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Current Electricity question

2024 · 29 Jan · Shift 1 · Q78
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  5. /2024 · 29 Jan · Shift 1 · Q78

Current Electricity question

2024 · 29 Jan · Shift 1 · Q78

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The electric current through a wire varies with time as I=I0+βtI=I_0+\beta tI=I0​+βt, where I0=20 AI_0=20 \mathrm{~A}I0​=20 A and β=3 A/s\beta=3 \mathrm{~A} / \mathrm{s}β=3 A/s. The amount of electric charge crossed through a section of the wire in 20 s20 \mathrm{~s}20 s is :
  1. A
    80 C
  2. B
    800 C
  3. C
    1000 C
  4. D
    1600 C
View written solutionFree

Correct answer: C

  1. The current varies with time as I(t)=I0+βtI(t)=I_0+\beta tI(t)=I0​+βt with I0=20 A,β=3 A/s.I_0=20\,\text{A}, \qquad \beta=3\,\text{A/s}.I0​=20A,β=3A/s.

  2. The total charge crossing a section in time t=20 st=20\,\text{s}t=20s is Q=∫020I(t) dt.Q=\int_0^{20} I(t)\,dt.Q=∫020​I(t)dt. So, Q=∫020(I0+βt) dt.Q=\int_0^{20} (I_0+\beta t)\,dt.Q=∫020​(I0​+βt)dt.

  3. Substitute the given values: Q=∫020(20+3t) dt.Q=\int_0^{20} (20+3t)\,dt.Q=∫020​(20+3t)dt.

  4. Integrate: Q=[20t+3t22]020.Q=\left[20t+\frac{3t^2}{2}\right]_0^{20}.Q=[20t+23t2​]020​.

  5. Put the limits: Q=20(20)+3(20)22Q=20(20)+\frac{3(20)^2}{2}Q=20(20)+23(20)2​ =400+3⋅4002=400+\frac{3\cdot 400}{2}=400+23⋅400​ =400+600=400+600=400+600 =1000 C.=1000\,\text{C}.=1000C.

  6. Therefore, the correct option is: C: 1000 C\boxed{\text{C: }1000\,\text{C}}C: 1000C​

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