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Units and Measurements question

2023 · 15 Apr · Shift 1 · Q56
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Units and Measurements question

2023 · 15 Apr · Shift 1 · Q56

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
The speed of a wave produced in water is given by v=λagbρcv=\lambda^{a} g^{b} \rho^{c}v=λagbρc. Where λ,g\lambda, gλ,g and ρ\rhoρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a,ba, ba,b and ccc respectively, are :
  1. A
    12,0,12\frac{1}{2}, 0, \frac{1}{2}21​,0,21​
  2. B
    1,1,01,1,01,1,0
  3. C
    1,−1,01,-1,01,−1,0
  4. D
    12,12,0\frac{1}{2}, \frac{1}{2}, 021​,21​,0
View written solutionFree

Correct answer: D

  1. Write the given relation

    The wave speed is given by v=λagbρcv = \lambda^a g^b \rho^cv=λagbρc

    where:

    • vvv = speed of wave
    • λ\lambdaλ = wavelength
    • ggg = acceleration due to gravity
    • ρ\rhoρ = density of water
  2. Write dimensions of each quantity

    • Speed: [v]=LT−1[v] = LT^{-1}[v]=LT−1
    • Wavelength: [λ]=L[\lambda] = L[λ]=L
    • Acceleration due to gravity: [g]=LT−2[g] = LT^{-2}[g]=LT−2
    • Density: [ρ]=ML−3[\rho] = ML^{-3}[ρ]=ML−3
  3. Substitute dimensions into the equation

    [v]=[λ]a[g]b[ρ]c[v] = [\lambda]^a [g]^b [\rho]^c[v]=[λ]a[g]b[ρ]c

    So, LT−1=(L)a(LT−2)b(ML−3)cLT^{-1} = (L)^a (LT^{-2})^b (ML^{-3})^cLT−1=(L)a(LT−2)b(ML−3)c

    Expanding, LT−1=McLa+b−3cT−2bLT^{-1} = M^c L^{a+b-3c} T^{-2b}LT−1=McLa+b−3cT−2b

  4. Compare powers of MMM, LLL, and TTT

    • For mass MMM: c=0c = 0c=0

    • For time TTT: −2b=−1⇒b=12-2b = -1 \Rightarrow b = \frac{1}{2}−2b=−1⇒b=21​

    • For length LLL: a+b−3c=1a + b - 3c = 1a+b−3c=1 Since c=0c=0c=0 and b=12b=\frac{1}{2}b=21​, a+12=1a + \frac{1}{2} = 1a+21​=1 a=12a = \frac{1}{2}a=21​

  5. Final values

    a=12,b=12,c=0a = \frac{1}{2}, \quad b = \frac{1}{2}, \quad c = 0a=21​,b=21​,c=0

  6. Check options

    This matches: Option D: (12,12,0)\left(\frac{1}{2}, \frac{1}{2}, 0\right)(21​,21​,0)

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