
The edge of a cube is measured using a vernier callipers [9 divisions of the main scale is equal to 10 divisions of the vernier scale and 1 main scale division is 1 mm]. The main scale division reading is 10 and 1 division of vernier scale was found to be coinciding with the main scale. The mass of the cube is 2.736 gm. Calculate the density in gm/cc up to correct significant figures.
Answer
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Hint: The least count of the vernier callipers is described as the difference between 1 main scale division reading and 1 vernier scale division reading. From this we can calculate least count. Then, the total reading is given as:
Total reading= [main scale divisions x 1 main scale reading] + [least count x the coinciding vernier scale division].This is equal to the cube edge from which we can find the volume. The mass is given and hence the density can be calculated.
Complete step by step answer:
The least count of the vernier callipers is described as the difference between 1 main scale division reading and 1 vernier scale division reading. Given,
1 main scale reading = 1mm
1 vernier scale reading = 9/10 x 1 mm=0.9mm
Least count = (1-0.9) mm = 0.1mm= 0.01cm
Then, the total reading is given as:
Total reading= [main scale divisions x 1 main scale reading] + [least count x the coinciding vernier scale division]
Total reading = [10x1] mm + [1x0.1] mm = 10.1 mm = 1.01 cm
The length of the cube edge is 1.01 cm. Therefore, the volume is expressed as:
\[
V = {a^3} = {(1.01)^3}c{m^3} \\
V = 1.03cc \\
\]
The mass is given as 2.736 gm. Therefore, the density is given as
\[
d = \dfrac{{mass}}{{volume}} = \dfrac{{2.736}}{{1.03}}gm/cc \\
d = 2.656gm/cc \\
\]
The density of the cube is 2.656 gm/cc
Note:The conversion of unit (from mm to cm) is important here else the whole answer will be wrong. It is a subjective question and therefore there is no chance to redeem yourself with the help of options. The only typical part where you have to be careful is while calculating the least count and thus the reading of vernier callipers.
Total reading= [main scale divisions x 1 main scale reading] + [least count x the coinciding vernier scale division].This is equal to the cube edge from which we can find the volume. The mass is given and hence the density can be calculated.
Complete step by step answer:
The least count of the vernier callipers is described as the difference between 1 main scale division reading and 1 vernier scale division reading. Given,
1 main scale reading = 1mm
1 vernier scale reading = 9/10 x 1 mm=0.9mm
Least count = (1-0.9) mm = 0.1mm= 0.01cm
Then, the total reading is given as:
Total reading= [main scale divisions x 1 main scale reading] + [least count x the coinciding vernier scale division]
Total reading = [10x1] mm + [1x0.1] mm = 10.1 mm = 1.01 cm
The length of the cube edge is 1.01 cm. Therefore, the volume is expressed as:
\[
V = {a^3} = {(1.01)^3}c{m^3} \\
V = 1.03cc \\
\]
The mass is given as 2.736 gm. Therefore, the density is given as
\[
d = \dfrac{{mass}}{{volume}} = \dfrac{{2.736}}{{1.03}}gm/cc \\
d = 2.656gm/cc \\
\]
The density of the cube is 2.656 gm/cc
Note:The conversion of unit (from mm to cm) is important here else the whole answer will be wrong. It is a subjective question and therefore there is no chance to redeem yourself with the help of options. The only typical part where you have to be careful is while calculating the least count and thus the reading of vernier callipers.
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