If L has the dimensions of length- V that of potential and -0 is the permittivity of free space then quantity LV -0 has the dimensions of-

The correct option is B charge L → #160;M^0L^1T^0 V → ML^2I^ 1T^ 3 ϵ_0→ M^ 1 L^ 3 I^2T^4 LV ϵ_0→ M^0 L^0 I^1T^1 Charge → M^0L^0I^1T^1 ...

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Byju's Answer

Standard XII

Physics

Dimensional Analysis

If L has th...

Question

If $L$ has the dimensions of length, $V$ that of potential and $ϵ_{0}$ is the permittivity of free space then quantity $L V$ $ϵ_{0}$ has the dimensions of:

current

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charge

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resistance

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voltage

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Solution

The correct option is B charge
$L \to M^{0} L^{1} T^{0}$
$V \to M L^{2} I^{- 1} T^{- 3}$
$ϵ_{0} \to M^{- 1} L^{- 3} I^{2} T^{4}$
$L V ϵ_{0} \to M^{0} L^{0} I^{1} T^{1}$
Charge $\to M^{0} L^{0} I^{1} T^{1}$
Hence, $L V ϵ_{0}$ has dimension of charge.

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Similar questions

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If L has the dimensions of length, V that of potential and ϵ0 is the permittivity of free space then quantity LVϵ0 has the dimensions of:

The correct option is B chargeL→M0L1T0V→ML2I−1T−3ϵ0→M−1L−3I2T4LVϵ0→M0L0I1T1Charge →M0L0I1T1Hence, LVϵ0 has dimension of charge.

If $L$ has the dimensions of length, $V$ that of potential and $ϵ_{0}$ is the permittivity of free space then quantity $L V$ $ϵ_{0}$ has the dimensions of:

The correct option is B charge
$L \to M^{0} L^{1} T^{0}$
$V \to M L^{2} I^{- 1} T^{- 3}$
$ϵ_{0} \to M^{- 1} L^{- 3} I^{2} T^{4}$
$L V ϵ_{0} \to M^{0} L^{0} I^{1} T^{1}$
Charge $\to M^{0} L^{0} I^{1} T^{1}$
Hence, $L V ϵ_{0}$ has dimension of charge.