Solveeit Logo
Login

Question

Question: The length of a cube increases by \(0.1\% \) .What is the bulk strain? A. \(0.003\) B. \(0.006\)...

The length of a cube increases by 0.1%0.1\% .What is the bulk strain?
A. 0.0030.003
B. 0.0060.006
C. 0.0020.002
D. 0.0010.001

Explanation

Solution

This problem can be solved by applying the bulk strain formula. First of all calculate the new voltage with the help of the change in length as given in the question. The new length should be equal to origina lange added to change in length. Finally put all the values in the bulk strain formula.

Formula Used:
Bulk Strain=V′−VV = \dfrac{{V' - V}}{V}
Where, V’V’= New Volume and VV= Original Volume.

Complete step by step answer:
In the question it is given that the length of the cute is increased by 0.1%0.1\% . The above statement can be written as change in length divided by the original length whole multiplied by 100100 gives 0.10.1. Numerically represented as
ΔLL×100=0.1\dfrac{{\Delta L}}{L} \times 100 = 0.1
⇒ΔLL=0.1100\Rightarrow \dfrac{{\Delta L}}{L} = \dfrac{{0.1}}{{100}}
⇒ΔLL=0.001\Rightarrow \dfrac{{\Delta L}}{L} = 0.001
Where,
ΔL=\Delta L = Change in length
L=L = Length of cube
Hence the volume of the cube will be,
V=L3V = {L^3}

Let V′V' be the new volume of the cube due to increase in length of the cube
Now we can write that
New Volume= (origina length + change in length) whole cube
V′=(L+ΔL)3V' = {\left( {L + \Delta L} \right)^3}
Taking out LL as common term we will get,
V′=L3(1+ΔLL)3V' = {L^3}{\left( {1 + \dfrac{{\Delta L}}{L}} \right)^3}
Opening the cube basket by applying(a+b)3{\left( {a + b} \right)^3} formula, we get
V′=L3(1+3ΔLL+ΔL3L+3ΔL2L2)V' = {L^3}\left( {1 + \dfrac{{3\Delta L}}{L} + \dfrac{{\Delta {L^3}}}{L} + \dfrac{{3\Delta {L^2}}}{{{L^2}}}} \right)
As ΔL\Delta L itself is a very small value squaring and cubing ΔL\Delta L value will give very very small values hence these terms can be neglected so we get,
V′=L3(1+3ΔLL)V' = {L^3}\left( {1 + \dfrac{{3\Delta L}}{L}} \right)

Here we know that L3{L^3} is the original volume of the cube.
Putting VV in place ofL3{L^3} we will get,
V′=V(1+3ΔLL)V' = V\left( {1 + \dfrac{{3\Delta L}}{L}} \right)
As in the question it is given that, ΔLL=0.001\dfrac{{\Delta L}}{L} = 0.001
By putting this value in new volume we will get,
V′=V(1+3×0.001)V' = V\left( {1 + 3 \times 0.001} \right)
⇒V′=V(1+0.003)\Rightarrow V' = V\left( {1 + 0.003} \right)
By rearranging the above equation we get,
V′V=1+0.003\dfrac{{V'}}{V} = 1 + 0.003
Subtracting both sides with 11 ,we get
V′V−1=1+0.003−1\dfrac{{V'}}{V} - 1 = 1 + 0.003 - 1
⇒V′−VV=0.003\Rightarrow \dfrac{{V' - V}}{V} = 0.003
Where, Bulk Strain=V′−VV = \dfrac{{V' - V}}{V}
Hence, Bulk Strain=0.003 = 0.003

Therefore the correct option is (A)\left( A \right).

Note: Bulk strain has no unit because it is the ratio of volumes. Remember every rational term is unitless. After calculating V’ from that step you can directly use the bulk strain formula that should be easier for you to understand and to solve the problem.