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AIIMS 2005
Exam
AIIMS
Year
2005
Subject
Physics
Chapter
Electrostatics

AIIMS 2005 · Physics · Electrostatics

Two concentric conducting thin spherical shells A and B having...

Two concentric conducting thin spherical shells AA and BB having radii rAr_A and r_B\left(r_B>r_A\right)arechargedtoare charged toQ_Aandand-Q_B\left(\left|Q_B\right|>\left|Q_A\right|\right). The electrical field along a line, (passing through the centre is)Two concentric conducting thin spherical shells A and B having radii r_A and r_ B(r_B>r_A) are charged to Q_A andTwo concentric conducting thin spherical shells A and B having radii r_A and r_ B(r_B>r_A) are charged to Q_A andTwo concentric conducting thin spherical shells A and B having radii r_A and r_ B(r_B>r_A) are charged to Q_A andTwo concentric conducting thin spherical shells A and B having radii r_A and r_ B(r_B>r_A) are charged to Q_A and

Options

  1. ATwo soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S
  2. BTwo soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S
  3. CTwo soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S
  4. DTwo soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S

Correct answer

C. Two soap bubbles coalesce to form a single bubble. If V is the subsequent change in volume of contained air and S

Step-by-step solution

∣QB∣>∣QA∣(given).Inside\left|Q_B\right| \gt \left|Q_A\right| (given). Inside the shellB,thepotentialisaconstantandnegative.But, the potential is a constant and negative. ButV_Ainsideinsider_Aispositiveandgreaterthanthatofis positive and greater than that ofBbecausebecauser_A \lt r_B.Upto. Uptor_B,oneashencepositivepotential.Beyond, one as hence positive potential. Beyondr_A, V_A \propto(1 / r),thisis, this is \gt V_B,onegetsthecourse, one gets the course\propto 1 / r.At. AtB,thenegativepotentialismorethereforethepotentiallineisstillnegative.(c)explainsallthesefactors.Note:Howeverthequestionaskedisthefieldandthediagramsgivenareforthepotential.Thisisanerror., the negative potential is more therefore the potential line is still negative. (c) explains all these factors. Note: However the question asked is the field and the diagrams given are for the potential. This is an error. Resistance = potential difference  current =ML2 T−3 A−1 A=[ML2 T−3 A−2]. \begin{aligned}& \text{ Resistance }=\frac{\text{ potential difference }}{\text{ current }}\\ & =\frac{\mathrm{ML}^2\mathrm{~T}^{-3}\mathrm{~A}^{-1}}{\mathrm{~A}}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\mathrm{~A}^{-2}\right] \text{. }\\ & \end{aligned}$
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