Ask a Doubt
Request a call back

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

CBSE Class 12-science Answered

A simple pendulum is having bob of mass m carrying a charge q and is hanging from ceiling. A uniform electric field E is applied in downward direction then time period of simple pendulum is.
question image
Asked by ruchiaggarwal083 | 12 May, 2021, 08:47: AM
answered-by-expert Expert Answer
Period of simple pendulam in absence of electric field is given as
 
begin mathsize 14px style T space equals space 2 pi square root of l over g end root end style
where l is length of simple pendulam and g is acceleration due to gravity.
 
When the electric field is present , in addition to gravitational force ,
we have force (qE ) due to electric field in the same direction of gravitational force.
Hence downward acceleration due to this electric field is force / mass  = [ ( qE ) / m ]
 
where q is the charge on the bob of pendulam , E is electric field intensity and m is mass of pendulam
 
Hence change in period T of simple pendulam is given as
 
begin mathsize 14px style T space equals space 2 pi square root of fraction numerator l over denominator g apostrophe end fraction end root space equals space 2 pi square root of fraction numerator l over denominator open parentheses g space plus space begin display style fraction numerator q E over denominator m end fraction end style close parentheses end fraction end root end style
---------------------------------------------------------------
 
Second question
 
Force F between two charges of equal magnitude q is given as
 
begin mathsize 14px style F space equals space fraction numerator q squared over denominator 4 pi epsilon subscript o d squared end fraction end style.................................(1)
Since force F is directly proportional to square of charge magniude,
charge magnitude has to be made as 2q to get force 4F .
 
Hence increase in charge = q
 
From eqn.(1), we get q as

begin mathsize 14px style q space equals space square root of 4 pi epsilon subscript o d squared end root end style..........................(2)
if e is charge on electron, number of electrons n is to be added is determined as
 
begin mathsize 14px style q space equals space n space e space equals space space square root of 4 pi epsilon subscript o d squared end root end style
Hence we get ,  begin mathsize 14px style n space equals space square root of fraction numerator 4 pi epsilon subscript o F d squared over denominator e squared end fraction end root end style
Answered by Thiyagarajan K | 12 May, 2021, 02:19: PM

Concept Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
CBSE 12-science - Physics
what is electric field intensity? What is its S.I unit?
Asked by glorialyanzaw | 13 Oct, 2023, 05:53: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
answer
question image
Asked by r01raunak | 23 May, 2023, 10:24: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
an infinite line charge produces a field of 9×10^4 N/C at a distance of 2 cm calculate the linear charge
Asked by srishtijha944 | 15 Jul, 2022, 09:55: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
Three charges \( 1 \mathrm{C},-1 \mathrm{C} \), and \( 2 \mathrm{C} \) are kept on \( X \) axis \( -5 \mathrm{~cm}, x=0 \mathrm{~cm} \) amd \( x=10 \mathrm{~cm} \) respectively find net force one each of them
Asked by marshelojigas | 28 Jun, 2022, 06:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
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
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
A beam of electrons, moving with avelocity of 10^7m/s, enters midway between two horizontal parallel plates in a directions parallel to the plates. Each plate is 5 cm long. These plates are kept 2 cm apart and a potential difference of 90 V is applied between, them. Calculate the velocity of the electron- beam with which it just grazes the edge of the positive plate. (e  /m=1.8×10^{11}  c/kg)
Asked by arjunsah797 | 18 Mar, 2022, 12:06: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
An oil drop of mass 3.25*10^(-15) kg falls vertically with uniform velocity , through the air between vertical parallel plates which are 2 m apart. When a p.d. of 1000 V is applied to the plates the drop moves to the positively charged plate , being inclined at 45 degrees to the vertical. Calculate the charge on the drop. 
Asked by arjunsah797 | 15 Mar, 2022, 09:58: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
Two point charges 2 C and 3 uC are placed at two comers of an equilateral triangle of side 20 cm in free space. Calculate the magnitude of resultant electric field at the third corner of the triangle. If an & - particle is placed at the third corner, what is the force acting on it? (Charge on & – particle is 3.2 x 10-19 €).
Asked by vedhachethan | 19 Sep, 2021, 01:20: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
Asked by avaneesh1162 | 14 Jun, 2021, 08:03: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 12-science - Physics
State Gauss's theorem ? Derive the expression for electric field due Spherical hollow conductor and infinitely charged sheet and linearly charged conductors.
Asked by avaneesh1162 | 13 Jun, 2021, 08:56: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt
Grab Our Exam Special Offer-Enrol now!
×