If Surface tension $(S)$, Moment of Inertia $(I)$ and Planck’s constant $(h)$, were to be taken as the fundamental units, the dimensional formula for linear momentum would be
If Surface tension $(S)$, Moment of Inertia $(I)$ and Planck’s constant $(h)$, were to be taken as the fundamental units, the dimensional formula for linear momentum would be
- [JEE MAIN 2019]
- A
$S^{1 /2} I^{1 /2} h^0$
- B
$S^{1 /2} I^{3 /2} h^{-1}$
- C
$S^{3 /2} I^{1 /2} h^0$
- D
$S^{1 /2} I^{1 /2} h^{-1}$
Similar Questions
A function $f(\theta )$ is defined as $f(\theta )\, = \,1\, - \theta + \frac{{{\theta ^2}}}{{2!}} - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + ...$ Why is it necessary for $f(\theta )$ to be a dimensionless quantity ?
A function $f(\theta )$ is defined as $f(\theta )\, = \,1\, - \theta + \frac{{{\theta ^2}}}{{2!}} - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + ...$ Why is it necessary for $f(\theta )$ to be a dimensionless quantity ?
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express time in terms of dimensions of these quantities.
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express time in terms of dimensions of these quantities.
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of $L$, which of the following statement ($s$) is/are correct ?
$(1)$ The dimension of force is $L ^{-3}$
$(2)$ The dimension of energy is $L ^{-2}$
$(3)$ The dimension of power is $L ^{-5}$
$(4)$ The dimension of linear momentum is $L ^{-1}$
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of $L$, which of the following statement ($s$) is/are correct ?
$(1)$ The dimension of force is $L ^{-3}$
$(2)$ The dimension of energy is $L ^{-2}$
$(3)$ The dimension of power is $L ^{-5}$
$(4)$ The dimension of linear momentum is $L ^{-1}$
- [IIT 2019]
Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
- [JEE MAIN 2022]
The dimensions of the area $A$ of a black hole can be written in terms of the universal gravitational constant $G$, its mass $M$ and the speed of light $c$ as $A=G^\alpha M^\beta c^\gamma$. Here,
The dimensions of the area $A$ of a black hole can be written in terms of the universal gravitational constant $G$, its mass $M$ and the speed of light $c$ as $A=G^\alpha M^\beta c^\gamma$. Here,
- [KVPY 2015]