Physics-
General
Easy

Question

The load versus extension graph for four wires of same material is shown. The thinnest wire is represented by the line

  1. OA    
  2. OB    
  3. OC    
  4. OD    

The correct answer is: OA

Related Questions to study

General
physics-

A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
This charge will remain constant after switch is shifted from position 1 to position 2.
U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.

A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

physics-General
q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
This charge will remain constant after switch is shifted from position 1 to position 2.
U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.
General
physics-

What is the potential difference between points A a n d B in the circuit shown?

Consider the charge distribution as shown. Considering the branch on upper side, we have

fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
From Eqs. (i) and (ii), we get
fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
therefore V subscript A end subscript equals 4 v o l t
Similarly for the lower side branch
fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
From Eqs. (iii) and (iv)
fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
therefore blank V subscript B end subscript equals 2 blank v o l t
therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t

What is the potential difference between points A a n d B in the circuit shown?

physics-General
Consider the charge distribution as shown. Considering the branch on upper side, we have

fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
From Eqs. (i) and (ii), we get
fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
therefore V subscript A end subscript equals 4 v o l t
Similarly for the lower side branch
fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
From Eqs. (iii) and (iv)
fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
therefore blank V subscript B end subscript equals 2 blank v o l t
therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t
General
Maths-

If m and n are order and degree of the equation open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 plus 4 times fraction numerator open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed over denominator fraction numerator d cubed y over denominator d x cubed end fraction end fraction plus fraction numerator d cubed y over denominator d x cubed end fraction equals x squared minus 1 then

open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 plus 4 times fraction numerator open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed over denominator fraction numerator d cubed y over denominator d x cubed end fraction end fraction plus open parentheses fraction numerator d cubed y over denominator d x cubed end fraction close parentheses equals x squared minus 1
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 fraction numerator d cubed y over denominator d x cubed end fraction plus 4 open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed plus open parentheses fraction numerator d cubed y over denominator d x cubed end fraction close parentheses squared equals open parentheses x squared minus 1 close parentheses fraction numerator d cubed y over denominator d x cubed end fraction
N 0 w space w e space c a n space s a y space t h a t space o r d e r comma m equals 3 space a n d space d e g r e e comma n equals 2 space

If m and n are order and degree of the equation open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 plus 4 times fraction numerator open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed over denominator fraction numerator d cubed y over denominator d x cubed end fraction end fraction plus fraction numerator d cubed y over denominator d x cubed end fraction equals x squared minus 1 then

Maths-General
open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 plus 4 times fraction numerator open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed over denominator fraction numerator d cubed y over denominator d x cubed end fraction end fraction plus open parentheses fraction numerator d cubed y over denominator d x cubed end fraction close parentheses equals x squared minus 1
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 fraction numerator d cubed y over denominator d x cubed end fraction plus 4 open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed plus open parentheses fraction numerator d cubed y over denominator d x cubed end fraction close parentheses squared equals open parentheses x squared minus 1 close parentheses fraction numerator d cubed y over denominator d x cubed end fraction
N 0 w space w e space c a n space s a y space t h a t space o r d e r comma m equals 3 space a n d space d e g r e e comma n equals 2 space
parallel
General
physics-

What is the potential difference across 2muF capacitor in the circuit shown?


Net emf in the circuit here
E equals E subscript 2 end subscript minus E subscript 1 end subscript equals 16 minus 6 equals 10 volt
While the equivalent capacity
C equals blank fraction numerator C subscript 1 end subscript C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction equals blank fraction numerator 2 cross times 3 over denominator 2 plus 3 end fraction equals fraction numerator 6 over denominator 5 end fraction mu F
Charge on each capacitor
q equals C V equals fraction numerator 6 over denominator 5 end fraction cross times 10 equals 12 mu C
therefore Potential difference across 2 mu F capacitor
V subscript 1 end subscript equals fraction numerator q over denominator C subscript 1 end subscript end fraction equals fraction numerator 12 over denominator 2 end fraction equals 6 blank v o l t

What is the potential difference across 2muF capacitor in the circuit shown?

physics-General

Net emf in the circuit here
E equals E subscript 2 end subscript minus E subscript 1 end subscript equals 16 minus 6 equals 10 volt
While the equivalent capacity
C equals blank fraction numerator C subscript 1 end subscript C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction equals blank fraction numerator 2 cross times 3 over denominator 2 plus 3 end fraction equals fraction numerator 6 over denominator 5 end fraction mu F
Charge on each capacitor
q equals C V equals fraction numerator 6 over denominator 5 end fraction cross times 10 equals 12 mu C
therefore Potential difference across 2 mu F capacitor
V subscript 1 end subscript equals fraction numerator q over denominator C subscript 1 end subscript end fraction equals fraction numerator 12 over denominator 2 end fraction equals 6 blank v o l t
General
physics-

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V subscript 2 end subscript minus V subscript 1 end subscript equals 20V. (i e, plate 2 is at a higher potential). The plates are separated by d equals 0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6cross times 10 to the power of negative 19 end exponent C comma m subscript 0 end subscript equals 9.11 cross times 10 to the power of negative 31 end exponent k g right parenthesis

Since V subscript 2 end subscript greater than V subscript 1 end subscript comma so electric field will point from plate 2 to plate 1.
The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.

Use work-energy theorem to find speed of electron when it strikes the plate 2.
fraction numerator m subscript e end subscript v to the power of 2 end exponent over denominator 2 end fraction minus 0 equals e left parenthesis V subscript 2 end subscript minus V subscript 1 end subscript right parenthesis
Where v is the required speed.
therefore blank fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 2 end fraction v to the power of 2 end exponent equals 1.6 cross times 10 to the power of negative 19 end exponent cross times 20
rightwards double arrow blank v equals blank square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 40 over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root equals 2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V subscript 2 end subscript minus V subscript 1 end subscript equals 20V. (i e, plate 2 is at a higher potential). The plates are separated by d equals 0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6cross times 10 to the power of negative 19 end exponent C comma m subscript 0 end subscript equals 9.11 cross times 10 to the power of negative 31 end exponent k g right parenthesis

physics-General
Since V subscript 2 end subscript greater than V subscript 1 end subscript comma so electric field will point from plate 2 to plate 1.
The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.

Use work-energy theorem to find speed of electron when it strikes the plate 2.
fraction numerator m subscript e end subscript v to the power of 2 end exponent over denominator 2 end fraction minus 0 equals e left parenthesis V subscript 2 end subscript minus V subscript 1 end subscript right parenthesis
Where v is the required speed.
therefore blank fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 2 end fraction v to the power of 2 end exponent equals 1.6 cross times 10 to the power of negative 19 end exponent cross times 20
rightwards double arrow blank v equals blank square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 40 over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root equals 2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent
General
physics-

Calculate the force ' ' F that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m. Radius of the wheel is 1m and its mass is open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

applying principle of moments
F (0.6)=100(0.8)
ÞF=133.3N

Calculate the force ' ' F that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m. Radius of the wheel is 1m and its mass is open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

physics-General
applying principle of moments
F (0.6)=100(0.8)
ÞF=133.3N
parallel
General
physics-

A wheel of radius ‘r’ and mass ‘m’ stands in front of a step of height 'h’ The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

Applying the condition of rotational equilibrium,
F left parenthesis r minus h right parenthesis equals m g x
But r to the power of 2 end exponent equals x to the power of 2 end exponent plus left parenthesis r minus h right parenthesis to the power of 2 end exponent rightwards double arrow x equals square root of h left parenthesis 2 r minus h right parenthesis end root
therefore F equals fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction

A wheel of radius ‘r’ and mass ‘m’ stands in front of a step of height 'h’ The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

physics-General
Applying the condition of rotational equilibrium,
F left parenthesis r minus h right parenthesis equals m g x
But r to the power of 2 end exponent equals x to the power of 2 end exponent plus left parenthesis r minus h right parenthesis to the power of 2 end exponent rightwards double arrow x equals square root of h left parenthesis 2 r minus h right parenthesis end root
therefore F equals fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction
General
Maths-

The order of differential equation  isopen parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent

open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 6 equals 1 plus fraction numerator d y over denominator d x end fraction
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 6 minus fraction numerator d y over denominator d x end fraction equals 1
order = 2

The order of differential equation  isopen parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent

Maths-General
open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 6 equals 1 plus fraction numerator d y over denominator d x end fraction
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 6 minus fraction numerator d y over denominator d x end fraction equals 1
order = 2
General
physics-

Two condensers, one of capacity C and the other of capacity fraction numerator C over denominator 2 end fraction, are connected to a V volt battery , as shown. The work done in charging fully both the condensers is

The two capacitor the circuit are in parallel order, hence
C to the power of ´ end exponent equals C plus fraction numerator C over denominator 2 end fraction equals fraction numerator 3 C over denominator 2 end fraction
The work done in charging the equivalent capacitor is stored in the form of potential energy.
Hence, W equals U equals fraction numerator 1 over denominator 2 end fraction C to the power of ´ end exponent V to the power of 2 end exponent
equals fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 3 C over denominator 2 end fraction close parentheses V to the power of 2 end exponent
equals fraction numerator 3 over denominator 4 end fraction C V to the power of 2 end exponent

Two condensers, one of capacity C and the other of capacity fraction numerator C over denominator 2 end fraction, are connected to a V volt battery , as shown. The work done in charging fully both the condensers is

physics-General
The two capacitor the circuit are in parallel order, hence
C to the power of ´ end exponent equals C plus fraction numerator C over denominator 2 end fraction equals fraction numerator 3 C over denominator 2 end fraction
The work done in charging the equivalent capacitor is stored in the form of potential energy.
Hence, W equals U equals fraction numerator 1 over denominator 2 end fraction C to the power of ´ end exponent V to the power of 2 end exponent
equals fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 3 C over denominator 2 end fraction close parentheses V to the power of 2 end exponent
equals fraction numerator 3 over denominator 4 end fraction C V to the power of 2 end exponent
parallel
General
physics-

A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction ‘m’ A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

applying the condition of rotational equilibrium
F left parenthesis L right parenthesis equals m g open square brackets fraction numerator L over denominator 2 end fraction close square brackets semicolon rightwards double arrow F equals fraction numerator m g over denominator 2 end fraction

A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction ‘m’ A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

physics-General
applying the condition of rotational equilibrium
F left parenthesis L right parenthesis equals m g open square brackets fraction numerator L over denominator 2 end fraction close square brackets semicolon rightwards double arrow F equals fraction numerator m g over denominator 2 end fraction
General
physics-

If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m

If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

physics-General
T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m
General
Maths-

Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 equals y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals y
N o w space w e space c a n space s e e space t h a t space o r d e r space equals 2 space a n d space d e g r e e equals 4

Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

Maths-General
fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 equals y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared
rightwards double arrow open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 4 minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared equals y
N o w space w e space c a n space s e e space t h a t space o r d e r space equals 2 space a n d space d e g r e e equals 4
parallel
General
Maths-

The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x
rightwards double arrow x squared fraction numerator d y over denominator d x end fraction minus x y equals 2 cos squared fraction numerator y over denominator 2 x end fraction
rightwards double arrow fraction numerator 1 over denominator cos squared begin display style fraction numerator y over denominator 2 x end fraction end style end fraction open square brackets x squared. fraction numerator d y over denominator d x end fraction minus x y close square brackets equals 2
rightwards double arrow 1 half s e c squared fraction numerator y over denominator 2 x end fraction open square brackets fraction numerator x. begin display style fraction numerator d y over denominator d x end fraction end style minus y over denominator x squared end fraction close square brackets equals 1 over x cubed
rightwards double arrow fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction equals 1 over x cubed
O n space i n t e g r a t i n g space w e space g e comma
integral fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction d x equals integral 1 over x cubed d x
tan fraction numerator y over denominator 2 x end fraction equals C minus fraction numerator 1 over denominator 2 x squared end fraction

The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

Maths-General
x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x
rightwards double arrow x squared fraction numerator d y over denominator d x end fraction minus x y equals 2 cos squared fraction numerator y over denominator 2 x end fraction
rightwards double arrow fraction numerator 1 over denominator cos squared begin display style fraction numerator y over denominator 2 x end fraction end style end fraction open square brackets x squared. fraction numerator d y over denominator d x end fraction minus x y close square brackets equals 2
rightwards double arrow 1 half s e c squared fraction numerator y over denominator 2 x end fraction open square brackets fraction numerator x. begin display style fraction numerator d y over denominator d x end fraction end style minus y over denominator x squared end fraction close square brackets equals 1 over x cubed
rightwards double arrow fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction equals 1 over x cubed
O n space i n t e g r a t i n g space w e space g e comma
integral fraction numerator d over denominator d x end fraction tan fraction numerator y over denominator 2 x end fraction d x equals integral 1 over x cubed d x
tan fraction numerator y over denominator 2 x end fraction equals C minus fraction numerator 1 over denominator 2 x squared end fraction
General
physics-

For the given figure, calculate zero correction.

For the given figure, calculate zero correction.

physics-General
General
physics-

Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
Since, V is the voltage of the battery, so charge on this system
q equals C V
q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
therefore New charge stored
q to the power of ´ end exponent equals C ´ V
q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
Change in total charge
increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets

Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

physics-General
Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction