In the relation - dy-dt-2- sin- t - - the dimensional formula for - t - - is

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Question

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

[MLT]

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$[M L T^{0}]$

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$[M^{1} L^{0} T^{0}]$

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$[M^{0} L^{0} T^{0}]$

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Solution

The correct option is D
$[M^{0} L^{0} T^{0}]$

$H e r e (ω t + ϕ)$ is dimensionless because it is an argument of a trigonometric function.

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In the relation : dydt=2ω sin(ωt+ϕ), the dimensional formula for (ωt+ϕ) is

The correct option is D [M0L0T0] Here(ωt+ϕ) is dimensionless because it is an argument of a trigonometric function.

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 correct option is D
$[M^{0} L^{0} T^{0}]$

$H e r e (ω t + ϕ)$ is dimensionless because it is an argument of a trigonometric function.