Question

The side of a cube is measured by Vernier calipers (10 divisions of the vernier scale coincide with 9 divisions of the main scale, where 1 division of the main scale is 1mm) . The main scale reads 10 mm and the first division of vernier scale coincides with the main scale. The mass of the cube is 2.736g. Find the density of cube in appropriate significant figures.

Answer 1

Vernier calipers are used to measure the length of any substance much more precisely than a normal scale. For any instrument, the least count(L.C) gives the value of the minimum unit that can be measured precisely using that instrument. For Vernier calipers that L.C is in one-tenth of a millimeter range.

Formula Used:
The least count of Vernier calipers is given by:
$L.C. = \dfrac{{1M.S.D.}}{N}$ (1)
Where,
L.C. denotes the least count,
M.S.D. denotes the value of one main scale division,
N is the number of divisions in the Vernier scale.

Reading using Vernier calipers is given by:
$l = M.S.R + L.C. \times V.S.R$ (2)
Where,
l denotes the total length,
M.S.R. is the main scale reading,
V.S.R is Vernier scale reading.

The density of a cube can be calculated as:
$d = \dfrac{M}{{{l^3}}}$ (3)
Where,
d is the density of the cube,
M is the mass of the cube.

Complete step by step answer:
Given:

1. Number of divitions in Vernier scale is N=10.
2. One main scale division is 1 mm, i.e. $1M.S.D = 1mm = {10^{ - 3}}m$.
3. Here, main scale reading is given as$M.S.R. = 10mm = {10^{ - 2}}m$.
4. First division of Vernier scale coincides with main scale, so V.S.R = 1.
   5)Mass of the cube is, $M = 2.736{\text{ }}g = 2.736 \times {10^{ - 3}}kg$.

To find the Density of the cube.

Step 1
Use eq.(1) and the given values to get the least count of Vernier calipers as:
$L.C. = \dfrac{{1M.S.D.}}{N} \\\ \therefore L.C. = \dfrac{{{{10}^{ - 3}}m}}{{10}} = {10^{ - 4}}m \\\$ (4)

Step 2
Now, use this value of L.C. from eq.(4) and other given values of M.S.R and V.S.R values to get the length of the cube l:
$l = {10^{ - 2}}m + 1 \times {10^{ - 4}}m \\\ \therefore l = 0.0101m \\\$ (5)

Step 3
Substitute M and l in eq.(3) to get the value of density d as:
$d = \dfrac{{2.736 \times {{10}^{ - 3}}kg}}{{{{\left( {0.0101m} \right)}^3}}} \\\ \therefore d = 2655.535kg{m^{ - 3}} \\\$

Final answer:
Density of the cube is $2655.535kg{m^{ - 3}}$.

Note: Many students can’t understand how V.C. works. 10 vernier scale division matches with 9 main scale divisions. So, 10 V.C division lags 10 main scale divisions by 1 main scale division i.e. 1 mm. So, for each step 1 Vernier scale division lags 1 main scale division by L.C. amount. Now, when we measure something often the exact length is a bit greater than some main scale marking. To measure that ‘excess’ part after the main scale mark, we see how much vernier scale reading is needed to match one main scale. For each step, the gap decreases by the L.C. amount. So, we get the actual reading just by adding the product of L.C. and Vernier scale reading with the main scale reading.