A) \[10,5,\,\frac{5}{2},\,...\]
B) \[20,\,\frac{20}{3},\,\frac{20}{5},\,...\]
C) 30, 20, 10, ?
D) 35, 25, 15, ?
Correct Answer: B
Solution :
As in \[\frac{Q}{4\pi {{R}^{2}}}=\frac{q}{4\pi {{r}^{2}}}=\sigma \] \[\therefore \] \[V=\frac{1}{{{\varepsilon }_{0}}}\,[\sigma R+\sigma r]=\frac{\sigma }{{{\varepsilon }_{0}}}[R+r]\] Path difference \[{{N}_{S}}=\left( \frac{\rho S}{\rho P} \right)\times {{N}_{P}}\] \[=\frac{(4.4\times {{10}^{3}})\,\times 100}{220}=2000\] \[{{\rho }_{S}}\] \[{{\rho }_{P}}\] \[-h=-u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}\] But at A, the wave suffers reflection at the surface of rigid/fixed end or denser medium, hence the wave must suffer an additional path change of \[-h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}\] or a phase change of \[0=u({{t}_{2}}-{{t}_{1}})+\frac{1}{2}g(t_{1}^{2}-t_{2}^{2})\]. Net path difference \[u=\frac{1}{2}g({{t}_{1}}+{{t}_{2}})\] For maximum, net path difference \[h=\frac{g{{t}_{1}}{{t}_{2}}}{2}\]. \[\sqrt{2}\,mv\] \[\sqrt{2}\,mv'=(2m)\,v\] or \[\Rightarrow \] \[v'\frac{v}{\sqrt{2}}\] \[E=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}(2m){{v}^{2}}\] \[=mv{{'}^{2}}+m{{v}^{2}}\] \[\Rightarrow \]You need to login to perform this action.
You will be redirected in
3 sec