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Force F is given in terms of time t and distance ‘x’ by F = A sin Ct + B cos Dx. Then dimensions of A/B and C/D are

(a) [M0L0T0], [M0L0T–1]

(b) [MLT–2], [M0L–1T0]

(c) [M0L0T0], [M0LT-1]

(d) [M0L–1T-1], [M0L0T0]

Shane Macguire , 11 Years ago
Grade upto college level
anser 2 Answers
Simran Bhatia

Last Activity: 11 Years ago

(a)

    A/B=Force/Force [M^o L^o T^o ]

    ct = angle

    C = angle/time=1/T=T^(-1)

    Dx = angle

    D = angle/distance=1/L = L–1

    C/D=T^(-1)/L^(-1) =[M^o LT^(-1) ]

ankit singh

Last Activity: 4 Years ago

A/B= no dimensions, C/D= LT^(-1)

Explanation:

obviously...Ct & Dx indicate the angles

Ct = radians

Dx = radians

C = radians/sec

D = radians/metre

since sinCt & sinDx are simply numericals hence A & B indicate Force components

A = Newtons

B = Newtons

Now,

C/D = [radions/sec]/radions/metre]= metre/sec =LT^(-1)

A/B = Newtons/Newtons is a ratio and no dimensions

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