A homogeneous solid cylinder of length L (L

Question

(A) For a floating object,the weight of the object is equal to the total buoyant force exerted by the liquids.Let $V$ be the total volume of the cylin

Vedclass · Accepted Answer

$\frac{5}{4}d$

Vedclass · Answer

(A) For a floating object,the weight of the object is equal to the total buoyant force exerted by the liquids.Let $V$ be the total volume of the cylinder,$V = (A/5) 	imes L$.The weight of the cylinder is $W = V 	imes D 	imes g = (A/5) 	imes L 	imes D 	imes g$.The buoyant force is the sum of the forces from the two liquids.The length of the cylinder in the upper liquid (density $d$) is $L - L/4 = 3L/4$.The length of the cylinder in the lower liquid (density $2d$) is $L/4$.The buoyant force $F_B = V_{upper} 	imes d 	imes g + V_{lower} 	imes 2d 	imes g$.$F_B = (A/5 	imes 3L/4) 	imes d 	imes g + (A/5 	imes L/4) 	imes 2d 	imes g$.Equating weight and buoyant force: $(A/5) 	imes L 	imes D 	imes g = (A/5) 	imes (3L/4) 	imes d 	imes g + (A/5) 	imes (L/4) 	imes 2d 	imes g$.Dividing both sides by $(A/5) 	imes L 	imes g$,we get:$D = (3/4)d + (1/4) 	imes 2d = (3/4)d + (2/4)d = 5d/4$.

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