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A 100 volt AC source of frequency 500 hertz is connected to a LCR circuit with L = 8.1 millihenry, C = 12.5 microfarad and R = 10 ohm, all connected in series. The potential difference across the resistance will be:

Solution: Impedance of series RLC circuit is given by: \(\begin{array}{l}Z=\sqrt{R^{2} + ((2(\pi)fL)- \frac{1}{2(\pi)fC)}})^{2}\end{array} \) \(\begin{array}{l}Z=\sqrt{10^{2} + ((2(\pi)500*8.1*10^{-3})- \frac{1}{2(\pi)500*1.25*10^{-6})}})^{2}\end{array} \) \(\begin{array}{l}Z=10A... View Article

Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in the air. The refractive index of the medium is nearly:

Solutions: \(\begin{array}{l}y_{8}=y_{5}\end{array} \) \(\begin{array}{l}\frac{8\lambda _{m}D}{d}\end{array} \) \(\begin{array}{l}\frac{2({5-1})}{2}\frac{\lambda D}{d}\end{array} \) \(\begin{array}{l}\frac{8\lambda }{\mu }\frac{D}{d}\end{array} \) = \(\begin{array}{l}\frac{9}{2}\frac{\lambda D}{d}\end{array} \) \(\begin{array}{l}\lambda=1.78\end{array} \)

The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux linked with the primary coil is given by ϕ=ϕ0+4t, where ϕ is in webers, t is time in seconds and ϕ0 is a constant, the output voltage across the secondary coil is:

Solution: \(\begin{array}{l}E_{p}=\frac{d\phi }{dt}=\frac{d(\phi_{0} +4t)}{dt} = 4 Volts\end{array} \) Induced emf in secondary coil is given by \(\begin{array}{l}\frac{E_{s}}{E_{p}}=\frac{N_{s}}{N_{p}}=\frac{E_{s}}{4}\end{array} \) \(\begin{array}{l}=1500/50= 120... View Article

Two identical capacitors each of capacitance 5μF are charged to potential 2 KV and 1 KV respectively. There negative ends are connected together, when the positive ends are also connected together, the loss of the energy of the system is

Solution: \(\begin{array}{l}V= V1-V2 = (2-1) =1KV C_{eq}=\frac{5*5*10}{C_{1}+C_{2}} C_{eq}=\frac{5*5*10^{-12}}{10*10^{-6}} C_{eq}=2.5*10^{-6} F \Delta E\end{array} \) =\(\begin{array}{l} \frac{1}{2}C_{eq}V^{2} \Delta E = \frac{1}{2}*2.5*10^{-6}*(10^{3})^{2} \Delta... View Article

A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill should be between

Solution: \(\begin{array}{l}mg=\frac{mv^{2}}{R}\end{array} \) \(\begin{array}{l}v=\sqrt{Rg}\end{array} \) \(\begin{array}{l}v=v=\sqrt{20*10}\end{array} \) \(\begin{array}{l}v=\sqrt{200}\end{array} \) \(\begin{array}{l}v=14.1 m/s\end{array} \) The speed of the car at the top... View Article