The force $F$ is given in terms of time $t$ and displacement $x$ by the equation $F = A\,cos\,Bx + C\,sin\,Dt.$ The dimensional formulae of $D/B$ is
The force $F$ is given in terms of time $t$ and displacement $x$ by the equation $F = A\,cos\,Bx + C\,sin\,Dt.$ The dimensional formulae of $D/B$ is
- A
${M^0}{L^0}{T^0}$
- B
${M^0}{L^0}{T^{ - 1}}$
- C
${M^0}{L^{ - 1}}{T^0}$
- D
${M^0}{L^1}{T^{ - 1}}$
Similar Questions
The force is given in terms of time $t$ and displacement $x$ by the equation
${F}={A} \cos {Bx}+{C} \sin {Dt}$
The dimensional formula of $\frac{{AD}}{{B}}$ is -
${F}={A} \cos {Bx}+{C} \sin {Dt}$
The dimensional formula of $\frac{{AD}}{{B}}$ is -
- [JEE MAIN 2021]
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.
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.
If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be
If velocity $v$, acceleration $A$ and force $F$ are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of $v,\,A$ and $F$ would be
Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
- [JEE MAIN 2022]
If pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force is
If pressure $P$, velocity $V$ and time $T$ are taken as fundamental physical quantities, the dimensional formula of force is