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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) [M 0 L 0 T 0 ], [M 0 L 0 T –1 ] (b) [MLT –2 ], [M 0 L –1 T 0 ] (c) [M 0 L 0 T 0 ], [M 0 LT -1 ] (d) [M 0 L –1 T -1 ], [M 0 L 0 T 0 ]



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]


Grade:upto college level

2 Answers

Simran Bhatia
348 Points
7 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
askIITians Faculty 614 Points
one year 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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