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Question
Consider a vernier caliper. Least count of main scale is 1 mm. There are 20 equispaced marks on vernier scale. When nothing is put between the jaws, zero of main scale lies between zero and first division of vernier scale and 15th division of vernier exactly coincide with one of the main scale division. When zero of vernier is matched with zero of main scale, 20th division of vernier exactly coincides with 25th division of main scale. To measure the side of a cuboid, it is kept between the jaws of vernier caliper, it is now observed that 18th division of vernier exactly coincides with 32th division of main scale. Choose the correct option(s).
Consider a vernier caliper. Least count of main scale is 1 mm. There are 20 equispaced marks on vernier scale. When nothing is put between the jaws, zero of main scale lies between zero and first division of vernier scale and 15th division of vernier exactly coincide with one of the main scale division. When zero of vernier is matched with zero of main scale, 20th division of vernier exactly coincides with 25th division of main scale. To measure the side of a cuboid, it is kept between the jaws of vernier caliper, it is now observed that 18th division of vernier exactly coincides with 32th division of main scale. Choose the correct option(s).
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Solution
The correct options are
A Least count of vernier scale is 0.25 mm
C Side of cuboid is 10.25 mm
When, zero of both the scale coincided then 20th division of vernier exactly coincides with 25th division of main scale.So Vernier scale division will be given by 2520×1 mm=1.25 mm
Since, vernier scale division is larger (these vernier are called retrogate vernier), for least count we have to use formula of least count as
L.C = V.S.D-M.S.D=1.25−1=0.25 mm
Now, when zero of both the scale were not coinciding there will be zero error and as main scale zero lies between zero and 1 of vernier, so zero error will be of negative type. Let X divison of main scale coincide with 15th vernier scale division. Since, 15 V.S.D = 15×1.25=17.25 mm and as main scale zero was ahead, so, 18th division of main scale will coincide with 15th division of vernier scale.
So, zero error will be given by 15 VSD - X MSD
Zero error = 17.25−18=−0.75 mm
Since, to measure the side of a cuboid,18th division of vernier exactly coincides with 32th division of main scale.So difference between reading =32−18×1.25=9.5 mmwhich on inculcating factor of zero error will give actual length of side of cuboid.Therefore,
Actual length of side of cuboid =Main scale reading-zero error =9.50 mm−(−0.75 mm)=10.25 mm
A Least count of vernier scale is 0.25 mm
C Side of cuboid is 10.25 mm
When, zero of both the scale coincided then 20th division of vernier exactly coincides with 25th division of main scale.So Vernier scale division will be given by 2520×1 mm=1.25 mm
Since, vernier scale division is larger (these vernier are called retrogate vernier), for least count we have to use formula of least count as
L.C = V.S.D-M.S.D=1.25−1=0.25 mm
Now, when zero of both the scale were not coinciding there will be zero error and as main scale zero lies between zero and 1 of vernier, so zero error will be of negative type. Let X divison of main scale coincide with 15th vernier scale division. Since, 15 V.S.D = 15×1.25=17.25 mm and as main scale zero was ahead, so, 18th division of main scale will coincide with 15th division of vernier scale.
So, zero error will be given by 15 VSD - X MSD
Zero error = 17.25−18=−0.75 mm
Since, to measure the side of a cuboid,18th division of vernier exactly coincides with 32th division of main scale.So difference between reading =32−18×1.25=9.5 mmwhich on inculcating factor of zero error will give actual length of side of cuboid.Therefore,
Actual length of side of cuboid =Main scale reading-zero error =9.50 mm−(−0.75 mm)=10.25 mm