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Time period T of a simple pendulum may depend upon the mass of the bob m;length of the simple pendulum l and g,the acceleration due to gravity,i.e., T

directly proportional to mlg(raised to powers a.b,c),the values ofa,b,c are
a)-1/2, 0,1/2
b)1/2,0, -1/2
d)0,1/2 1/2
choose the correct answer with derivation also


T = k*m^a*l^b*g^c           (where k is dimensionless constant)
[M0L0T1] = [M]^a[L]^b[LT-2]^c
[M0L0T1] = [M^a*L^(b+c)*T^-2c]
compare the power of the both side
a = 0
b+c = 0
-2c = 1
c = -1/2
b = 1/2
 hence the value of a,b,c is 0,1/2,-1/2.

0 0 0
i also got the same answer as yours.but there is no option as same as ours.so what's the probabililty of the answer will be??
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We use the dimension analysis to derive the formula.

As\ given\ T \alpha \ m^x\ L^y\ g^z \\ \\ \ [ T^1 ] \alpha \ [ M^x ] [ L^y ] [ L^z T^{-2z} ] \\ \\ = [ M^{x} L^{y+z} T^{-2z} ] \\ \\ \ equating\ the\ powers\ of\ dimensions\ on\ both\ sides\ \\ \\ x = 0,\ \ y+z = 0, y = -z, \ \ -2z = 1\ \ \ \ z = -\frac{1}{2} \\ T \alpha \ \ m^0 L^\frac{1}{2} T^{-\frac{1}{2}} \\
1 5 1
there is no option as your answer sir!
what's the probability of the answer be?
don't mind. there are some printing mistakes in the text books etc. Being confident about our derivation, is good and needed. In fact, T = 2 Pi root (L/g). The answer is clear from there. good if u got the answer urself.
ok.then what's the option give not accurate,but a precise answer to our question sir?any guess
thanks and welcome
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