1.In Young's double-slit experiment, the central point on the screen is A) bright B) dark C) First bright and then dark D) first dark and then bright View Answer Report DiscussAnswer: Option AExplanation:In the normal adjustment of YDSE, Path difference between the waves at the central location is always zero. So maxima is obtained at the central position. Central point on the screen is bright.
2.In Young's double-slit experimental setup if the wavelength alone is doubled, the bandwidth β becomes A) $\frac{\beta}{2}$ B) $2\beta$ C) $3\beta$ D) $\beta$ View Answer Report DiscussAnswer: Option BExplanation:Fringe width, $\beta= \frac{\lambda D}{d}$ When λ alone is doubled then fringe width becomes $\beta '= \frac{\left(2\lambda\right) D}{d}$ = 2β
3. The work done by an uniform magnetic field on a moving charge is A) zero because $\overrightarrow{F}$ acts parallel to $\overrightarrow{v}$ B) positive because $\overrightarrow{F}$ acts perpendicular to $\overrightarrow{v}$ C) zero because $\overrightarrow{F}$ acts perpendicular to $\overrightarrow{v}$ D) negative because $\overrightarrow{F}$ acts parallel to $\overrightarrow{v}$ View Answer Report DiscussAnswer: Option CExplanation:Force on moving charge while moving in a magnetic field is; $\overrightarrow{F} = q (\overrightarrow{v} \times \overrightarrow{B})$ where $\overrightarrow{F}$ is perpendicular to $\overrightarrow{v}$ . Work done/sec = $\overrightarrow{F}$.$\overrightarrow{v}$ = Fv cos90º= 0
4.The current in the 1Ω resistor shown in the circuit is A) $\frac{2}{3}$ A B) 3A C) 6A D) 2A View Answer Report DiscussAnswer: Option DExplanation:Two 4Ω resistors are in parallel combination. Their equivalent resistance = $\frac{4×4}{4+4}$ = $\frac{16}{8}$ = 2Ω .'. Total resistance of the network = 2 + 1 = 3Ω .'. Cunent through 1Ω resistor, i= $\frac{6}{3}$ = 2A
5.Tube A has both ends open while tube B has one end closed, otherwise, they are identical. The ratio of the fundamental frequency of tube 'A and B' is A) 1 : 2 B) 1 : 4 C) 2 : 1 D) 4 : 1 View Answer Report DiscussAnswer: Option CExplanation:Fundamental freq. for, Tube A = fA = $\frac{v}{2L}$ Tube B = fB = $\frac{v}{4L}$ Now, $\frac{f_{A}}{f_{B}}$ = $\frac{v}{2L} \times \frac{4L}{v}$ = $\frac{2}{1}$ = fA: fB = 2 : 1
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