Question

13. A rubber ball is taken to depth 1 km inside water so that its volume reduces by 0.05%. What is the bulk
modulus for rubber?
(1) 2 x 10 N/m2
(2) 2 x 10 N/m2
(3) 2x 1010 N/m2
(4) 2 x 1011 N/m2​

Answer 1

Answer:

• The Bulk modulus (B) of rubber is 2 × 10¹⁰ N / m²

Given:

1. Given Depth (h) = 1 Km = 1000 m
2. Volume Decreases (Δ V / V) = 0.05 %

Explanation:

$\rule{300}{1.5}$

From the formula we know,

⇒ P = P₀ + h ρ g

Where,

• P Denotes Pressure.
• P₀ Denotes atmospheric pressure.
• h Denotes Height.
• ρ Denotes Density.
• g Denotes Acceleration due to gravity.

Now,

⇒ P = P₀ + h ρ g

Substituting the values,

⇒ P = P₀ + 1000 × 1000 × 10

∵ [ h = 1000 m; ρ = 1000 Kg/m³ ]

⇒ P = P₀ + 10⁷

⇒ P - P₀ = 10⁷

∵ [ P - P₀ = Δ P ]

⇒ Δ P = 10⁷

⇒ Δ P = 10⁷ N/m².

∴ We got the Pressure Difference.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

The volume reduces to 0.05 %[In Percentage]

Therefore,

⇒ Δ V / V = 0.05 %

⇒ Δ V / V = 0.05 / 100

⇒ Δ V / V = 5 × 10⁻² / 100

⇒ Δ V / V = 5 × 10⁻² × 10⁻²

⇒ Δ V / V = 5 × 10⁻⁴

⇒ Δ V / V = 5 × 10⁻⁴.

∴ We got the change in volume.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the formula we know,

⇒ B = Δ P / ( Δ V / V )

Substituting the values,

⇒ B = 10⁷ / 5 × 10⁻⁴

⇒ B = 10⁷ × 10⁴ / 5

⇒ B = 0.2 × 10⁽⁷⁺⁴⁾

⇒ B = 0.2 × 10¹¹

⇒ B = 2 × 10¹⁰

⇒ B = 2 × 10¹⁰ N / m².

∴ The Bulk modulus (B) of rubber is 2 × 10¹⁰ N / m².

$\rule{300}{1.5}$

Answer 2

$</p><p>\tt{\pink{\huge{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

• Depth = 1 km

• Volume reduction = 0.05%

$</p><p>\tt{\orange{Step-By-Step-Explaination \: - }}$

• Pressure at 1 km depth inside water :

P = > 1000×10×1000 N

= > 10^7 N / m^2

Let us assume the volume of the ball as V.

Hence,

$\tt{ \blue{ \frac{0.05}{100} \times v \: = { \delta}v}}$

$\tt{ \purple{ { \beta} \: = \dfrac{ - PV}{{ \delta}v } = \dfrac{ - {10}^{7} v}{ \dfrac{0.05V}{100} } = 2 \times {10}^{10} N / m^2 }}$

Hence option 3 is correct.