Question

Consider the following equation

P = A sin (Bt + Ct2) + x

Here t is time and x is length. The dimensions of AB/C will be

¤ [M°LT1]
¤ [MºLT-2]
¤ [MLT-1]
¤ [MºL2T-2]​

Answer 1

Given : P = A sin (Bt + Ct2) + x  , t is time and x is length

To Find  : The dimensions of AB/C

Solution

P = A sin (Bt + Ct²) + x

sin (Bt + Ct²) is unit less

Hence unit of A sin (Bt + Ct²)  is same as of A

its added with x

Hence unit of A is same as of  x

x is length

Hence Dimension  of A = L

Bt   and  Ct²  are of same unit

Dividing both by t

Hence B  & Ct  are of same unit

Dividing both by C

=> B / C   has unit of  t  time

=>  Dimension  of of B / C = T

The dimensions of AB/C will be     LT

= M⁰L¹T¹

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Answer 2

The dimensions of AB/C is [M°LT]

Therefore, option (1) is correct.

Explanation:

Given:

Equation

$P=A\sin (Bt+Ct^2)+x$

Where t is time and x is length

To find out:

Dimensions of AB/C

Solution:

Since x is length therefore, P and $A\sin (Bt+Ct^2)$ will also have the dimensions of length

Also because the sine of something will be a dimensionless quantity

Therefore,

The dimensions of A will be that of the dimensions of length i.e. [L]

Also the quantity whose sine is been taken should be dimensionless

Therefore,

B will have dimensions of inverse of time i.e. [T⁻¹]

And C will have dimensions of inverse of square of time i.e. [T⁻²]

Thus,

Dimensions of AB/C

= [L][T⁻¹]/[T⁻²]

= [LT]

= [M°LT]

Hope this answer is helpful.

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