(a) By principle of dimensional homogenity $\left[ {\frac{a}{{{V^2}}}} \right] = \left[ P \right]$ $\therefore $ $[a] = [P]\;[{V^2}] = [M{L^{ - 1}}

(a) By principle of dimensional homogenity $\left[ {\frac{a}{{{V^2}}}} \right] = \left[ P \right]$ $\therefore $ $[a] = [P]\;[{V^2}] = [M{L^{ - 1}}{T^{ - 2}}]\; \times [{L^6}]$ = $[M{L^5}{T^{ - 2}}]$

1.Units, Dimensions and MeasurementMedium

The equation of state of some gases can be expressed as $\left( {P + \frac{a}{{{V^2}}}} \right)\,(V - b) = RT$. Here $P$ is the pressure, $V$ is the volume, $T$ is the absolute temperature and $a,\,b,\,R$ are constants. The dimensions of $'a'$ are

A
$M{L^5}{T^{ - 2}}$
B
$M{L^{ - 1}}{T^{ - 2}}$
C
${M^0}{L^3}{T^0}$
D
${M^0}{L^6}{T^0}$

Std 11
Physics

To determine the Young's modulus of a wire, the formula is $Y = \frac{FL}{A\Delta L};$ where $L$ = length, $A = $area of cross-section of the wire, $\Delta L = $change in length of the wire when stretched with a force $F$. The conversion factor to change it from $CGS$ to $MKS$ system is .............. $10^{-1}\mathrm{N/m}^{2}$

Medium

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A new system of units is proposed in which unit of mass is $\alpha $ $kg$, unit of length $\beta $ $m$ and unit of time $\gamma $ $s$. How much will $5\,J$ measure in this new system ?

Difficult

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Time $(T)$, velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

Difficult

[JEE MAIN 2017]

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If mass is written as $\mathrm{m}=\mathrm{kc}^{\mathrm{p}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $\mathrm{k}$ a dimensionless constant)

Difficult

[JEE MAIN 2024]

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Match the following two coloumns

Column $-I$ Column $-II$

$(A)$ Electrical resistance $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$

$(B)$ Electrical potential $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$

$(C)$ Specific resistance $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$

$(D)$ Specific conductance $(s)$ None of these

Medium

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Column $-I$	Column $-II$
$(A)$ Electrical resistance	$(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$
$(B)$ Electrical potential	$(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$
$(C)$ Specific resistance	$(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$
$(D)$ Specific conductance	$(s)$ None of these

The equation of state of some gases can be expressed as $\left( {P + \frac{a}{{{V^2}}}} \right)\,(V - b) = RT$. Here $P$ is the pressure, $V$ is the volume, $T$ is the absolute temperature and $a,\,b,\,R$ are constants. The dimensions of $'a'$ are

Similar Questions

A new system of units is proposed in which unit of mass is $\alpha $ $kg$, unit of length $\beta $ $m$ and unit of time $\gamma $ $s$. How much will $5\,J$ measure in this new system ?

Time $(T)$, velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

If mass is written as $\mathrm{m}=\mathrm{kc}^{\mathrm{p}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $\mathrm{k}$ a dimensionless constant)

Match the following two coloumns Column $-I$ Column $-II$ $(A)$ Electrical resistance $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$ $(B)$ Electrical potential $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$ $(C)$ Specific resistance $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$ $(D)$ Specific conductance $(s)$ None of these

Match the following two coloumns

Column $-I$ Column $-II$

$(A)$ Electrical resistance $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$

$(B)$ Electrical potential $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$

$(C)$ Specific resistance $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$

$(D)$ Specific conductance $(s)$ None of these