Physics Q94 – AP-EAMCET 2026

Home Physics Mechanical Properties of SolidsQuestion #280

A rubber hose

50

cm long and internal diameter

1

cm is stretched to

60

cm. The internal diameter of stretched hose is (Poisson's ratio of the rubber

σ = 0.5

)

Clue 1 of 3: Elastic Properties

This problem involves the elastic property of a material known as Poisson's ratio, which describes the transverse contraction of a material when subjected to longitudinal extension.

🧠 Full Solution Path

Step 1: Identify Given Parameters✦ Active

The problem provides the initial length ( $L_{0}$ ), final length ( $L$ ), initial internal diameter ( $D_{0}$ ), and Poisson's ratio ( $σ$ ) of a rubber hose. We need to find the final internal diameter ( $D$ ) after stretching.

L_{0} = 50 cm

L = 60 cm

D_{0} = 1 cm = 10 mm

σ = 0.5

💡 Teacher's Secret Hint

Always convert all units to a consistent system (e.g., SI or CGS) or ensure they cancel out properly. Here, we can work with cm until the final conversion to mm.

Step 2: Calculate Longitudinal Strain○ Expand

Longitudinal strain ( $ϵ_{L}$ ) is the fractional change in length. It is calculated as the change in length divided by the original length.

ϵ_{L} = \frac{Δ L}{L_{0}} = \frac{L - L_{0}}{L_{0}}

ϵ_{L} = \frac{60 cm - 50 cm}{50 cm} = \frac{10 cm}{50 cm} = 0.2

Step 3: Calculate Lateral Strain using Poisson's Ratio○ Expand

Poisson's ratio ( $σ$ ) is defined as the negative ratio of lateral strain ( $ϵ_{D}$ ) to longitudinal strain ( $ϵ_{L}$ ). We can use this to find the lateral strain.

σ = - \frac{ϵ_{D}}{ϵ_{L}}

ϵ_{D} = - σ \times ϵ_{L}

ϵ_{D} = - 0.5 \times 0.2 = - 0.1

The negative sign indicates that the diameter decreases when the material is stretched longitudinally (positive longitudinal strain).

💡 Teacher's Secret Hint

Remember the negative sign in the Poisson's ratio formula. It's crucial for indicating whether the transverse dimension contracts or expands.

Step 4: Calculate the Final Internal Diameter○ Expand

Lateral strain ( $ϵ_{D}$ ) is also defined as the fractional change in diameter. We can use this to find the final diameter ( $D$ ).

ϵ_{D} = \frac{Δ D}{D_{0}} = \frac{D - D_{0}}{D_{0}}

D - D_{0} = ϵ_{D} \times D_{0}

D = D_{0} + (ϵ_{D} \times D_{0})

D = D_{0} (1 + ϵ_{D})

D = 1 cm (1 + (- 0.1)) = 1 cm (0.9) = 0.9 cm

Finally, convert the diameter to millimeters as per the options.

D = 0.9 cm \times 10 mm/cm = 9 mm

💡 Teacher's Secret Hint

Always check the units required in the options. Often, the calculated answer needs a final conversion.

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Physics Question 94 – AP-EAMCET 2026

Clue 1 of 3: Elastic Properties

🧠 Full Solution Path