If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t=√r^2bs^cd^a/2. It is observed that the time period is directly proportional to √d/s. The value of b should therefore be

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(3) 3/2. Idea: square both sides first, then match each base. t²=r^2bs^cd^a/2, with r=[L], s=[MT⁻²], d=[ML⁻³]. The stated proportionality to √d/s fixes a/2=1 and c=-1. Mass: 0=c+a/2=-1+1, which holds. Length: 0=2b-(3a)/2=2b-3, so b=3/2.

If the time period t of the oscillation of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t=√r^2bs^cd^a/2. It is observed that the time period is directly proportional to √d/s. The value of b should therefore be

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