Qstnid -Iv-1-the Amount Of Charge That Passes In Time T Through A Cross-section Of A Wire Is Q(t) = At<sup>2</sup> + Bt + C. (A) Write The Dimensional Formulae For A, B And C. (B) If The Numerical Values Of A, B And C Are 5, 3 And 1, Respectively, In S.i Units, Find The Value Of The Current At T = 5 S. (A) Write The Dimensional Formulae For A, B And C. (B) If The Numerical Values Of A, B And C Are 5, 3 And 1, Respectively, In S.i Units, Find The Value Of The Current At T = 5 S. - 32: Electric Current In Conductors

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NEET-XII-Physics

32: Electric Current in Conductors

with Solutions - page 4

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#1
The amount of charge that passes in time t through a cross-section of a wire is
Q(t) = At² + Bt + C. (a) Write the dimensional formulae for A, B and C. (b) If the numerical values of A, B and C are 5, 3 and 1, respectively, in S.I units, find the value of the current at t = 5 s. (a) Write the dimensional formulae for A, B and C. (b) If the numerical values of A, B and C are 5, 3 and 1, respectively, in S.I units, find the value of the current at t = 5 s.
Ans : (a) Amount of charge,
Q(t) = At² + Bt + C
We can only add the terms with the same dimensions. So, all the individual terms will have dimensions equal to the dimensions of the charge.
Comparing the dimensions of each term separately, we get:
`` A{t}^{2}=Q``
`` \Rightarrow A=\frac{Q}{{t}^{2}}``
`` Q=It``
`` \Rightarrow A=\frac{I}{t}=\left[{\,\mathrm{\,AT\,}}^{-1}\right],``
where I = current through the wire
Now,
`` Bt=Q``
`` \Rightarrow B=\frac{Q}{t}=I=\left[\,\mathrm{\,A\,}\right]``
Also,
`` C=Q``
`` \Rightarrow C=\left[\,\mathrm{\,AT\,}\right]`` (b) Current I = rate of flow of charge
`` \Rightarrow I=\frac{\,\mathrm{\,d\,}Q}{\,\mathrm{\,d\,}t}=\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,d\,}t}\left(A{t}^{2}+Bt+C\right)``
`` \Rightarrow I=2At+B``
`` A=5,B=3\,\mathrm{\,and\,}t=5\,\mathrm{\,s\,}``
`` \Rightarrow I=2\times 5\times 5+3=53\,\mathrm{\,A\,}``
Page No 198: (a) Amount of charge,
Q(t) = At² + Bt + C
We can only add the terms with the same dimensions. So, all the individual terms will have dimensions equal to the dimensions of the charge.
Comparing the dimensions of each term separately, we get:
`` A{t}^{2}=Q``
`` \Rightarrow A=\frac{Q}{{t}^{2}}``
`` Q=It``
`` \Rightarrow A=\frac{I}{t}=\left[{\,\mathrm{\,AT\,}}^{-1}\right],``
where I = current through the wire
Now,
`` Bt=Q``
`` \Rightarrow B=\frac{Q}{t}=I=\left[\,\mathrm{\,A\,}\right]``
Also,
`` C=Q``
`` \Rightarrow C=\left[\,\mathrm{\,AT\,}\right]`` (b) Current I = rate of flow of charge
`` \Rightarrow I=\frac{\,\mathrm{\,d\,}Q}{\,\mathrm{\,d\,}t}=\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,d\,}t}\left(A{t}^{2}+Bt+C\right)``
`` \Rightarrow I=2At+B``
`` A=5,B=3\,\mathrm{\,and\,}t=5\,\mathrm{\,s\,}``
`` \Rightarrow I=2\times 5\times 5+3=53\,\mathrm{\,A\,}``
Page No 198:

#1-a
Write the dimensional formulae for A, B and C.
Ans : Amount of charge,
Q(t) = At² + Bt + C
We can only add the terms with the same dimensions. So, all the individual terms will have dimensions equal to the dimensions of the charge.
Comparing the dimensions of each term separately, we get:
`` A{t}^{2}=Q``
`` \Rightarrow A=\frac{Q}{{t}^{2}}``
`` Q=It``
`` \Rightarrow A=\frac{I}{t}=\left[{\,\mathrm{\,AT\,}}^{-1}\right],``
where I = current through the wire
Now,
`` Bt=Q``
`` \Rightarrow B=\frac{Q}{t}=I=\left[\,\mathrm{\,A\,}\right]``
Also,
`` C=Q``
`` \Rightarrow C=\left[\,\mathrm{\,AT\,}\right]``

#1-b
If the numerical values of A, B and C are 5, 3 and 1, respectively, in S.I units, find the value of the current at t = 5 s.
Ans : Current I = rate of flow of charge
`` \Rightarrow I=\frac{\,\mathrm{\,d\,}Q}{\,\mathrm{\,d\,}t}=\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,d\,}t}\left(A{t}^{2}+Bt+C\right)``
`` \Rightarrow I=2At+B``
`` A=5,B=3\,\mathrm{\,and\,}t=5\,\mathrm{\,s\,}``
`` \Rightarrow I=2\times 5\times 5+3=53\,\mathrm{\,A\,}``
Page No 198: