Tag: making measurements

Questions Related to making measurements

The main scale of a vernier callipers reads 4.7 cm, the 3 rd division on the vernier scale coincides with a main scale division while measuring the length of a rod. The least count of the vernier callipers 0.1 mm. What is the length of the rod?

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 4.62 cm

  2. 4.61 cm

  3. 4.72 cm

  4. 4.73 cm

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Correct Option: D
Explanation:

$LC = 0.01 mm = 0.01 cm$
Length of the rod $=$ MSR + (VCD $\times$ LC)
$=4.7 cm + (3 \times 0.01)$
$=(4.7 + 0.03) cm$
$=4.73  cm$

A vernier having a positive zero error of +5 is used to measure the side of a 2 cm cube. The length as measure by the vernier will be : (use L.C $=$ 0.01 cm)

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 1.95 cm

  2. 2 cm

  3. 2.05 cm

  4. Cannot say

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Correct Option: C
Explanation:

Correction $=-0.505 cm (-5 \times 0.01 cm)$
Corrected reading $=$ Observed reading + correction
2 $=$ Observed reading + (-0.05) cm.
Observed reading $=$ 2.05 cm

If the length of a vernier scale having 25 divisions corresponding to 24 main scale divisions and given that 1 MSD $=$ 1 mm, then the least count of vernier calipers is

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 0.004 cm

  2. 4 micrometer

  3. 0.04 mm

  4. 0.04 m

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Correct Option: A,C
Explanation:

Given
N $=$ 25
25 VSD $=$ 24 MSD
1 MSD $=$ 1mm
$\therefore LC \displaystyle = \frac{MSD}{No.   of   VSD} = \frac{1mm}{25}$
$=0.04 mm$
$=0.04 \times 10^{-1} cm$
$=0.004 cm$
$=0.004 \times 10^{-2} m$
$=40  \mu m$

A vernier has a negative zero error. When the jaws $J _1$ and $J _2$ are brought in contact the zero of the vernier must :

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. Coincide with the zero of the main scale.

  2. Be to the right of the zero of the main scale.

  3. Be to the left of the zero of the main scale.

  4. None of the above.

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Correct Option: C
Explanation:

A vernier caliper is said to have a negative zero error when the zero of the vernier lies to left to the zero of the main scale if both the jaws are brought together.

In vernier callipers, if $L.C = l$ and pitch is P then $\displaystyle \frac{l}{P}$ is :

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. Always less than 1

  2. Always greater than 1

  3. Equal to 1

  4. None of these

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Correct Option: A
Explanation:

L.C $=$ 0.1 mm $=l$
pitch $=1$ mm  $=$ p
$\displaystyle \frac{l}{p} = \frac{0.1}{1} = 0.1  <  1$

The least count of a vernier callipers is 0.01 cm. It has an error of +0.02 cm while measuring the radius of a cylinder. The main scale reading is 3.60 cm and the 8th vernier scale division coincides with main scale, then what will be the correct radius of the cylinder?

physics making measurements vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

  1. 1.72

  2. 1.75

  3. 1.83

  4. 1.87

Show answer
Correct Option: C
Explanation:

L.C. $=$ 0.01 cm
Error $=$ + 0.02 cm $\Rightarrow$ Positive zero
error $=$ + 0.02 cm
main scale reading (MSR) $=$ 3.60 cm 8$^{th}$ vernier scale coincides with main scale
$\Rightarrow$ vernier coincidence (VC) $=$ 8
We know that
Correct diameter $=$ [main scale reading+ (L.C $\times$ V.C)] - (error)
$=$ [3.60 + (0.01 $\times$ 8)] - [+ 0.002]
$=$ (3.60 + 0.08)- (+0.02)
$=$ 3.68 - 0.02
$=$ 3.66 cm
$\therefore correct radius = \displaystyle \frac{diameter}{2} = \frac{3.66}{2}$
$=1.83 cm$