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A block of density $$2000 kg/m^{3}$$ and mass $$10 kg$$ is suspended by a  spring of stiffness $$100 N/m$$. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density $$1000 kg/m^{3}$$ . If the block is in equilibrium position.


A
the elongation of the spring is 1cm.
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B
the magnitude of buoyant force acting on the block is 50N.
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C
the spring potential energy is 12.5J.
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D
magnitude of spring force on the block is greater than the weight of the block.
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Solution

The correct options are
A the magnitude of buoyant force acting on the block is $$50 N.$$
B the spring potential energy is $$12.5 J.$$
There are three forces when the block is in equilibrium.
1. weight of the block downward.
2. buoyancy force due to the liquid upward
3. force due to the spring upward

Weight of block: $$m_bg=V_b\rho_bg=V_b\times 2000\times 10$$
Buoyancy force upward: $$F_{buoyancy}=V_b\rho_lg=V_b\times 1000\times 10$$
Upward force due to spring: $$F_{spring}=kx=100x$$
Equating the upward force with the downward force,
$$ V_b\times 1000\times 10 + 100x =V_b\times 2000\times 10$$
$$ \therefore 100x = V_b \times  10000= \dfrac{10}{2000} \times  10000$$
$$ \therefore x= \dfrac{1}{2}\:m$$
Potential Energy stored in the spring: $$PE = \dfrac{1}{2}kx^2= \dfrac{1}{2} \times  100 \times ( \dfrac{1}{2})^2= 12.5\: Nm = 12.5 \:J $$

134054_71878_ans.JPG

Physics

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