Physics, asked by aditya2020222003, 1 year ago

what is the forbidden language.??



btw please ans the question in the pic​

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ravi9848267328: same problem

Answers

Answered by Anonymous
23

⭐《ANSWER》

\huge\mathfrak\red {aNsWeR}

↗Actually welcome to the concept of the WAVE MOTION

↪Basically here talking about the Melde's experiment on how there is derivation of the Wave velocity along a Stretched String

↪Let here , the denotions be follwed as

↪T = Tension , Vw = wave velocity, myu ( u) = mass per unit length ,

↪so applying the above denotions we get as ,

↪Vw is directly proportional to the Tension and inversely proportional to the mass per unit length , so we get as

〽Vw = underoot T / u

↪or we can say as

〽Vw = ( T/u ) ^1/2

↪Also stating the "u" = M/L

↪so we get as

〽Vw = ( F*L / M ) ^1/2

Answered by Anonymous
19

 \textsf{\underline {\Large{Deducing Relations}}} :

Given, Velocity  \mathsf{(\:m{s} ^{-1}\:)} depends upon length L, tension F and mass M.

Let  \mathsf{v\:{\propto{\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}}}

Now, \boxed{\mathsf{v\:=\:k\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}} ↪️( i )

Here, K is a constant which is dimensionless.

Dimension of v =  \mathsf{[L{T} ^{-1}]}

Dimension of l =  \mathsf{[L]}

Dimension of F = \mathsf{[ML{T} ^{-2}]}

Dimension of M per unit length = \mathsf{[M{L} ^{-1}]}

Substituting these dimensions in equation ( i ),

 \mathsf{v\:=\:{l} ^{a} \:{F} ^{b} \:{M{L} ^{-1}} ^{c}}

 \mathsf{[{M} ^{0}L{T} ^{-1}]\:=\:{[L]}^{a} \:{[ML{T} ^{-2}]} ^{b} \:{[M{L} ^{-1}]} ^{c}}

 \mathsf{[{M}^{0}L{T} ^{-1}]\:=\: {[L]}^{a} \:[{M} ^{b} {L} ^{b} {T} ^{-2b}]\:{[M]} ^{c}{[{L} ^{-1}]} ^{c}}

 \mathsf{[{M}^{0}L{T} ^{-1}]\:=\: [{M} ^{b\:+\:c}] \:[{L} ^{a\:+\:b\:-\:c}] \:[{T} ^{-2b}]}

Now, Comparing powers of M, L and T in both side,

 \mathsf{[{M}^{0}] \:=\:[{M} ^{b\:+\:c}]}

↪️ b + c = 0

↪️ b = - c --> ( 1 )

 \mathsf{[{L}^{1}] \:=\:[{L}^{a\:+\:b\:-\:c}]}

↪️ 1 = a + b - c

↪️ 1 = a + 2b --> ( 2 )

 \mathsf{[{T}^{-1}] \:=\:[{T}^{-2b}]}

↪️ - 1 = - 2b

➡️ b =  \mathsf{\dfrac{1}{2}}

Putting value of ' b ' in equation ( 2 ),

↪️ 1 = 2b + a

↪️ 1 = a + 2  \mathsf{\dfrac{1}{2}}

↪️ a = 1 - 1

➡️ a = 0

Putting value of ' a ' in equation ( 1 ),

↪️ b = - c

↪️ c = - b

➡️ c = -  \mathsf{\dfrac{1}{2}}

Equating these valued of 'a', ' b', ' c ' , in equation ( i ),

 \mathsf{v\:=\:k\:{l} ^{a} \:{F} ^{b} \:{M} ^{c}}

 \mathsf{v\:=\:k\:{l} ^{\tiny{0}} \:{F} ^{\tiny{\dfrac{1}{2}}} \:{M} ^{\tiny{\dfrac{-1}{2}}}}

➡️  \boxed{\mathsf{v\:=\:k\:{\sqrt{\dfrac{F}{M}}}}}

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