The following MCQ which appeared in KEAM 2007 (Engineering) question paper can claim to be one different from the usual type:

**In a closed circuit, the current I (in ampere) at an instant of time t (in second) is given by I = 4** **– 0.08t. The number of electrons flowing in 50 s through the cross section of the conductor is**

**(a) 1.25****×10 ^{19} **

**(b) 6.25**

**×10**

^{20}**(c) 5.25**

**×10**

^{19}**(d) 2.55****×10 ^{20} **

**(e) 4.25**

**×10**

^{20}

To obtain the electrons flowing in 50 s, you have to calculate the total charge flowing in 50 s.

Total charge Q = _{ }∫Idt = ∫(4–0.08t)dt = [4t–(0.08t^{2}/2)].

Since the limits of integration are 0 and 50 seconds, Q = 4×50–(0.08×50^{2}/2) = 100 coulomb.

Since the electronic charge is 1.6×10^{–19} coulomb, the number of electrons flowing in 50 s is 100/(1.6×10^{–19}) = 6.25×10^{20}.

_{}

## No comments:

## Post a Comment