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Q.

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M=mass, L=length, T=time and A=electric current, then:

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a

[ε0]=[M1L3T4A2]

b

[ε0]=[M1L2T1A2]

c

[ε0]=[M1L2T1A]

d

[ε0]=[M1L3T2A]

answer is A.

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Detailed Solution

Density  ρ=ML3
 n1u1=n2u2         
    128[M1L13]=n2[M2L23] 
    n2=128[M1M2][L2L1]3=128×[100050]×[25100]3=128×20×164=40 
                                      
 

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