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Force F is given in terms of time t and ...

Force `F` is given in terms of time `t` and distance `x` by `F = A sin Ct + B cos Dx`. Then the dimensions of `A//B` and `C//D` are

A

`["M"^(0)"L"^(0)"T"^(0)],["M"^(0)"LT"^(-1)]`

B

`["MLT"^(-2)],["M"^(0)"L"^(0)"T"^(-1)]`

C

`["MLT"^(-2)],["M"^(0)"L"^(-1)"T"^(0)]`

D

`["M"^(0)"LT"^(-1)],["M"^(0)"L"^(0)"T"^(0)]`

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `F=AsinCt+BcosDx`
where, `t`= time and `x` = distance
As, we know that trigonometric ratios are dimensionless.
This implies
`sin Ct` = dimensionless and `cosDx` = dimensionless
Also,`" "[C]=[(1)/(t)]=["T"^(-1)]and [D]=[(1)/(x)]=["L"^(-1)]` As, Eq. (i) represents the force. So, `A//B` is dimensionless i.e. `["M"^(0)"L"^(0)"T"^(0)]`
While, `" "[(C)/(D)]=[("T"^(-1))/("L"^(-1))]=["M"^(0)"LT"^(-1)]`
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Knowledge Check

  • Force F is given in terms of time t and distance x by F= A sinCt+ 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)LT^(-1)], [M^(0)L^(0)T^(0)]`

  • The force F is given in terms of time t and displacement x by the equation F=A cos Bx+Csin Dt . The dimensions of (D)/(B) are

    A
    `M^(@)L^(@)T^(@)`
    B
    `M^(@)L^(@)T^(-1)`
    C
    `M^(@)L^(-1)T^(@)`
    D
    `M^(@)L^(1)T^(-1)`

  • The velocity (V) of a particle (in cm/s) is given in terms of time (t) in sec by the equation V=at+(b)/(c+t) . The dimensions of a, b and c are

    A
    `{:(a,b,c),(L^(2),T,L^(1)T^(-2)):}`
    B
    `{:(a,b,c),(L^(1)T^(2),L^(1)T,L^(1)):}`
    C
    `{:(a,b,c),(L^(1)T^(-2),L^(1),T^(1)):}`
    D
    `{:(a,b,c),(L^(1),L^(1)T,T^(2)):}`

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