The switch is closed at `t = 0`. The time after which the rate of dissipation for energy in the resistor is equal to rate at which energy is being stored in the inductor is : (A) `ln2` (B) `(1)/(2) ln(2)` (C) `(1)/(4) ln(2)` (D) `2ln2`

Consider the RL circuit in Fig. When the switch is closed in position 1 and opens in position 2 , electrical work must be performed on the inductor and on the resistor. The energy stored in the inductor is for the resistor energy appears as heat. a. What is the ratio of P_(L)//P_(R ) of the rate at which energy is stored in the inductor to the rate at which energy is dissipated in the resistor? b. Express the ratio P_(L)//P_(R ) as a function of time. c. If the time constant of circuit is t , what is the time at which P_(L) = P_(R ) ?

The graph between log t_(1//2) and log a at a given temperature is Rate of this reaction will …….with passage of time

Find the domain of f(x)=(log)_(10)(log)_2(log)_(2/pi)(t a n^(-1)x)^(-1)

int_2^4((log)_x2-(((log)_2 2)^2)/(ln2))dx= 0 (b) 1 (c) 2 (d) 4

The value of int_(sqrt(ln2))^(sqrt(ln3))(xsinx^2)/(sinx^2+sin(ln6-x^2))dx is (A) 1/4 ln(3/2) (B) 1/2 ln(3/2) (C) ln(3/2) (D) 1/6ln(3/2)

A graph between log t_((1)/(2)) and log a (abscissa), a being the initial concentration of A in the reaction For reaction Ato Product, the rate law is :

A graph between log t_((1)/(2)) and log a (abscissa), a being the initial concentration of A in the reaction For reaction Ato Product, the rate law is :

The number of solutions of (log)_4(x-1)=(log)_2(x-3) is (a) 3 (b) 1 (c) 2 (d) 0

If p >1 and q >1 are such that log(p+q)=logp+logq , then the value of log(p-1)+log(q-1) is equal to (a) 0 (b) 1 (c) 2 (d) none of these

If (log)_4 5=aa n d(log)_5 6=b , then (log)_3 2 is equal to 1/(2a+1) (b) 1/(2b+1) (c) 2a b+1 (d) 1/(2a b-1)