Question Number 38675 by rahul 19 last updated on 28/Jun/18
![](https://www.tinkutara.com/question/4522.png)
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18
![A→Q,B→Q](https://www.tinkutara.com/question/Q38763.png)
Commented by rahul 19 last updated on 30/Jun/18
![Correct Ans. is A→R , B→Q , C→Q , D→P](https://www.tinkutara.com/question/Q38819.png)
Commented by rahul 19 last updated on 30/Jun/18
![(MrW3 , Ajfour) sir pls help us ......](https://www.tinkutara.com/question/Q38820.png)
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jun/18
![q=q_0 (1−e^(−(t/(CR))) )= A)energy=(1/2)(q^2 /c) B)heat=joules heat =((i^2 Rt)/j) i=(dq/dt)=q_0 (0−e^((−t)/(cR)) ×((−1)/(cR)))=q_0 ((1/(cR))e^((−t)/(cR)) ) heat=(q_0 ^2 /(c^2 R^2 ))e^(−2(t/(cR))) ×R×(t/j) C)energy_ =(1/2)C×(pd)^2 } =(1/2)×c((q^2 /c^2 ))=(1/(2c))q_0 ^2 ×(1−e^((−t)/(cR)) )^2 D)w=(1/2)×(q^2 /c)=((q_0 ^2 (1−e^((−t)/(cR)) )^7 )/(2c)) energy=(q^2 /(2c)) q=q_o (1−e^(−(t/(cR))) )≈{1−(1−(t/(cR)))} so q≈q_0 (t/(cR)) energy=((q_0 ^2 t^2 )/(2c^3 R^2 )) so the graph alike y=x^2](https://www.tinkutara.com/question/Q38702.png)
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18
![i am tryiny to corelate with graph](https://www.tinkutara.com/question/Q38758.png)