Question

The edge of the cube is measured by a scale
of least count 1 mm. The measured value of
L is 1.2 cm. The volume of the cube should
be recorded as​

Answer 1

Answer:

The measured length is

a=(1.2±0.1)cm, where

Δa=0.1 cm

is the least count of the scale.

The volume of the cube is

V=a³

=(1.2)³

=1.728 cm³

The error in volume is ΔV=(3Δa/a)V

=0.432 cm³

Thus, volume should be reported as (1.7±0.4) cm³ up to two significant digits.

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Answer 2

The volume of the cube recorded as $1.728$×$10^{-6}$  

Explanation:

• Let the edge of the cube be denoted by 'a'
• The measured value of L is 1.2 cm is given=1.2×10−2 m (given)
• Volume is given as $a^{3}$
• when we substitute the value of a in volume  = ($1.2$ × $10^{-2} )^{3}$
• The volume of the cube recorded as $1.728$×$10^{-6}$  
• Least count of Vernier calipers is the difference between the main scale and Vernier scale division.
• It is the difference between the main scale division and Vernier scale division. reading.
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