Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 12-science Answered

ln a Millikan’s oil drop experiment, the horizontal plates are 1.5 cm apart. With the electric field switch off an oil drop is observed to fall with a steady velocity of 2.5×10^{-2}  cm  s^{-1}.  When the field is switched on the upper plate being positive, the drop just remains stationary when the   p.d.   between the plates is 1500 V. Calculate the radius of drop (b) How many electronic charges does it carry? (c)If the   p.d. between the two plates remains unchanged, with what velocity will the drop move when it has collected two more as a result of exposure to ionising radiation? (Oil density, rho=900 kgm^{3} , Viscosity of air,   eta =1.8×10^{-5} Nsm^{-2}, Density of air,  σ=1.293  kg  m^{-3})

Asked by arjunsah797 | 01 Apr, 2022, 07:31: AM
Expert Answer
When oil drop moves down with constant terminal velocity , force of gravity Fg on oil is balanced
by buoyancy force Fb and viscous drag Fd on oil .
 
Fg = Fb + Fd
 
begin mathsize 14px style 4 over 3 pi space r cubed space rho subscript o space g space equals space 4 over 3 pi space r cubed space rho subscript a i r end subscript space g space plus space left parenthesis space 6 space pi space r space eta space v space right parenthesis end style
 
where r is radius of oil drop ,  ρo is oil density , ρair is density of air , g is acceleration due to gravity ,
η is coefficient of viscosity and v is terminal velocity .
 
From above expression, we get relation for radius of drop as
 
begin mathsize 14px style r space equals space square root of 9 over 2 cross times space eta space cross times fraction numerator v over denominator open parentheses rho subscript o minus rho subscript a i r end subscript close parentheses g end fraction end root end style
 
begin mathsize 14px style r space equals space square root of 4.5 space cross times 1.8 space cross times 10 to the power of negative 5 to the power of space end exponent cross times fraction numerator 2.5 cross times 10 to the power of negative 4 end exponent over denominator open parentheses 900 minus 1.293 close parentheses cross times 9.8 end fraction end root space equals space 1.516 space cross times 10 to the power of negative 6 end exponent space m end style
 
----------------------------------
 
when oil drop remains stationary, Charge on it  is obtained by balancing
the force of gravity Fg and electorstaic force Fe
 
Fg = Fe
 
m g = q E .......................(1)
 
where m is mass of oil drop , g is acceleration due to gravity , q is charge on oil drop
and E is electric field intensity
 
m = (4/3) π r3 ρo g = (4/3) π ( 1.516 × 10-6 )3 × 900 × 9.8 = 1.287 × 10-13 kg
 
E = V / d = 1500 / 0.015 = 105 V/m
 
Where V is potential difference between plates and d is separation distance between plates
 
q = ( 1.287 × 10-13 × 9.8 ) / 105  = 1.261 × 10-17 C
 
Number of electronic charges = q / e = ( 1.261 × 10-17 ) / ( 1.602 × 10-19 ) ≈ 79
 
------------------------------------------------
 
when excess charge of 2 electrons are added to oil drop , oil drop will move upwards due to electrostatic force
and attains terminal velocity v again due to drag force .
 
Force balance equation is given as
 
Fe + Fb = Fg + Fd
 
[ ( q+2) eE ] + [ (4/3) π r3 ρair g ]  = [ m g ] + [ 6π r η v ]
Using eqn.(1) , above expression is rewritten as
 
[ 2 e E ] + [ (4/3) π r3 ρair g ] = [ 6π r η v ]
 
LHS = [ 2 × 1.602 × 10-19 × 105 ] + [ (4/3) π × ( 1.516 × 10-6 )3 × 1.293 × 9.8 ]
 
LHS = 3.222 × 10-14
 
v = ( 3.222 × 10-14 ) / ( 6π × 1.516 × 10-6 × 1.8 × 10-5 ) ≈ 9.5 × 10-5 m/s
 
 
 



Answered by Thiyagarajan K | 01 Apr, 2022, 09:44: AM
CBSE 12-science - Physics
Asked by srishtijha944 | 15 Jul, 2022, 09:55: PM
ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by avaneesh1162 | 14 Jun, 2021, 08:03: AM
ANSWERED BY EXPERT
×