A 20 ,g copper block is suspended by a vertic | vedclass

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(D) Mass of block, $m = 20 \,g = 0.02 \,kg$.Initial elongation, $x = 1 \,cm = 0.01 \,m$.At equilibrium, the spring force equals the weight: $kx = mg$.

Vedclass · Accepted Answer

$0.89$

Vedclass · Answer

(D) Mass of block, $m = 20 \,g = 0.02 \,kg$.Initial elongation, $x = 1 \,cm = 0.01 \,m$.At equilibrium, the spring force equals the weight: $kx = mg$.Spring constant $k = \frac{mg}{x} = \frac{0.02 	imes 10}{0.01} = 20 \,N/m$.When the block is immersed in water, it experiences an upward buoyant force $F_B = ho_w V g$, where $V = \frac{m}{ho_c}$.The new equilibrium condition is $kx' + F_B = mg$, where $x'$ is the new elongation.$kx' = mg - ho_w \left(\frac{m}{ho_c}ight) g = mg \left(1 - \frac{ho_w}{ho_c}ight)$.$x' = \frac{mg}{k} \left(1 - \frac{ho_w}{ho_c}ight) = x \left(1 - \frac{1000}{9000}ight)$.$x' = 0.01 	imes \left(1 - \frac{1}{9}ight) = 0.01 	imes \frac{8}{9} \approx 0.00888 \,m$.$x' \approx 0.89 \,cm$.

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