Question

Find dimensions of a and b if f= at+bt^2 where f is force and t is time

Answer 1

Answer:

Given:

An equation is provided as follows

f = at + bt²

To Find:

Dimensions of "a" and "b"

Concept:

Note the following:

1. Quantities with same dimensions can be added.

2. Each of the individual quantities can be equated to the dimensions on the LHS.

Calculation:

Dimension of "a":

Since "a" can be equated to the LHS, it will have the same dimensions as that of

"f" i.e force

=> [a] = [ M L (T^-2)].

Dimensions of "b":

Note in this case , (bt²) will have same dimensions as that of "a"

(refer to the 1st rule in Concept)

[bt²] = [a] = [f]

=> [b T²] = [ M L (T^-2)]

=> [b] = [ M L (T^-4)].

Answer 2

Question :-

Find dimensions of a and b if

f = at + bt² ,where f is the force and t is time .

Answer :-

$\blue{\boxed{ \orange{ \implies}}} \: \sf{\: [b] = [M \: L\: {T}^{ - 4} ] }\\ \\ \green{\boxed{ \red{ \implies}} }\: \sf{ \: [a] \: =[ M \: L \: {T}^{ - 3}] }$

Used dimensions :-

$\star \: \: \sf{force \: = [M \: L \:{T}^{ - 2}]} \\ \\ \star \: \sf{ time \: = [T]}$

Explanation :-

Used conditions :-

• If dimensions of quantity is same then it can be added up .

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Explanation :-

We have ,

→ f = at + bt²

according to the given condition ,

Dimension of F =dimension of at

$\implies \sf{[M \: L\: {T}^{ - 2}] = [a] \:[ T]} \\ \\ \implies \sf{ \: [a] \: =[ M \: L \: {T}^{ - 2 - 1} ]} \\ \\ \implies \boxed{\sf{[a] \: = [M \: L\: {T}^{ - 3}] }}$

And also ,

dimension of f = dimension of bt²

$\implies \sf{ [M \: L \: {T}^{ - 2} ] =[ b ]\: [{T}^{2}] } \\ \\ \implies \sf{ [b] \: =[ M \: L\: {T}^{ - 2 - 2}] } \\ \\ \implies \boxed{\sf{[b] \: = [M \: L \: {T}^{ - 4} ]}}$

These are the required dimensions.

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