Question

Figure shows five identical opentop container filled up to the brim with water. In figure (b), toy duck is floating, in figure (c) a heavy toy duck sinks and touches the bottom, in figure (d) a heavy toy duck is suspended from a string and in figure (e) a light toy duck is tied with a mass less string. If all the five containers are put on weighing machine, the correct order of their reading will be :

• A. $\left(a\right)=\left(b\right)=\left(c\right)=\left(d\right)=\left(e\right)$
• B. $\left(e\right)<\left(a\right)=\left(b\right)=\left(d\right)<\left(c\right)$
• C. $\left(e\right)<\left(d\right)<\left(a\right)=\left(b\right)<\left(c\right)$
• D. $\left(a\right)=\left(b\right)=\left(d\right)=\left(e\right)<\left(c\right)$

Answer 1

The correct option is B $\left(e\right)<\left(a\right)=\left(b\right)=\left(d\right)<\left(c\right)$

(b) weight of the displaced water( upward thrust due to buoyancy) is equal to the weight of the duck. Thus the weighing machine will have the same reading as the case in (a) which has water in a container
(c) Since the heavy duck sinks inside water, its density is greater than that of water. the weighing machine reading will be greater than that in (a)
(d)The buoyancy force on duck is equal to the weight of the displaced water.So, weight of system (d) $=$ weight of system (a)
(e)As the density of the duck is less than that of water, so weight of system (e) is less that of system (a).