Question

Check whether the equation
T=2π√m/g is dimensionally correct
T= time period of a simple pendulam
m=mass of the boh

Answer 1

Answer:

[M1 L-1 T2]

Explanation:

T=2 π √m/g

There fore T

2=4 π

2

(

m

g

)

Dimension of LHS , [T2

] = MLT

Dimension of [RHS], m

g

=

M

1

L

1

T

−2

= [M

1

L

-1

T

2

]

Dimension of LHS and RHS are not equal. Thus according to principle of homogeneity the

equation is wrong.

Answer 2

Answer:

The given equation T=2π√(m/g)  is dimensionally incorrect.

Explanation:

We have given the equation fir time period:

$T=2\pi \sqrt{\frac{\big m}{\big g} }$

where, T is time period of simple pendulum

m is the mass of the bob

and, g is acceleration (9.8 ms⁻²) due to gravity.

The dimensional formula for time period = [M⁰ L⁰ T¹]

The dimensional formula for mass = [M¹ L⁰ T⁰]

The dimensional formula for 'g' = [M⁰ L¹ T⁻²]

Now, $\sqrt{\frac{m}{g} }=\frac{[M^{1}L^{o}T^{o}]^{\frac{1}{2} }}{[M^{o}L^{1}T^{-2}]^{\frac{1}{2} }}$

$\sqrt{\frac{m}{g} }=\frac{[M^{\frac{1}{2}}L^{o}T^{o}]}{[M^{o}L^{\frac{1}{2}}T^{-1}]}$

$\sqrt{\frac{m}{g} }=[M^{\frac{1}{2} }L^\frac{-1}{2} T^{1}]$

Now, the dimensional formula of time period (T) is not same as dimensional formula of $\sqrt{\frac{m}{g} }$.

Therefore, the given equation is dimensionally incorrect.