Tag: making measurements

Quizzes (5) Questions (6)

Questions Related to making measurements

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

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?

  1. 4.62 cm

  2. 4.61 cm

  3. 4.72 cm

  4. 4.73 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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)

  1. 1.95 cm

  2. 2 cm

  3. 2.05 cm

  4. Cannot say
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. 0.004 cm

  2. 4 micrometer

  3. 0.04 mm

  4. 0.04 m
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

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

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

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 :

  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.
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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?

  1. 1.72

  2. 1.75

  3. 1.83

  4. 1.87
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$\therefore correct radius = \displaystyle \frac{diameter}{2} = \frac{3.66}{2}$
$=1.83 cm$