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(B) Let $m$ be the mass of each coin. The initial submerged depth is $h_0 = 3 \, cm$ for $40$ coins.By Archimedes' principle,the weight of the contain

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let $m$ be the mass of each coin. The initial submerged depth is $h_0 = 3 \, cm$ for $40$ coins.By Archimedes' principle,the weight of the container equals the weight of the displaced water: $(40m)g = A \cdot h_0 \cdot \rho_w \cdot g$,where $A$ is the cross-sectional area and $\rho_w$ is the density of water.Thus,$A \rho_w = \frac{40m}{3}$.When $N$ additional coins are added,the total mass is $(40+N)m$. The new submerged depth $h'$ is given by $(40+N)m = A \cdot h' \cdot \rho_w$,so $h' = \frac{(40+N)m}{A \rho_w} = \frac{(40+N)m}{40m/3} = \frac{3(40+N)}{40} \, cm$.The centre of mass $(CM)$ of the system (measured from the base) is $CM = \frac{40m(0) + Nm(9)}{(40+N)m} = \frac{9N}{40+N} \, cm$.The geometric centre $(GC)$ of the submerged portion is at a distance $h'/2$ from the base: $GC = \frac{h'}{2} = \frac{3(40+N)}{80} \, cm$.For the transition from stable to unstable equilibrium,the $CM$ must coincide with the $GC$ $(CM = GC)$:$\frac{9N}{40+N} = \frac{3(40+N)}{80}$$720N = 3(40+N)^2$$240N = 1600 + 80N + N^2$$N^2 - 160N + 1600 = 0$Using the quadratic formula: $N = \frac{160 \pm \sqrt{160^2 - 4(1600)}}{2} = \frac{160 \pm \sqrt{25600 - 6400}}{2} = \frac{160 \pm \sqrt{19200}}{2} = 80 \pm 40\sqrt{3} \approx 80 \pm 69.28$.Since $N The value of $N$ is closest to $10$.

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