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1) Force F is given in terms of time t and distance ‘x’ by F = A sin Ct + B cosdx. 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 ]


1)       Force F is given in terms of time t and distance ‘x’ by F = A sin Ct + B cosdx. 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:10

1 Answers

Vikas TU
14149 Points
3 years ago

F=A cos Bx+C sin Dt

It is a common saying in mathematics that the elements inside cos and sin function is always constant.

So here in the above equation

B x. & Dt must be dimensionless

Since x=[L] &t=[T]

So B=[L]^-1 & D=[T]^-1

So,D/B=[LT^-1]

Thanks

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