Question

The force F is expressed in terms of distance x and time t as F= a√x + bt2.

The dimensions of a/b is​

Answer 1

Answer:

Dimensions of a/b is L^-1/2×T^2

Explanation:

Answer 2

Answer:

The dimensions of a/b are

$[ \frac{a}{b} ]= L {}^{ - \frac{1}{2} } T {}^{2}$

Explanation:

Given that,

The force F is expressed in terms of distance x and time t in the way as shown below.

$F=a \sqrt{x} + bt {}^{2}$

According to the principle of homogeneity,

The dimensions of the term present in the right side of the relation are equal to the dimensions of each term present on left side of relation.Hence,The dimensions of the force are equal to the dimensions of both the terms.Therefore,we have

$[F]=[a \sqrt{x} ] \\ = > MLT {}^{ - 2} = aL {}^{ \frac{1}{2} } \\ = > [a ] = \frac{MLT {}^{ - 2} }{L {}^{ \frac{1}{2} } } = ML {}^{ \frac{ 1}{2} } T {}^{ - 2}$

Similarly for the second term of relation

$[F]=[bt {}^{2} ] \\ = > MLT {}^{ - 2} = b T {}^{2} \\ = > [b ] = MLT {}^{ - 4}$

Now our required dimensions are found as

$[ \frac{a}{b} ]= \frac{ML {}^{ \frac{ 1}{2} } T {}^{ - 2} }{MLT {}^{ - 4} } = L {}^{ - \frac{1}{2} } T {}^{2}$

Therefore,the required dimensions are found to be

$[ \frac{a}{b} ]= L {}^{ - \frac{1}{2} } T {}^{2}$

#SPJ3