NEET-XII-Physics

exam-1 year:2016

with Solutions -
  • #14
    The charge flowing through a resistance R varies
    with time t as Q = at - ``bt^2``, where a and b are positive
    constants. The total heat produced in R is:
    (1)``\frac {a^3R}{6b}``
    (2)``\frac {a^3R}{3b}``
    (3) ``\frac {a^3R}{2b}``
    (4)``\frac {a^3R}{b}``
    digAnsr:   1
    Ans : 1
    Sol. =  2Q at bt
    i = a - 2bt { for i = 0  &implies; t =
    a
    2b
    }
    From joule's law of heating
    dH = i2Rdt
    = 
    a /2b
    2
    0
    H (a 2bt) Rdt
    H =
    
     
    a
    3 2b
    0
    (a 2bt) R
    3 2b
    =
    3a R
    6b