Question

A 20 g copper block is suspended by a vertical spring causing 1 cm elongation over the natural length of the spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is (Density of copper 9000 kg m\(^{-3}\), Density of water 1000 kg m\(^{-3}\), \( g = 10 \) ms\(^{-2}\)):

• A. \( 0.25 \, \text{cm} \)
• B. \( 0.15 \, \text{cm} \)
• C. \( 0.78 \, \text{cm} \)
• D. \( 0.89 \, \text{cm} \)

Hint:

Remember to consider all forces acting on the submerged object. The buoyant force effectively reduces the force stretching the spring, leading to a smaller elongation compared to when the object is in air. Pay close attention to unit conversions throughout the problem.

Answer 1

The Correct Option is D

Step 1: Determine the spring constant using the initial elongation in air.
When the copper block is suspended in air, the only downward force acting on it is its weight \( W \), which causes the spring to elongate by \( x_1 = 1 \, \text{cm} \). The mass of the copper block is \( m = 20 \, \text{g} \). To use SI units consistently, we convert this to kilograms: \[ m = 20 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.02 \, \text{kg} \] The weight of the block is given by \( W = mg \), where \( g \) is the acceleration due to gravity (\( 10 \, \text{ms}^{-2} \)): \[ W = (0.02 \, \text{kg})(10 \, \text{ms}^{-2}) = 0.2 \, \text{N} \] According to Hooke's Law, the force exerted by a spring is \( F = kx \), where \( k \) is the spring constant and \( x \) is the elongation. In the initial situation, the weight of the block is balanced by the spring force: \( W = kx_1 \). We also need to convert the elongation to meters: \[ x_1 = 1 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.01 \, \text{m} \] Now we can find the spring constant \( k \): \[ 0.2 \, \text{N} = k (0.01 \, \text{m}) \] \[ k = \frac{0.2 \, \text{N}}{0.01 \, \text{m}} = 20 \, \text{Nm}^{-1} \]
Step 2: Calculate the buoyant force acting on the copper block when immersed in water.
When the copper block is completely immersed in water, it experiences an upward buoyant force \( F_B \) according to Archimedes' Principle. The buoyant force is equal to the weight of the fluid (water) displaced by the block. To find this, we first need the volume of the copper block. The volume \( V \) is given by \( V = \frac{m}{\rho} \), where \( m \) is the mass and \( \rho \) is the density. The density of copper \( \rho_{cu} = 9000 \, \text{kg m}^{-3} \). \[ V_{cu} = \frac{0.02 \, \text{kg}}{9000 \, \text{kg m}^{-3}} = \frac{2 \times 10^{-2}}{9 \times 10^3} \, \text{m}^3 = \frac{2}{9} \times 10^{-5} \, \text{m}^3 \] The volume of water displaced \( V_w \) is equal to the volume of the copper block \( V_{cu} \). The density of water \( \rho_w = 1000 \, \text{kg m}^{-3} \). The buoyant force \( F_B \) is the weight of this displaced water: \[ F_B = \rho_w V_w g = (1000 \, \text{kg m}^{-3}) \left( \frac{2}{9} \times 10^{-5} \, \text{m}^3 \right) (10 \, \text{ms}^{-2}) \] \[ F_B = \frac{2 \times 10^4 \times 10^{-5}}{9} \, \text{N} = \frac{2 \times 10^{-1}}{9} \, \text{N} = \frac{0.2}{9} \, \text{N} \approx 0.0222 \, \text{N} \]
Step 3: Determine the net downward force acting on the spring when the block is immersed.
When the block is immersed, there are two vertical forces acting on it: the downward weight \( W \) and the upward buoyant force \( F_B \). The net downward force \( F_{net} \) that stretches the spring is the difference between the weight and the buoyant force: \[ F_{net} = W - F_B = 0.2 \, \text{N} - 0.0222 \, \text{N} = 0.1778 \, \text{N} \]
Step 4: Calculate the new elongation of the spring due to this net force.
Let the new elongation of the spring be \( x_2 \). According to Hooke's Law, the net force is related to the new elongation by \( F_{net} = kx_2 \). We know \( k = 20 \, \text{Nm}^{-1} \) and \( F_{net} = 0.1778 \, \text{N} \). We can solve for \( x_2 \): \[ x_2 = \frac{F_{net}}{k} = \frac{0.1778 \, \text{N}}{20 \, \text{Nm}^{-1}} = 0.00889 \, \text{m} \] Finally, we convert this elongation back to centimeters: \[ x_2 = 0.00889 \, \text{m} \times \frac{100 \, \text{cm}}{1 \, \text{m}} = 0.889 \, \text{cm} \] Rounding to two decimal places, the elongation of the spring when the copper block is completely immersed in water is approximately \( 0.89 \, \text{cm} \).