c) = - (Mean Zero Error)

= .......... cm

M.S.D=0.1cm, VSD=9/10 MSD

$=0.9\times 0.1= 0.09cm$

Least count $=MSD-VSD=0.1-0.09$

$=0.01cm$

Total reading=MS Reading $\times$ Vernier Coincidence $\times$ (Least count)

1_st Reading $=0.5\times 8\times 0.01=0.58 \:cm$

2_nd Reading $=0.5\times 4\times 0.01=0.54 \:cm$

3^rdReading $=0.5\times 6\times 0.01=0.56 \:cm$

Average Reading $=\frac{(0.58+0.54+0.56)}{3}=0.56$

Correct option is 1.

Option 1)

0.56 cm

Correct

Option 2)

0.59 cm

Incorrect

Option 3)

0.53 cm

Incorrect

Option 4)

0.52 cm

Incorrect

Posted by

prateek

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

correct option is 2)

this prateek dude forgot to add the zero error.

Posted by

Osama Binlad

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