# 413 - Physics Questions Answers

*Consider an electron trapped in an infinite well whose width is 98.5 pm. If it is in a state with n = 15, what are (i) its energy? (ii) the uncertainty in its momentum? (iii) the uncertainty in its position?*

**Asked By: ANKIT SONI**

**read solutions ( 1 ) | submit your answer**

**Joshi sir comment**

$Letwidth=b\phantom{\rule{0ex}{0ex}}Fornthstaten\frac{\lambda}{2}=b\Rightarrow \lambda =\frac{2b}{n}\phantom{\rule{0ex}{0ex}}NowE=\frac{{p}^{2}}{2m}=\frac{{\left({\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\right)}^{2}}{2m}\phantom{\rule{0ex}{0ex}}Solveafterputtingthevalueof\lambda \phantom{\rule{0ex}{0ex}}Uncer\mathrm{ta}intyinposition=b\phantom{\rule{0ex}{0ex}}Nowuse\u2206p.\u2206x=nh/2\mathrm{\pi}\mathrm{for}\mathrm{finding}\mathrm{uncertainty}\mathrm{in}\mathrm{momentum}$

A solid conducting sphere of radius R is placed in a uniform electric field E as shown in figure-1.441. Due to electric field non uniform surface charges are induced on the surface of the sphere. Consider a point A on the surface of sphere at a polar angle θ from the direction of electric field as shown in figure. Find the surface density of induced charges at point A in terms of electric field and polar angle θ.

**Asked By: DEBRAJ SAHA**

**submit your answer**

**Joshi sir comment**

To nullify the electric field inside, the conducting sphere creates a situation in which electric field inside due to charge on sphere would be same as that of given electric field. Now this is the case, which is same as Irodov 3.17.

See the video now

link:

Finally$E=\frac{{\sigma}_{0}}{3{\epsilon}_{0}}=\frac{\sigma}{3{\epsilon}_{0}\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}Nowsolve$

**Asked By: LUFFY**

**submit your answer**

**Joshi sir comment**

By superposition principle, it is solved in minimum time and effort, other method is very complex. So use superposition principle.

Divide current in point 1 and 2, considering symmetry throughout, we get the equivalent resistance of the whole network except resistor R between 1 and 2 is 2R

Now 2R and R in parallel between 1 and 2

Now solve

A cylindrical pipe of radius r is rolling towards a frog sitting on the horizontal ground. Center of the pipe is moving with velocity v. To save itself ,the frog jumps off and passes over the pipe touching it only at top. Denoting air time of frog by T, horizontal range of the jump by R and acceleration due to gravity by g. Find T and R.

**Asked By: MUDIT**

**submit your answer**

**Joshi sir comment**

$Accordingtogivencondition2r={{u}_{y}}^{2}/2g,\phantom{\rule{0ex}{0ex}}T=2{u}_{y}/g=2\sqrt{4gr}/g=4\sqrt{r/g}\phantom{\rule{0ex}{0ex}}Attoppointduetotouch,vecocityofthefrogbe{u}_{x}-2v.\phantom{\rule{0ex}{0ex}}R=[{u}_{x}{u}_{y}/g]+[{u}_{x}-2v){u}_{y}/g]=[{u}_{y}/g\left]\right(2{u}_{x}-2v)\phantom{\rule{0ex}{0ex}}=[2{u}_{y}/g\left]\right({u}_{x}-v)=T({u}_{x}-v)\phantom{\rule{0ex}{0ex}}Nowminimumpossiblevalueofvelocityoffrogattopto\phantom{\rule{0ex}{0ex}}crossthecylinderis\sqrt{gr}(Bytheconceptofverticalcircle)\phantom{\rule{0ex}{0ex}}nowsolvecarefullywithouterror$

**Asked By: BHASKAR JHA**

**submit your answer**

**Joshi sir comment**

$Let\alpha issmallangulardisplacement\phantom{\rule{0ex}{0ex}}Asshownx=l\mathrm{sin}\alpha andy=l\mathrm{cos}\alpha /\mu \phantom{\rule{0ex}{0ex}}nowuse{\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha =1$

A particle of mass m = 2kg is placed on the top of a hemisphere of M 4 kg. The hemisphere is placed on smooth ground. The particle is displaced gently. Then ratio of normal and pseudo force(seen from hemisphere frame) acting on particle when theta = 30 degree. Assume particle remain in contact with hemisphere.

**Asked By: GUDDU KUMAR**

**submit your answer**

**Joshi sir comment**

$Lettheaccelerationofhemisphereis{a}_{h}\phantom{\rule{0ex}{0ex}}N\mathrm{sin}\theta =M{a}_{h}\phantom{\rule{0ex}{0ex}}\Rightarrow N/m{a}_{h}=M/m\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}Herepseudoforceisconsideredduetoaccelerationofhemisphere$

a small ball of mass 1kg and a charge 2/3 uC is placed at the center of a uniformly charged sphere of radius 1m and charge 1/3 mC. a narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. the sphere is made to rotate about its vertical diameter at a constant rate of 1/2pi revolutions per second. find the spedd w.r.t ground with which the ball slide out from the groove. neglect any magnetic force acting on ball?

**Asked By: VISHNU VARDHAN**

**submit your answer**

**Joshi sir comment**

$Potentialdifferencefromcentretosurface=3V/2-V=V/2\phantom{\rule{0ex}{0ex}}hereV=kq/r(qischargeofsphere)\phantom{\rule{0ex}{0ex}}m{{v}_{r}}^{2}/2=VQ/2(Qischargeofparticle)\phantom{\rule{0ex}{0ex}}{v}_{r}isradialvelocity\phantom{\rule{0ex}{0ex}}Besides\mathrm{tan}gentialvelocityduetorotationofsphere=r\omega =r2\pi n\phantom{\rule{0ex}{0ex}}nowsolve\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

A small ball is thrown from foot of a wall with minimum possible velocity to hit a bulb B on the ground at a distance L away from the wall.Find the expression for height h of shadow of the ball on the wall as a function of time t.Acceleration due to gravity is g.

**Asked By: MUDIT**

**read solutions ( 1 ) | submit your answer**

**Joshi sir comment**