A) \[5\mu F\]cm
B) \[200\,\,V\] cm
C) \[300\,\,V\] cm
D) \[400\,\,V\]
Correct Answer: C
Solution :
Let the focal length in air be \[{{f}_{t}}=\frac{0.5}{0.125}\times 15\approx 60\,\,cm\] and that in liquid be\[e\]. Then \[e=-L\frac{\Delta i}{\Delta t}\] \[\frac{\Delta i}{\Delta t}\] Where, \[L=5H,\frac{\Delta i}{\Delta t}=-2A/s\] is refractive index of glass relative to air and \[\therefore \] is refractive of glass relative of liquid. \[e=-5\times \left( -2 \right)=+10\,V\] \[{{q}_{1}}\,\,and\,\,{{q}_{2}}\] Given, \[Q={{Q}_{1}}+{{Q}_{2}}\] \[C\,\,V={{C}_{1}}\,\,{{V}_{1}}+{{C}_{2}}\,\,{{V}_{2}}\] \[C={{C}_{1}}+{{C}_{2}}\] \[\therefore \]
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