Tag: poisson ratio

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Questions Related to poisson ratio

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

A metal wire of length L is loaded and an elongation of $\Delta L$ is produced. If the area of cross section of the wire is A, then the change in volume of the wire, when elongated is . Take Poisson's ratio as 0.25

  1. $\Delta V=(\Delta L)^2A/L$

  2. $\Delta V=(\Delta L)^2A/4L$

  3. $\Delta V=(\Delta L)^2A/2L$

  4. $\Delta V=(\Delta L)^2A/3L$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $\sigma =(\Delta A/A) / (\Delta L/L)=(\Delta V/V)/(\Delta L/L)^2 \implies \Delta V=\sigma (\Delta L/L)^2V=\sigma (\Delta L/L)^2(LA)=\sigma (\Delta L)^2A/L$

Substituting $\sigma=0.25$, we get, $\Delta V=(\Delta L)^2A/4L$

The correct option is (b)

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

The change in unit volume of a material under tension with increase in its poisson's ratio will be

  1. Increase

  2. Decrease

  3. Remains same

  4. Initially increases and then decreases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The poisson's ratio is related to modulus of elasticity as $Y = 3B(1-2 \sigma)$. Since stress is same for Y and B, we get, $dL/L=dV/3V(1-2 \sigma) \implies dV=3V (dL/L)(1-2 \sigma)$
As $\sigma$ is increased, $dV$ decreases. 

The correct option is (b)

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

A student measures the poisson's ratio to be greater than 1 in an experiment. The meaning of this statement would be

  1. An increase in length would also result in decrease in area of cross section of the wire

  2. An increase in length would also result in increase in area of cross section of the wire

  3. An decrease in length would also result in decrease in area of cross section of the wire

  4. An increase in length will not change the area of cross section of the wire
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Poisson's ratio = change in area /  change in length. If poisson's ratio >1, then change in area > change in length. Thus area expands when length increases

The option (b) is the correct option

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

A copper wire 3 m long is stretched to increase its length by 0.3 cm. Find the lateral strain produced in the wire , if poisson's ratio for copper is 0.25

  1. $5 \times 10^{-4}$

  2. $2.5 \times 10^{-4}$

  3. $5 \times 10^{-3}$

  4. $2.5 \times 10^{-3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Longitudinal strain = 0.3 cm/3 m = 0.0001

Lateral strain = poisson's ratio x longitudinal strain =$ 0.25 \times 0.0001 = 2.5 \times 10^{-4}$

The correct option is (b)

Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

The theoretical limits of poisson's ratio lies between -1 to 0.5 because

  1. Shear modulus and bulk's modulus should be positive

  2. Bulk's modulus is negative during compression

  3. Shear modulus is negative during compression

  4. Young's modulus should be always positive
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let Y, K, n and $\sigma$ be the Young's Modulus, Bulk modulus, Modulus of Rigidity and Poisson's Ratio, respectively. 
Y = 3K (1 - 2$\sigma$) [Standard formula] 
Y = 2n (1 + $\sigma$) [Standard formula] 
Hence, 3K (1 - 2$\sigma$) = 2n (1 + $\sigma$) 
Now K and n are always positive, so 
i) If $\sigma$ be +ve, then RHS is always +ve. So LHS must also be +ve. Therefore, 2$\sigma$ < 1 or $\sigma$ <1/2 
ii) If $\sigma$ be -ve, then LHS will always be +ve. Therefore, 1+$\sigma$ > 0 or $\sigma$ > -1 
Thus the limiting values of Poisson's ratio are -1 < $\sigma$ < 1/2

The correct option is (a)