Let us consider a system of units in which mass and angular momentum are dimensionless. If length ha

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1.Units, Dimensions and Measurement

easy

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 $L ^{-1}$

$1,2,4$

$1,2,3$

$1,2$

$1,3$

(IIT-2019)

Solution

Mass $= M ^0 L ^0 T ^0$

$MVr = M ^0 L ^0 T ^0$

$M ^0 \frac{ L ^1}{ T ^1} \cdot L ^1= M ^0 L ^0 T ^0$

$L ^2= T ^1$ $. . . . . . .(1)$

Force $= M ^1 L ^1 T ^{-2} \quad \text { (in SI) }$

$= M ^0 L ^1 L ^{-4}$(In new system from equation $(1)) $

$= L ^{-3}$

Energy $= M ^1 L ^2 T ^{-2} \quad \text { (In SI) }$

$= M ^0 L ^2 L ^{-4}$ (In new system from equation $ (1)) $

$= L ^{-2}$

Power $=\frac{\text { Energy }}{\text { Time }}$

$= M ^1 L ^2 T ^{-5} \quad \text { (in SI) }$

$=M^0 L^2 L^{-6} $ (In new system from equation $(1)) $

$= L ^4$

Linear momentum $= M ^1 L ^1 T ^{-1} \text { (in SI) }$

$=M^0 L^1 L^{-2} $ (In new system from equation $(1)) $

$= L ^{-1}$

Ans. $(1,2,4)$

Standard 11

Physics

The density of a material in $SI\, units$ is $128\, kg\, m^{-3}$. In certain units in which the unit of length is $25\, cm$ and the unit of mass $50\, g$, the numerical value of density of the material is

medium

(JEE MAIN-2019)

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 ?

hard

Match List $I$ with List $II$ and select the correct answer using the codes given below the lists :

List $I$ List $II$

$P.$ Boltzmann constant $1.$ $\left[ ML ^2 T ^{-1}\right]$

$Q.$ Coefficient of viscosity $2.$ $\left[ ML ^{-1} T ^{-1}\right]$

$R.$ Planck constant $3.$ $\left[ MLT ^{-3} K ^{-1}\right]$

$S.$ Thermal conductivity $4.$ $\left[ ML ^2 T ^{-2} K ^{-1}\right]$

Codes: $ \quad \quad P \quad Q \quad R \quad S $

hard

(IIT-2013)

If pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force is

medium

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 constant of gravitation and $e$ is charge $] $

hard

(NEET-2017)

List $I$	List $II$
$P.$ Boltzmann constant	$1.$ $\left[ ML ^2 T ^{-1}\right]$
$Q.$ Coefficient of viscosity	$2.$ $\left[ ML ^{-1} T ^{-1}\right]$
$R.$ Planck constant	$3.$ $\left[ MLT ^{-3} K ^{-1}\right]$
$S.$ Thermal conductivity	$4.$ $\left[ ML ^2 T ^{-2} K ^{-1}\right]$

Solution

Similar Questions

The density of a material in $SI\, units$ is $128\, kg\, m^{-3}$. In certain units in which the unit of length is $25\, cm$ and the unit of mass $50\, g$, the numerical value of density of the material is

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 pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force 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 constant of gravitation and $e$ is charge $] $

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