The time period of a simple pendulum is given by T=2πlg+kxT = 2\pi\sqrt{\frac{l}{g}} + kxT=2πgl+kx, where lll is the length, ggg is acceleration due to gravity, and xxx is displacement. The dimensions of kkk are:
[L⁻¹T]
[LT⁻¹]
[L⁻¹T⁻¹]
[L⁻²T²]
Related Questions
Which of the following is the smallest unit
Millimetre
Angstrom
Fermi
Metre
One femtometre is equivalent to
1015 m{10^{15}}{\rm{\;m}}1015m
10−15 m{10^{ - 15}}{\rm{\;m}}10−15m
10−12 m{10^{ - 12}}{\rm{\;m}}10−12m
1012 m{10^{12}}{\rm{\;m}}1012m
