In the relation dydx - 2-sin - t - - 0 the dimensional formula for - t - - 0 is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Dimensional Analysis

In the relati...

Question

In the relation $\frac{d y}{d x} = 2 ω sin (ω t + ϕ_{0})$ the dimensional formula for $ω t + ϕ_{0}$ is

MLT

No worries! We‘ve got your back. Try BYJU‘S free classes today!

MLT

^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{0}

^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{0}

^{0}

^{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D M $^{0}$ L $^{0}$ T $^{0}$
Trigonometric functions are dimensionless quantities.
$∴$ Dimension of $(ω t + ϕ_{0}) = M^{0} L^{0} T^{0}$

Suggest Corrections

Similar questions

Q. In the relation

\frac{d y}{d t} = 2 ω sin (ω t + ϕ_{0})

, the dimensional formula for

ω t + ϕ_{0}

In the relation : $\frac{d y}{d t} = 2 ω s i n (ω t + ϕ)$ , the dimensional formula for $(ω t + ϕ)$ is

The force F is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt. The dimensional formula of $\frac{D}{B}$ is:

The force F is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt. The dimensional formula of $\frac{D}{B}$ is

Given that the displacement of an oscillating particle is given by $y = A sin[Bx + Ct + D]$ . The dimension formula for ABCD is (where x is displacement and t is time)

Join BYJU'S Learning Program

In the relation dydx=2ωsin(ωt+ϕ0) the dimensional formula for ωt+ϕ0 is

The correct option is D M0L0T0Trigonometric functions are dimensionless quantities.∴ Dimension of (ωt+ϕ0)=M0L0T0

In the relation $\frac{d y}{d x} = 2 ω sin (ω t + ϕ_{0})$ the dimensional formula for $ω t + ϕ_{0}$ is

The correct option is D M $^{0}$ L $^{0}$ T $^{0}$
Trigonometric functions are dimensionless quantities.
$∴$ Dimension of $(ω t + ϕ_{0}) = M^{0} L^{0} T^{0}$