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A wooden block is floating on water kept...

A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to `g//2`. The block will then

A

sink

B

float with 10% above the water surface

C

float with 40% above the water surface

D

float with 70% above the water surface

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The correct Answer is:
To solve the problem, we need to analyze the situation step by step. ### Step 1: Understand the initial condition Initially, 40% of the wooden block is above the water surface, which means 60% of the block is submerged. - Let the total height of the block be \( h \). - Height above the water surface = \( 0.4h \) - Height submerged in water = \( 0.6h \) ### Step 2: Analyze the effect of the lift's acceleration The beaker is in a lift that accelerates upwards with an acceleration of \( \frac{g}{2} \). This changes the effective gravitational force acting on the block. - The effective gravitational acceleration \( g' \) when the lift accelerates upwards is given by: \[ g' = g + \frac{g}{2} = \frac{3g}{2} \] ### Step 3: Apply Archimedes' principle According to Archimedes' principle, the buoyant force acting on the block must equal the weight of the block for it to float. - The weight of the block \( W \) is given by: \[ W = mg \] - The buoyant force \( F_b \) is equal to the weight of the water displaced by the submerged part of the block. If \( V \) is the volume of the block, then: \[ F_b = \text{density of water} \times g \times \text{volume submerged} \] ### Step 4: Calculate the new conditions When the lift accelerates, the effective weight of the block changes due to the increased effective gravity. - The new effective weight of the block is: \[ W' = mg' = m \cdot \frac{3g}{2} \] ### Step 5: Set up the equilibrium condition For the block to float, the buoyant force must equal the new effective weight: \[ F_b = W' \] Substituting the expressions: \[ \text{density of water} \cdot g \cdot V_{submerged} = m \cdot \frac{3g}{2} \] ### Step 6: Determine the volume submerged Let \( V_{submerged} \) be the volume of the block that is submerged. Initially, 60% of the block was submerged, so: \[ V_{submerged} = 0.6V \] Substituting this into the buoyant force equation: \[ \rho \cdot g \cdot (0.6V) = m \cdot \frac{3g}{2} \] ### Step 7: Relate mass and volume Since \( m = \rho V \): \[ \rho \cdot g \cdot (0.6V) = \rho V \cdot \frac{3g}{2} \] ### Step 8: Simplify the equation Canceling \( \rho \) and \( g \) from both sides: \[ 0.6V = \frac{3}{2}V \] This implies: \[ 0.6 = \frac{3}{2} \cdot \text{fraction of volume submerged} \] Solving for the fraction of volume submerged: \[ \text{fraction of volume submerged} = \frac{0.6}{1.5} = 0.4 \] ### Step 9: Calculate the new height above water If 60% of the block was submerged initially, and now it is only 40% submerged, then: - Height above water = \( 100\% - 40\% = 60\% \) Thus, 10% of the block will be above the water surface. ### Conclusion The block will float with 10% of its height above the water surface.

To solve the problem, we need to analyze the situation step by step. ### Step 1: Understand the initial condition Initially, 40% of the wooden block is above the water surface, which means 60% of the block is submerged. - Let the total height of the block be \( h \). - Height above the water surface = \( 0.4h \) - Height submerged in water = \( 0.6h \) ...
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