21.Two point dipoles $p\hat{k}$ and $\frac{p}{2}\hat{k}$ are located at (0,0,0) and (1m,0,2m) respectively . The resultant electric field due to the two dipoles at the point (1m, 0,0) is A) $\frac{9 p}{32\pi\epsilon_{0}}\hat{k}$ B) $\frac{-7 p}{32\pi\epsilon_{0}}\hat{k}$ C) $\frac{7 p}{32\pi\epsilon_{0}}\hat{k}$ D) none of these View Answer Report DiscussAnswer: Option BExplanation: The given point is on axis of $\frac{\overrightarrow{p}}{2}$ dipole and at equatorial line of $\overrightarrow{p}$ dipole so that field at given point is ($\overrightarrow{E_{1}}+\overrightarrow{E_{2}}$) $\overrightarrow{E_{1}}=\frac{2K(p/2)}{2^{3}}=\frac{K_{p}}{8}(+\hat{k})$ $\overrightarrow{E_{2}}=\frac{K_{p}}{1}(-\hat{k})$ $\overrightarrow{E_{1}}+\overrightarrow{E_{2}}=-\frac{7}{8}K_{p}(-\hat{k})=\frac{-7 p}{32\pi\epsilon_{0}}\hat{k}$
22.A solid sphere of radius R carries a uniform volume charge density $\rho$. The magnitude of the electric field inside the sphere at a distance r from the centre is A) $\frac{r\rho}{3\epsilon_{0}}$ B) $\frac{R\rho}{3\epsilon_{0}}$ C) $\frac{R^{3}\rho}{2\epsilon_{0}}$ D) $\frac{R^{3}\rho}{3\epsilon_{0}}$ View Answer Report DiscussAnswer: Option AExplanation:$E= \frac{1}{4\pi\epsilon_{0}}\frac{qr}{R^{3}}=\frac{q}{\frac{4}{3}\pi R^{3}}\times\frac{r}{3\epsilon_{0}}=\frac{\rho\times r}{3\epsilon_{0}}$
23. Let the energy of an emitted photoelectron be E and wavelength of incident light be $\lambda$. What will be the change in E if $\lambda$ is doubled? A) E B) E/2 C) 2E D) E/4 View Answer Report DiscussAnswer: Option BExplanation:We have hv= W0+E, where E is the energy of emitted photoelectron $\Rightarrow \frac{hc}{\lambda}=W_{0}+E$ As hc and W0 are constant $E\propto \frac{1}{\lambda}$ Therefore, as $\lambda$ is doubled, E will become half
24.For what value od A, B, and C , the output Y=1 A) 0 0 1 B) 1 0 1 C) 1 0 0 D) 0 1 0 View Answer Report DiscussAnswer: Option BExplanation:Clearly, Y= (A+B).C= (A.C) +(B.C) For, A=0, B=0 & C=1 , Y= (0.1)+(0.1)=0 A=1, B=0 & C=1 , Y= (1.1)+(0.1)=1 A=1, B=0 & C=0 , Y= (0.1)+(0.0)=0 A=0, B=1 & C=0 , Y= (0.0)+(1.0)=0 So, option (b) is correct
25.If the wavelength is brought down from 6000Å to 4000Å in a photoelectric experiment then what will happen? A) The work function of the metal will increase B) The threshold frequency will decrease C) No change will take place D) Cut off voltage will increase View Answer Report DiscussAnswer: Option DExplanation:When wavelength decreases, frequency increase, Also we know that cut off voltage (or stopping potential) increases when frequency increases. Hence, option (d) is correct. Note that work function and threshold frequency are constant for a given metal.
