# 409 - Physics Questions Answers

Initially the spheres a and b are at potential be a and b respectively now sphere b is earthd by closing the switch the poential of a will become

**Asked By: ROMEO ROMI**

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**Joshi sir comment**

No meaning can be incurred by this language. Rectify it for getting answer

**Asked By: DARIUS**

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**Joshi sir comment**

$Sincethicknessissamesoforceperunitlengthshouldbesame\phantom{\rule{0ex}{0ex}}Forceperunitlength=F/2\pi R\phantom{\rule{0ex}{0ex}}ChargedensityinthesphereisproportionaltoE\phantom{\rule{0ex}{0ex}}soF\alpha qE\alpha \sigma AE\alpha {E}^{2}{R}^{2}\phantom{\rule{0ex}{0ex}}Thusforceperunitlength\alpha {E}^{2}{R}^{2}/2\pi R\alpha {E}^{2}R\phantom{\rule{0ex}{0ex}}Nowcomparethetwocases$

**Asked By: DARIUS**

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**Joshi sir comment**

$Atageneralpoint\phantom{\rule{0ex}{0ex}}-2T\mathrm{sin}\theta =ma(\theta ismeasuredfromhorizontal)\phantom{\rule{0ex}{0ex}}\Rightarrow a=-2Mg\theta /m=-2Mgy/ml\phantom{\rule{0ex}{0ex}}comparewitha=-{\omega}^{2}y\phantom{\rule{0ex}{0ex}}\omega =\sqrt{\frac{2Mg}{ml}}\left(1\right)\phantom{\rule{0ex}{0ex}}Nowformaxdisplacementofsmallm\phantom{\rule{0ex}{0ex}}mg(h+y)=2Mgxandx={y}^{2}/2l\phantom{\rule{0ex}{0ex}}onsolvingthisquadraticeq.fory/lweget\phantom{\rule{0ex}{0ex}}\frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}}(giventhat\frac{m}{M}\frac{h}{l})\phantom{\rule{0ex}{0ex}}sox=\frac{m}{2M}h\left(2\right)\phantom{\rule{0ex}{0ex}}so{v}_{max}=x\omega \phantom{\rule{0ex}{0ex}}nowsolve$

Two boys enter a running escalator at the ground floor of a shopping mall . The first boy repeatedly follows a cycle of p1=1 step up and then q1=2 steps down whereas the second boy repeatedly follows a cycle of p2=2 steps up and then q2=1 step down. Both of them move relative to escalator with a speed v=50cm/s . If the boys take t1=250s and t2=50s respectively to reach the first floor in complete number of cycles, how fast is the escalator running ?

**Asked By: JATIN KUMAR**

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**Joshi sir comment**

$letuandvarethespeedsofboyandescalator\phantom{\rule{0ex}{0ex}}accordingtogivencondition\phantom{\rule{0ex}{0ex}}\left[1\right(u+v)T-2(u-v\left)T\right]{n}_{1}=\left[2\right(u+v)T-1(u-v\left)T\right]{n}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]T{n}_{1}=\left[2\right(u+v)-1(u-v\left)\right]T{n}_{2}\phantom{\rule{0ex}{0ex}}GiventhatT{n}_{1}=250andT{n}_{2}=50\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]250=\left[2\right(u+v)-1(u-v\left)\right]50\phantom{\rule{0ex}{0ex}}nowsolveitforv,uisgiven$

A particle projected from the ground passes two points , which are at height 12m and 18m above the ground and a distance d=10m apart . What could be the minimum speed of projection. Acceleration due to gravity is g=10m/s^2 .

**Asked By: JATIN KUMAR**

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**Joshi sir comment**

SIR ANSWER IS given as C could u pls explain?

**Asked By: AYUSH PAINE**

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**Joshi sir comment**