Question

Q.31 For a body moving along x-axis, the distance travelled by
body from a reference point is given as function of timet
as x = até + b, where a and b are constants, then the
dimension of Vab is same as -
(A) Speed
(B) Distance travelled
(C)Acceleration
(D) None of these​

Answer 1

Correct Question:-

For a body moving along x-axis, the distance travelled by the body from a reference point is given as function of time $\displaystyle\sf {t}$ as $\displaystyle\sf {x=at^2+b,}$ where $\displaystyle\sf {a}$ and $\displaystyle\sf {b}$ are constants. Then the dimension of $\displaystyle\sf {\sqrt{ab}}$ is same as,

(A) Speed

(B) Distance travelled

(C) Acceleration

(D) None of these

Solution:-

Given,

$\displaystyle\longrightarrow\sf{x=at^2+b}$

According to Principle of Homogeneity,

$\displaystyle\longrightarrow\sf{[x]=[a][t]^2=[b]}$

Then,

$\displaystyle\longrightarrow\sf{[x]=[a][t]^2}$

$\displaystyle\longrightarrow\sf{[a]=[x][t]^{-2}}$

$\displaystyle\longrightarrow\sf{[a]=L\,T^{-2}}$

And,

$\displaystyle\longrightarrow\sf{[b]=[x]}$

$\displaystyle\longrightarrow\sf{[b]=L}$

Then,

$\displaystyle\longrightarrow\sf{\left[\sqrt{ab}\right]=[ab]^{\frac {1}{2}}}$

$\displaystyle\longrightarrow\sf{\left[\sqrt{ab}\right]=\left[L\,T^{-2}\,L\right]^{\frac {1}{2}}}$

$\displaystyle\longrightarrow\sf{\left[\sqrt{ab}\right]=\left[L^2\,T^{-2}\right]^{\frac {1}{2}}}$

$\displaystyle\longrightarrow\sf{\left[\sqrt{ab}\right]=L\,T^{-1}}$

This implies dimension of $\displaystyle\sf {\sqrt{ab}}$ is the same as that of speed.

Hence (A) is the answer.

Answer 2

Answer:

OPTION a is correct

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