Question

Match the following.

p.	Capacitance	i.	volt (ampere) $^{-1}$
q.	Magnetic induction	ii.	volt sec (ampere) $^{-1}$
r.	Inductance	iii.	newton (ampere) ^{-1}(metre) $^{-1}$
s.	Resistance	iv.	coulomb $^{2}$ (joule) $^{-1}$

• A. $(p) - (ii) , (q) - (iii) , (r) - (iv) , (s) - (i)$
• B. $(p) - (iv) , (q) - (iii) , (r) - (i)i , (s) - (i)$
• C. $(p) - (iii) , (q) - (iv) , (r) - (i) , (s) - (ii)$
• D. $(p) - (iv) , (q) - (i) , (r) - (ii) , (s) - (iii)$

Answer 1

Answer: (B)

Capacitance $=\frac{q}{V}=\frac{q^{2}}{V q}=(\text { coulomb })^{2}$ joule $^{-1}$ $\therefore p -( iv )$
Magnetic induction $=\frac{F}{q v}=($ newton $)(\text { ampere })^{-1}(\text {metre })^{-1}$ $\therefore q -( iii )$
Induced $\operatorname{emf}|\varepsilon|=\frac{L d I}{d t}$ or Inductance, $L=\frac{|\varepsilon| d t}{d I}$
$\therefore \quad r-(i i)$
$=\frac{(\text { volt })(\text { second })}{\text { ampere }}=$ volt sec(ampere) $^{-1}$
Resistance $=\frac{V}{I}=$ volt (ampere) $^{-1}$
$\therefore s-(i)$