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

The Poisson's ratio $\sigma$ should satisfy the relation :

  1. -1< $\sigma $ < 0.5

  2. -0.5 < $\sigma $ < 1.0

  3. 0.5 < $\sigma $ < 1.0

  4. -1.0 < $\sigma $ < -0.5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Poisson's ratio is the ratio of transverse contradiction strain to longitudinal extension strain in the direction of stretching force.

The Poisson's ratio $\sigma$ should satisfy the relation,
$-1<\sigma <0.5$

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

A metallic wire of young's modulus Y and poisson's ratio $\sigma$, length L and area of cross section A is stretched by a load of W kg. The increase in volume of the wire is:

  1. $\sigma (W^2 L/2AY^2)$

  2. $\sigma (W^2 L/AY^2)$

  3. $\sigma (W^2 L/4AY^2)$

  4. $\sigma (2W^2 L/AY^2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

We also know $Y=(W/A)/(\Delta L/L) \implies (\Delta L/L)=W/AY$

Substituting this value in the previous expression, we get, $\Delta V=\sigma (W/AY)^2V=\sigma (W^2 L/AY^2)$

The correct option is (b)

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

Poisson' ratio is defined as the ratio of 

  1. longitudinal stress and longitudinal strain

  2. longitudinal stress and lateral stress

  3. lateral stress and longitudinal stress

  4. lateral stress and lateral strain
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Poisson' ratio is defined as the ratio of lateral stress and longitudinal stress

The correct option is (c)

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)