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Standard XII

Physics

Dimensional Analysis

The amount of...

Question

The amount of charge passed in time $t$ through a cross section of a wire is
$Q (t) = A t^{2} + B t + C$
(a) Write the dimensional formulae for $A, B$ and $C$
(b) If the numerical vlaues of $A, B$ and $C$ are $5, 3$ and $1$ respectively in SI units, find the value of the current at $t = 5 s$

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Solution

$Q (t) = A t^{2} + B t + c$
a) $A t^{2} = Q$
$\Rightarrow A = \frac{Q}{t^{2}} = \frac{A^{'} T^{'}}{T^{2}} = A^{1} T^{- 1}$
b) $B t = Q$
$\Rightarrow B = \frac{Q}{T} - \frac{A^{'} T^{'}}{T} = A$
c) $C = [Q]$
$\Rightarrow C = A^{'} T^{'}$
d) Current $t = \frac{d Q}{d t} = \frac{d}{d t} (A t^{2} + B t + C)$

$i_{t = 5} = 2 A t + B = 2 \times 5 \times 5 + 3 = 53 A$ .

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The amount of charge passed in time t through a cross section of a wire isQ(t)=At2+Bt+C(a) Write the dimensional formulae for A,B and C(b) If the numerical vlaues of A,B and C are 5,3 and 1 respectively in SI units, find the value of the current at t=5s

Q(t)=At2+Bt+ca) At2=Q⇒A=Qt2=A′T′T2=A1T−1b) Bt=Q⇒B=QT−A′T′T=Ac) C=[Q]⇒C=A′T′d) Current t=dQdt=ddt(At2+Bt+C)it=5=2At+B=2×5×5+3=53A.

The amount of charge passed in time $t$ through a cross section of a wire is
$Q (t) = A t^{2} + B t + C$
(a) Write the dimensional formulae for $A, B$ and $C$
(b) If the numerical vlaues of $A, B$ and $C$ are $5, 3$ and $1$ respectively in SI units, find the value of the current at $t = 5 s$