A) 1.95V
B) 1.5V
C) 2V
D) 1.8V
Correct Answer: A
Solution :
Key Idea: When cell is giving current then the potential difference across its plates is less than its emf by potential drop across internal resistance. When a cell of emf E is connected to a resistance of \[_{g}{{O}^{16}}\] then the emf E of the cell remains constant, while voltage V goes on decreasing on taking more and more current from the cell \[=16-8=8\] \[_{2}H{{e}^{4}}\] where r is internal resistance. Also current \[=4-2=2\] \[_{26}F{{e}^{56}}\] \[=56-26=30\] Putting the numerical values, we have \[_{92}{{U}^{235}}\] \[\text{m=}\frac{\text{image}\,\text{distance}}{\text{object}\,\text{distance}}\text{=-}\frac{\text{v}}{\text{u}}\text{=2}\] \[\Rightarrow \]You need to login to perform this action.
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