(A) The argument of an exponential function must be dimensionless. Therefore,the dimension of $\alpha t$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$,we

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(A) The argument of an exponential function must be dimensionless. Therefore,the dimension of $\alpha t$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$,we have $[\alpha] [T] = [1]$,which gives $[\alpha] = [T^{-1}]$.In the expression $x(t) = (v_0 / \alpha) (1 - e^{-\alpha t})$,the term $(1 - e^{-\alpha t})$ is dimensionless.Therefore,the dimension of $x(t)$ must be equal to the dimension of $(v_0 / \alpha)$.$[x] = [L]$,so $[v_0 / \alpha] = [L]$.$[v_0] = [L] \times [\alpha] = [L] \times [T^{-1}] = [L T^{-1}]$.Thus,the dimensions of $v_0$ and $\alpha$ are $[M^0 L^1 T^{-1}]$ and $[T^{-1}]$ respectively.

Class 11 Physics Units, Dimensions and Measurement Dimensional Analysis, Uses and Limitations

Medium

The position of a particle at time $t$ is given by the relation $x(t) = \left( \frac{v_0}{\alpha} \right) (1 - e^{-\alpha t})$,where $v_0$ is a constant and $\alpha > 0$. The dimensions of $v_0$ and $\alpha$ are respectively:

A
$M^0 L^1 T^{-1}$ and $T^{-1}$
B
$M^0 L^1 T^0$ and $T^{-1}$
C
$M^0 L^1 T^{-1}$ and $L T^{-2}$
D
$M^0 L^1 T^{-1}$ and $T$

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Units, Dimensions and Measurement

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Dimensional Analysis, Uses and Limitations

Velocities $(V)$ and accelerations $(a)$ in two systems of units $1$ and $2$ are related as $V_2 = \frac{n}{m^2} V_1$ and $a_2 = \frac{a_1}{mn}$ respectively. Here $m$ and $n$ are constants. Dimensionally, the relations between distances ($S_1$ and $S_2$) and times ($t_1$ and $t_2$) in the two systems are respectively:

EasyTS EAMCET 2023

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The number of particles crossing a unit area perpendicular to the $X$-axis in unit time is given by $n = -D \frac{n_2 - n_1}{x_2 - x_1}$,where $n_1$ and $n_2$ are the number of particles per unit volume at positions $x_1$ and $x_2$ respectively. Find the dimensions of $D$,which is known as the diffusion constant.

Medium

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The work done by a gas molecule in an isolated system is given by $W = \alpha \beta^{2} e^{-\frac{x^{2}}{\alpha kT}}$,where $x$ is the displacement,$k$ is the Boltzmann constant,$T$ is the temperature,and $\alpha$ and $\beta$ are constants. Then the dimension of $\beta$ will be:

DifficultJEE MAIN 2021

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If $\varepsilon_0$ is the permittivity of free space,$e$ is the charge of a proton,$G$ is the universal gravitational constant,and $m_p$ is the mass of a proton,then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is:

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$A$ physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^2}{4\pi \varepsilon_0}$ is $[c$ is velocity of light,$G$ is the universal gravitational constant and $e$ is charge$]$.

DifficultNEET 2017

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The position of a particle at time $t$ is given by the relation $x(t) = \left( \frac{v_0}{\alpha} \right) (1 - e^{-\alpha t})$,where $v_0$ is a constant and $\alpha > 0$. The dimensions of $v_0$ and $\alpha$ are respectively:

Similar Questions

The work done by a gas molecule in an isolated system is given by $W = \alpha \beta^{2} e^{-\frac{x^{2}}{\alpha kT}}$,where $x$ is the displacement,$k$ is the Boltzmann constant,$T$ is the temperature,and $\alpha$ and $\beta$ are constants. Then the dimension of $\beta$ will be:

If $\varepsilon_0$ is the permittivity of free space,$e$ is the charge of a proton,$G$ is the universal gravitational constant,and $m_p$ is the mass of a proton,then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is:

$A$ physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^2}{4\pi \varepsilon_0}$ is $[c$ is velocity of light,$G$ is the universal gravitational constant and $e$ is charge$]$.

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