A) 5
B) 1
C) \[\sqrt{15}\]
D) None of these
Correct Answer: C
Solution :
It is clear that f(x) has a definite and unique value for each \[\Delta {{V}_{p}}=-\mu \times \Delta {{V}_{s}}\] Thus, for every point in the interval [1, 5], the value of f(x) exists. So, f(x) is continuous in the interval [1,5]. Also, \[=-50\left( -20 \right)=10V\] which clearly exists for all x in an open interval (1, 5). Hence, f?(x) is differentiable in (1, 5). So, there must be a value c \[E=\Delta m{{c}^{2}}\] such that \[=0.5\times {{10}^{-3}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}\] \[=4.5\times {{10}^{13}}\] But \[E=\frac{4.5\times {{10}^{13}}}{3.6\times {{10}^{6}}}\] \[=1.25\times {{10}^{7}}kWh\] \[i=\frac{5}{20+30}=\frac{5}{50}A\] \[=5\times {{10}^{3}}W\] \[=5000kW\] \[=\frac{mgh}{t}\] \[=7.5\times 9.8\times 4.7\] \[=3454.5kW\] \[\text{=}\frac{\text{Power used}}{\text{Power consumed}}\text{ }\!\!\times\!\!\text{ 100}\] \[=\frac{3454.5}{5000}\times 100=69%\] \[E=V+{{l}_{r}}\] \[{{\lambda }_{red}}>{{\lambda }_{violet}}\] \[\lambda \] \[r=\frac{mV}{B}=\frac{V}{\left( \frac{e}{m} \right)B}\] \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\]
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