$1$ $joule$ of energy is to be converted into new system of units in which length is  measured in $10\, m$, mass in $10\, kg$ and time in $1$ $minute$ then numerical value of  $1\, J$ in the new system is 

  • A
    $36 \times 10^{-4}$
  • B
    $36 \times 10^{-3}$
  • C
    $36 \times 10^{-2}$
  • D
    $36 \times 10^{-1}$

Similar Questions

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as  $E = mc^2$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in $MeV$, where $1\,MeV = 1.6\times 10^{-13}\,J$ ; the masses are measured i unified atomicm mass unit (u) where, $1\,u = 1.67 \times 10^{-27}\, kg$

$(a)$ Show that the energy equivalent of $1\,u$ is $ 931.5\, MeV$.

$(b)$ A student writes the relation as $1\,u = 931.5\, MeV$. The teacher points out that the relation  is dimensionally incorrect. Write the correct relation.

The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $V\, = \,\frac{{\pi p{r^4}}}{{8\eta l}}$ where $p$ is the pressure difference between the two ends of the pipe and $\eta $ is coefficent of viscosity of the liquid having dimensional formula $[M^1L^{-1}T^{-1}] $. Check whether the equation is dimensionally correct.

If speed $V,$ area $A$ and force $F$ are chosen as fundamental units, then the dimension of Young's modulus will be :

  • [JEE MAIN 2020]

An artificial satellite is revolving around a planet of mass $M$ and radius $R$ in a circular orbit of radius $r$. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis that $T\, = \,\frac{k}{R}\sqrt {\frac{{{r^3}}}{g}} $, where $k$ is dimensionless constant and $g$ is acceleration due to gravity.

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]