Potential energy of a particle varies with distance x from a fixed origin as U, = ,frac{{

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1.Units, Dimensions and Measurement

medium

The potential energy of a particle varies with distance $x$ from a fixed origin as $U\, = \,\frac{{A\sqrt x }}{{{x^2} + B}}$ Where $A$ and $B$ are dimensional constants then find the dimensional formula for $A/B$

${M^2}{L^1}{T^{ - 2}}$

${M^1}{L^{3/2}}{T^{ - 2}}$

${M^0}{L^{1/5}}{T^{ - 3}}$

${M^2}{L^{2/2}}{T^{ - 3}}$

Solution

$[\mathrm{B}]=\left[\mathrm{x}^{2}\right]=\mathrm{L}^{2}$

$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}]}{[\mathrm{x}]^{3 / 2}} \Rightarrow[\mathrm{A}]=[\mathrm{u}][\mathrm{x}]^{3 / 2}$

$=\mathrm{MI}^{2} \mathrm{T}^{-2} \times \mathrm{L}^{3 / 2}$

$[\mathrm{A}]=\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}$

$\left[\frac{\mathrm{A}}{\mathrm{B}}\right]=\frac{\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}}{\mathrm{L}^{2}}$

$[\mathrm{A} / \mathrm{B}]=\mathrm{ML}^{3 / 2} \mathrm{T}^{-2}$

Standard 11

Physics

$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-

easy

An expression of energy density is given by $u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$, where $\alpha, \beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be.

hard

(JEE MAIN-2022)

The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$

where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for

medium

The potential energy of a point particle is given by the expression $V(x)=-\alpha x+\beta \sin (x / \gamma)$. A dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is

hard

(KVPY-2012)

If force $(F)$, length $(L) $ and time $(T)$ are assumed to be fundamental units, then the dimensional formula of the mass will be

medium

The potential energy of a particle varies with distance $x$ from a fixed origin as $U\, = \,\frac{{A\sqrt x }}{{{x^2} + B}}$ Where $A$ and $B$ are dimensional constants then find the dimensional formula for $A/B$

Solution

Similar Questions

$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-

An expression of energy density is given by $u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$, where $\alpha, \beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be.

The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$ where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for

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The Bernoulli's equation is given by $p +\frac{1}{2} \rho v ^{2}+ h \rho g = k$

where $p =$ pressure, $\rho =$ density, $v =$ speed, $h =$ height of the liquid column, $g=$ acceleration due to gravity and $k$ is constant. The dimensional formula for $k$ is same as that for