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 $=\frac{q}{V}=\frac{q^2}{Vq}=(\text{ coulomb }{)}^2$ joule ${}^{-1}$ $\therefore p-(iv)$
Magnetic induction $=\frac{F}{qv}=($ newton $)(\text{ ampere }{)}^{-1}(\text{metre }{)}^{-1}$ $\therefore q-(iii)$
Induced $\mathrm{emf}|\epsilon |=\frac{LdI}{dt}$ or Inductance, $L=\frac{|\epsilon |dt}{dI}$
$\therefore r-(ii)$
$=\frac{(\text{ volt })(\text{ second })}{\text{ ampere }}=$ volt sec(ampere) ${}^{-1}$
Resistance $=\frac{V}{I}=$ volt (ampere) ${}^{-1}$
$\therefore s-(i)$