The SI unit of surface tension is

(1) Dyne/cm

(2) Newton/cm

(3) Newton/metre

(4) Newton-metre

E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the dimension of $\frac{E{l}^2}{m^5{G}^2}$ are

(1) Angle

(2) Length

(3) Mass

(4) Time

From the equation $\tan \theta =\frac{rg}{v^2}$, one can obtain the angle of banking θ for a cyclist taking a curve (the symbols have their usual meanings). Then say, it is

(1) Both dimensionally and numerically correct

(2) Neither numerically nor dimensionally correct

(3) Dimensionally correct only

(4) Numerically correct only

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity η flowing per second through a tube of radius r and length l and having a pressure difference p across its end, is

(1) $V=\frac{\pi p{r}^4}{8\eta l}$

(2) $V=\frac{\pi \eta l}{8p{r}^4}$

(3) $V=\frac{8p\eta l}{\pi r^4}$

(4) $V=\frac{\pi p\eta }{8l{r}^4}$

The velocity v (in cm/sec) of a particle is given in terms of time t (in sec) by the relation $v=at+\frac{b}{t+c}$; the dimensions of a, b and c are

(1) $a=L^2,b=T,c=L{T}^2$

(2) $a=L{T}^2,b=LT,c=L$

(3) $a=L{T}^{-2},b=L,c=T$

(4) $a=L,b=LT,c=T^2$

From the dimensional consideration, which of the following equation is correct ?

(1) $T=2\pi \sqrt{\frac{R^3}{GM}}$

(2) $T=2\pi \sqrt{\frac{GM}{R^3}}$

(3) $T=2\pi \sqrt{\frac{GM}{R^2}}$

(4) $T=2\pi \sqrt{\frac{R^2}{GM}}$

The dimensions of $\frac{a}{b}$ in the equation $P=\frac{a-t^2}{bx}$, where P is pressure, x is distance and t is time, are 

(1) MT^–2

(2) M²LT^–3

(3) ML³T^–1

(4) LT^–3

Dimensions of $\frac{1}{{\mu }_0{\epsilon }_0}$, where symbols have their usual meaning, are 

(1) [LT^–1]

(2) [L^–1T]

(3) [L^–2T²]

(4) [L²T^–2]

The dimensions of $e^2/4\pi {\epsilon }_{0}hc$, where $e,{\epsilon }_0,h$ and c are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively 

(1) $\left[M^0{L}^0{T}^0\right]$

(2) $\left[M^1{L}^0{T}^0\right]$

(3) $\left[M^0{L}^1{T}^0\right]$

(4) $\left[M^0{L}^0{T}^1\right]$

If radius of the sphere is (5.3 ± 0.1)cm. Then percentage error in its volume will be 

(1) $3+6.01\times \frac{100}{5.3}$

(2) $\frac{1}{3}\times 0.01\times \frac{100}{5.3}$

(3) $\left(\frac{{\displaystyle 3\times 0.1}}{{\displaystyle 5.3}}\right)\times 100$

(4) $\frac{0.1}{5.3}\times 100$