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1.Units, Dimensions and Measurement
medium
If $L$ and $R$ are respectively the inductance and resistance, then the dimensions of $\frac{L}{R}$ will be
A${M^0}{L^0}{T^{ - 1}}$
B${M^0}L{T^0}$
C${M^0}{L^0}T$
DCannot be represented in terms of $M,\,L$ and T
Solution
$E =\frac{1}{2} L i ^{2} \Rightarrow L =\frac{2 E }{ i ^{2}}$
dimension of $L=\frac{M L T^{-2}}{Q^{2}} \cdot T^{2}=M L T^{0} Q^{-2}$
$E = i ^{2} R T \Rightarrow R =\frac{ E }{ i ^{2} t }$
dimension of $R =\frac{ M L T ^{-2}}{ Q ^{2}} \cdot T ^{-1}= M L T ^{-1} Q ^{-2}$
$\left(\frac{L}{R}\right)=\frac{M L T^{0} Q^{-2}}{M L T^{-1} Q^{-2}}=[T]$
dimension of $L=\frac{M L T^{-2}}{Q^{2}} \cdot T^{2}=M L T^{0} Q^{-2}$
$E = i ^{2} R T \Rightarrow R =\frac{ E }{ i ^{2} t }$
dimension of $R =\frac{ M L T ^{-2}}{ Q ^{2}} \cdot T ^{-1}= M L T ^{-1} Q ^{-2}$
$\left(\frac{L}{R}\right)=\frac{M L T^{0} Q^{-2}}{M L T^{-1} Q^{-2}}=[T]$
Standard 11
Physics
Similar Questions
Match List $I$ with List $II$
| LIST$-I$ | LIST$-II$ |
| $(A)$ Torque | $(I)$ $ML ^{-2} T ^{-2}$ |
| $(B)$ Stress | $(II)$ $ML ^2 T ^{-2}$ |
| $(C)$ Pressure of gradient | $(III)$ $ML ^{-1} T ^{-1}$ |
| $(D)$ Coefficient of viscosity | $(IV)$ $ML ^{-1} T ^{-2}$ |
