Force $F$ is given in terms of time $t$ and distance $x$ by $F = a\sin ct + b\cos \,dx$, then the dimension of $a/b$ is:-

$\left[ {{M^0}{L^1}{T^{ - 1}}} \right]$

$\left[ {{M^1}{L^1}{T^{ - 1}}} \right]$

$\left[ {{M^1}{L^0}{T^{ - 1}}} \right]$

Back to UNITS & MEASUREMENTS

JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Analyzing +4 | -1

Force

F

is given in terms of time

t

and distance

x

F = a\sin ct + b\cos \,dx

, then the dimension of

a/b

is:-

View answer and explanation

Correct option: C

\left[ {{M^0}{L^0}{T^0}} \right]

Option C ([M⁰L⁰T⁰]) is correct because the arguments of trigonometric functions are dimensionless, so the dimensions of a and b must be the same as F, making a/b dimensionless.

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →

Assertion :

\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

has the unit

\mathrm{ms}^{-1} \text {. }

Reason: Acceleration has the dimensions of

\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

The method of dimensional analysis can be used to derive which of the following relations? Statement-I: Dimensions of constant of proportionality of constants can be derived from dimensional method
Statement-II: Numerical value of constant of proportionality can be found from experiments only The dimensions of torque are The physical quantity having the dimensions

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right]

is A quantity X is given by

{\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;

, where

{\varepsilon _0}

is the permittivity of free space, L is a length,

{\Delta V}

is a potential difference and

{\Delta t}

is a time interval. The dimensional formula for X is the same as that of