Four identical beakers contain same amount of water as shown below. Beaker 'a' contains only water. A lead ball is held submerged in the beaker 'b' by string from above. A same sized plastic ball, say a table tennis (TT) ball, is held submerged in beaker 'c' by a string attached to a stand from outside. Beaker 'd' contains same sized TT ball which is held submerged from a string attached to the bottom of the beaker. These beakers (without stand) are placed on weighing pans and register reading $\mathrm{W}_{\mathrm{a}}, \mathrm{W}_{\mathrm{b}}, \mathrm{W}_{\mathrm{c}}$ and $\mathrm{W}_{\mathrm{d}}$ for $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and d, respectively. (Effects of the mass and volume of the stand and string are to be neglected)

• A

  $W_{\mathrm{a}}=\mathrm{W}_{\mathrm{b}}=\mathrm{W}_{\mathrm{c}}=\mathrm{W}_{\mathrm{d}}$
• Correct answer
  B

  $W_{b}=W_{c}>W_{d}>W_{a}$
• C

  $W_{\mathrm{b}}=\mathrm{W}_{\mathrm{c}}>\mathrm{W}_{\mathrm{a}}>\mathrm{W}_{\mathrm{d}}$
• D

  $\mathrm{W}_{\mathrm{b}}>\mathrm{W}_{\mathrm{c}}>\mathrm{W}_{\mathrm{d}}>\mathrm{W}_{\mathrm{a}}$

Solution

(A)
$\mathrm{N}_{\mathrm{A}}=\mathrm{Mg}$ (B)
$\begin{aligned}\left(\mathrm{N}_{\mathrm{B}}\right.&=\left(\mathrm{M}+\mathrm{M}_{\mathrm{L}}\right) \mathrm{g}-\mathrm{T} \\ \mathrm{N}_{\mathrm{B}} &=\mathrm{Mg}+\mathrm{B} \end{aligned}$ (C)
$\mathrm{N}_{\mathrm{C}}=\mathrm{T}+\left(\mathrm{M}+\mathrm{m}_{\mathrm{T}}\right) \mathrm{g}$ Therefore $\mathrm{N}_{\mathrm{L}}=\mathrm{B}-\mathrm{m}_{\mathrm{T}} \mathrm{g}+\left(\mathrm{M}+\mathrm{m}_{\mathrm{T}}\right) \mathrm{g}$ $\mathrm{N}_{\mathrm{L}}=\mathrm{B}+\mathrm{Mg}$ (D)
$\mathrm{N}_{\mathrm{D}}=\left(\mathrm{M}+\mathrm{m}_{\mathrm{T}}\right) \mathrm{g}$

Hence, the correct answer is option: 'B'.