1.

Derive an expression for time period (t) of a simple pendulem, which may depend upon : mass of bob (m), length of pendulum (I) and acceleration due to gravity(g).

Answer» Let `t prop m^a l^b g^c` where a, b, c are the dimensions. Or `t = k m^a l^b g^c …(4)`
where k is dimensionless constant of proportionality.
Writing the dimensions in terms of M, L, T on either side of (4), we get
`[M^0 L^0 T^1] = M^a L^b (LT^(-2))^c = M^a L^(b + c) T^(-2c)`
Applying the principle of homogeneity of dimensions, we get
`a = 0 b+c =0 ....(5)`
` -2 c= 1 or c = -(1)/(2)`
From (5),` b=- c = -(-(1)/(2)) = (1)/(2)`
`"Putting the value of a,b,c, in"` (4), `"we get"`
`t = km^0 I^(1//2) g^(-1//2)`
` t = ksqrt(I)/(g)`
Using other methoud, we calculate the value of dimensionless constant, ` k =2pi :, t = 2 pi sqrt(I//g)`


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