A) 1 : 2 : 3
B) 1 : 1 : 1
C) 3 : 2 : 2
D) 2 : 3 : 3
Correct Answer: B
Solution :
We know, current flowing through the wire, \[l=\frac{dq}{dt}=\frac{Q}{t}\] or \[Q=l\times t\] Thus, area under the graph between current and time will give charge flowing through the wire. For first time interval,\[({{t}_{1}}=2-1=1\,s)\] \[{{Q}_{1}}={{l}_{1}}\times {{t}_{1}}=2\times 1=2C\] For second time interval, \[({{t}_{2}}=5-3=2\,s)\] \[{{Q}_{2}}={{l}_{2}}\times {{t}_{2}}=1\times 2=2C\] For third time interval, \[({{t}_{3}}=8-6=2s)\] \[{{Q}_{3}}=\frac{1}{2}\times {{l}_{3}}\times {{t}_{3}}=\frac{1}{2}\times 2\times 2=2C\] \[\therefore \]Ratio of charges flowing through the wire at different intervals is \[{{Q}_{1}}:{{Q}_{2}}:{{Q}_{3}}=2:2:2:\,=1:1:1\]
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