A) 100 W lamp will fuse
B) 40 W lamp will fuse
C) Both lamps will fuse
D) Neither lamp will fuse
Correct Answer: B
Solution :
When current i flows, across potential V, then power = \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 10\times {{(10)}^{2}}=500kg.m/s.\] and \[v=10m/s,\mu =0.5,\,m=10kg,\,g=10m/{{s}^{2}}\] (Ohms law) The currents required by the two lamps for their normal brightness are \[s=\frac{\frac{1}{2}\times m\times {{(10)}^{2}}}{\mu mg}\] \[=\frac{50}{0.5\times 10}=10m.\] \[F=\mu R\] The resistance of the filaments are \[R=w=250N.\] \[\mu =0.3\] \[\therefore \] \[F=0.3\times 250=75N.\] The current in each lamp when connected in series with a 40 V supply is \[\overrightarrow{\mathbf{F}}\] \[\Delta t\] \[\overrightarrow{\mathbf{F}}\,\times \Delta t\]. Thus, \[\overrightarrow{\mathbf{F}}\,=m\,\overrightarrow{\mathbf{a}}\] and \[m=150g=150\times {{10}^{-3}}kg,a=20m/{{s}^{2}}\] Thus, the 40 W lamp will fuse, while the 100 W lamp will light dim. Alternative: Resistance of 40 W lamp is \[\therefore \] Resistance of 100 W lamp will be \[F=0.15\times 20=3N\] Now the current through series combination is given by \[=F\,\Delta t=3\times 0.1=0.3N.s.\] Potential drop across 40 W lamp will be given by \[{{v}^{2}}={{u}^{2}}-2gh\] \[\therefore \] and across 100 W lamp \[v=0\] \[\Rightarrow \] Therefore, 40 W bulb will fuse because the lamp can tolerate only 220 V.You need to login to perform this action.
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