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Question-38675




Question Number 38675 by rahul 19 last updated on 28/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18
A→Q,B→Q
AQ,BQ
Commented by rahul 19 last updated on 30/Jun/18
Correct Ans. is   A→R , B→Q , C→Q , D→P
CorrectAns.isAR,BQ,CQ,DP
Commented by rahul 19 last updated on 30/Jun/18
(MrW3 , Ajfour) sir pls help us ......
(MrW3,Ajfour)sirplshelpus
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
q=q0(1etCR)=A)energy=12q2cB)heat=joulesheat=i2Rtji=dqdt=q0(0etcR×1cR)=q0(1cRetcR)heat=q02c2R2e2tcR×R×tjC)energy=12C×(pd)2}=12×c(q2c2)=12cq02×(1etcR)2D)w=12×q2c=q02(1etcR)72cenergy=q22cq=qo(1etcR){1(1tcR)}soqq0tcRenergy=q02t22c3R2sothegraphalikey=x2
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18
i am tryiny to corelate with graph
iamtryinytocorelatewithgraph

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