Question

If the time period of a bar of mass M. length L and Young's modulus Y is given by T = 2pi
root(ML^3/3Yx)
then find the dimensional formula of x.​

Answer 1

Answer:

X=M^-1/2L^5/2T¹

Explanation:

time period =T¹,Young's modulus= m¹l-1t-2,mass=m,length=l

T¹=(ML³)^1/2/x*(ML^-1T^-2)

T¹x= M^1/2 L^3/2*(M-1L1T2)

x=M^-1/2L^5/2T²/T¹

x=M^-1/2L^5/2T¹

Hope it works:)

Answer 2

Answer:

The dimensional formula of x is [L⁴].

Explanation:

Given,

The mass of the bar = M

The length of the bar= L

The Young's modulus = Y

The time period is given by the equation:

• $T = 2\pi \sqrt{\frac{ML^{3} }{3Yx} }$

The dimensional formula of x =?

Solve the equation of the time period for x:

• $[\frac{T}{2\pi } ]^{2} = \frac{ML^{3} }{3Yx}$
• $\frac{T^{2} }{4\pi ^{2} } = \frac{ML^{3} }{3Yx}$
• $x = \frac{ML^{3}4\pi ^{2} }{3YT^{2} }$       ------- equation (1)

Now, let the dimensions for:

• Mass = M
• Length = L
• Time period= T

Now, we have to solve the dimensions for Y:

As we know,

• $Y = \frac{\sigma}{E}$

Here,

• σ = $\frac{force}{area}$
• E = $\frac{change in length}{length}$

Therefore,

• $Y = \frac{F/A}{dl/l}$
• $Y = \frac{F\times l}{A\times dl}$         ------- equation (2)

As we know,

• Force = mass × acceleration

And,

• Acceleration = $\frac{Velocity}{time}$ = $ms^{-2}$ unit

In terms of dimensions,

• F = M × L × T⁻² = MLT⁻²

Now, after putting these dimensions of force in equation (2), we get:

• $Y = \frac{MLT^{-2} L}{L^{2} L}$= MT⁻²L⁻¹

Now, after putting the dimensions of Y in equation (1), we get:

• $x = \frac{ML^{3}4\pi ^{2} }{3MT^{-2}L^{-1} T^{2} }$

Let the terms $\frac{4\pi }{3}$ = 1

• $x = \frac{ML^{3} }{MT^{-2}L^{-1} T^{2} }$
• x = [L⁴]

Hence, the dimensional formula of x =  [L⁴].