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Chapter 12: Problem 1755

Formula for current flowing through a wire is \(I=6 t^{2}+4 t-2\) here \(\mathrm{t}\) is in second and I is an ampere. In this wire, what is the quantity of electric charge passing in time interval \(1 \mathrm{sec}\) to 2 sec? (A) \(8 \mathrm{C}\) (B) \(18 \mathrm{C}\) (C) \(20 \mathrm{C}\) (D) \(24 \mathrm{C}\)

Short Answer

Expert verified
The quantity of electric charge passing through the wire in the time interval from 1 second to 2 seconds is \(18 \mathrm{C}\).

Step by step solution

01

Write down the given formula

The formula for the current flowing through the wire is given as: \[ I(t) = 6t^2 + 4t - 2 \]
02

Integrate the current function with respect to time

To find the electric charge passing through the wire, we need to integrate the current function with respect to the time interval, between 1 second and 2 seconds. The integral can be written as: \[Q = \int_1^2 I(t) dt = \int_1^2 (6t^2 + 4t - 2) dt \]
03

Evaluate the integral

Now, we can evaluate the integral as follows: \[\begin{align*} Q &= \left[ 2t^3 + 2t^2 - 2t \right]_1^2 \\ &= \left[ 2(2)^3 + 2(2)^2 - 2(2) \right] - \left[ 2(1)^3 + 2(1)^2 - 2(1) \right] \\ &= (16 + 8 - 4) - (2 + 2 - 2) \\ &= 20 - 2 \\ &= 18 \mathrm{C} \end{align*}\]
04

State the answer

The quantity of electric charge passing through the wire in the time interval from 1 second to 2 seconds is 18 Coulombs. Thus, the correct answer is (B) 18 C.

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Most popular questions from this chapter

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)

n identical cells each of e.m.f \(E\) and internal resistance \(r\) are connected in series An external resistance \(R\) is connected in series to this combination the current through \(\mathrm{R}\) is. (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\) (B) \(\\{(\mathrm{nE}) /(\mathrm{n} R+\mathrm{r})\\}\) (C) \(\\{\mathrm{E} /(\mathrm{R}+\mathrm{nr})\\}\) (D) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{r})\\}\)

Resistors \(P\) and \(Q\) connected in the gaps of the meter bridge. the balancing point is obtained \(1 / 3 \mathrm{~m}\) from the zero end. If a \(6 \Omega\) resistance is connected in series with \(\mathrm{p}\) the balance point shifts to \(2 / 3 \mathrm{~m}\) form same end. \(\mathrm{P}\) and \(\mathrm{Q}\) are. (A) 4,2 (B) 2,4 (C) both (a) and (b) (D) neither (a) nor (b)

Which is the dimensional formula for conductance from the give below? (A) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (B) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
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