A spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta $. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v)$ is called time constant $(\tau )$. Dimensionally $\tau $ can be represented by
A spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta $. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v)$ is called time constant $(\tau )$. Dimensionally $\tau $ can be represented by
- [AIIMS 1987]
- A
$\frac{{m{r^2}}}{{6\pi \eta }}$
- B
$\sqrt {\left( {\frac{{6\pi mr\eta }}{{{g^2}}}} \right)} $
- C
$\frac{m}{{6\pi \eta rv}}$
- D
None of the above
Similar Questions
Match List $-I$ with List $-II$
List $-I$
List $-II$
$A$.
Coefficient of Viscosity
$I$.
$[M L^2T^{–2}]$
$B$.
Surface Tension
$II$.
$[M L^2T^{–1}]$
$C$.
Angular momentum
$III$.
$[M L^{-1}T^{–1}]$
$D$.
Rotational Kinetic energy
$IV$.
$[M L^0T^{–2}]$
| List $-I$ | List $-II$ | ||
| $A$. | Coefficient of Viscosity | $I$. | $[M L^2T^{–2}]$ |
| $B$. | Surface Tension | $II$. | $[M L^2T^{–1}]$ |
| $C$. | Angular momentum | $III$. | $[M L^{-1}T^{–1}]$ |
| $D$. | Rotational Kinetic energy | $IV$. | $[M L^0T^{–2}]$ |
- [JEE MAIN 2024]
In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is
In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is
If the time period $t$ of the oscillation of a drop of liquid of density $d$, radius $r$, vibrating under surface tension $s$ is given by the formula $t = \sqrt {{r^{2b}}\,{s^c}\,{d^{a/2}}} $ . It is observed that the time period is directly proportional to $\sqrt {\frac{d}{s}} $ . The value of $b$ should therefore be
- [JEE MAIN 2013]
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position $]=\left[X^\alpha\right] ;[$ speed $]=\left[X^\beta\right]$; [acceleration $]=\left[X^{ p }\right]$; [linear momentum $]=\left[X^{ q }\right]$; [force $]=\left[X^{ I }\right]$. Then -
$(A)$ $\alpha+p=2 \beta$
$(B)$ $p+q-r=\beta$
$(C)$ $p-q+r=\alpha$
$(D)$ $p+q+r=\beta$
- [IIT 2020]
$\left(P+\frac{a}{V^2}\right)(V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^2}{a}$, will be
$\left(P+\frac{a}{V^2}\right)(V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^2}{a}$, will be
- [JEE MAIN 2023]