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Asked by ankitnayak652 | 03 Mar, 2019, 14:50: PM
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Let the density be p, the radius is r and surface tension be σ
 
Thus, 
Time of oscillation t can be proved dimesionally as follows:
 
t depends upon p, r, σ 
 
Thus, we have 
 
open square brackets t close square brackets space equals space open square brackets p close square brackets to the power of x space open square brackets r close square brackets to the power of y space open square brackets sigma close square brackets to the power of z space T h e space d i m e n s i o n space o f space t h e space q u a n t i t i e s space a r e space a s space f o l l o w s comma open square brackets t close square brackets space equals open square brackets M to the power of 0 L to the power of 0 T to the power of 1 close square brackets open square brackets p close square brackets equals open square brackets M to the power of 1 L to the power of negative 3 end exponent T to the power of 0 close square brackets open square brackets r close square brackets space equals space open square brackets M to the power of 0 L to the power of 1 T to the power of 0 close square brackets open square brackets sigma close square brackets space equals space open square brackets M to the power of 1 L to the power of 0 T to the power of negative 2 end exponent close square brackets T h u s comma space open square brackets M to the power of 0 L to the power of 0 T to the power of 1 close square brackets space equals space open square brackets M to the power of 1 L to the power of negative 3 end exponent T to the power of 0 close square brackets to the power of x space open square brackets M to the power of 0 L to the power of 1 T to the power of 0 close square brackets to the power of y space open square brackets M to the power of 1 L to the power of 0 T to the power of negative 2 end exponent close square brackets to the power of z open square brackets M to the power of 0 L to the power of 0 T to the power of 1 close square brackets space equals space open square brackets M to the power of x L to the power of negative 3 x end exponent T to the power of 0 close square brackets space open square brackets M to the power of 0 L to the power of y T to the power of 0 close square brackets space open square brackets M to the power of z L to the power of 0 T to the power of negative 2 z end exponent close square brackets therefore space open square brackets M to the power of 0 L to the power of 0 T to the power of 1 close square brackets space equals space open square brackets M to the power of x plus z end exponent L to the power of negative 3 x plus y end exponent T to the power of negative 2 z end exponent close square brackets space H e n c e comma space c o m p a r i n g space t h e space d i m e n s i o n s space x plus z space equals 0
 
 
negative 3 x space plus space y space equals space 0 minus 2 z space equals space 1 space T h u s space z space equals negative space 1 half T h u s comma space x space equals space 1 half & space y space equals space 3 over 2 H e n c e comma space t space equals space p to the power of 1 half end exponent space r to the power of 3 over 2 end exponent space sigma to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent space W h i c h space c a n space b e space w r i t t e n space a s comma t space equals space square root of fraction numerator p r cubed over denominator sigma end fraction end root H e n c e space p r o v e d
 
Answered by Shiwani Sawant | 03 Mar, 2019, 15:17: PM

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