Maths-
General
Easy

Question

Solution of differential equation fraction numerator d y over denominator d x end fraction plus a y equals e to the power of m x end exponent is

  1. left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c
  2. y e to the power of a x end exponent equals m e to the power of m x end exponent plus c
  3. y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent
  4. left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent

The correct answer is: left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent

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Related Questions to study

General
physics-

In given circuit when switch S has been closed then charge on capacitor A and B respectively are

The circuit is given as

Let q subscript 1 end subscript a n d q subscript 2 end subscript be the charge after switch S has been closed.
Then, V equals fraction numerator q subscript 1 end subscript over denominator 6 C end fraction equals fraction numerator q subscript 2 end subscript over denominator 3 C end fraction
rightwards double arrow blank fraction numerator q subscript 1 end subscript over denominator 2 end fraction equals q subscript 2 end subscript
rightwards double arrow blank q subscript 1 end subscript equals 2 q subscript 2 end subscript…(i)
But we know that, charge is conserved
q subscript 1 end subscript plus q subscript 2 end subscript equals 3 q plus 6 q
or q subscript 1 end subscript plus q subscript 2 end subscript equals 9 q…(ii)
On putting the value of q subscript 1 end subscriptEq. (ii)
2 q subscript 2 end subscript plus q subscript 2 end subscript equals 9 q
rightwards double arrow 3 q subscript 2 end subscript equals 9 q
q subscript 2 end subscript equals 3 q
Now, from Eq. (i)
q subscript 1 end subscript equals 2 cross times 3 q
rightwards double arrow q subscript 1 end subscript equals 6 q
Hence,q subscript 1 end subscript equals 6 q comma blank q subscript 2 end subscript equals 3 q

In given circuit when switch S has been closed then charge on capacitor A and B respectively are

physics-General
The circuit is given as

Let q subscript 1 end subscript a n d q subscript 2 end subscript be the charge after switch S has been closed.
Then, V equals fraction numerator q subscript 1 end subscript over denominator 6 C end fraction equals fraction numerator q subscript 2 end subscript over denominator 3 C end fraction
rightwards double arrow blank fraction numerator q subscript 1 end subscript over denominator 2 end fraction equals q subscript 2 end subscript
rightwards double arrow blank q subscript 1 end subscript equals 2 q subscript 2 end subscript…(i)
But we know that, charge is conserved
q subscript 1 end subscript plus q subscript 2 end subscript equals 3 q plus 6 q
or q subscript 1 end subscript plus q subscript 2 end subscript equals 9 q…(ii)
On putting the value of q subscript 1 end subscriptEq. (ii)
2 q subscript 2 end subscript plus q subscript 2 end subscript equals 9 q
rightwards double arrow 3 q subscript 2 end subscript equals 9 q
q subscript 2 end subscript equals 3 q
Now, from Eq. (i)
q subscript 1 end subscript equals 2 cross times 3 q
rightwards double arrow q subscript 1 end subscript equals 6 q
Hence,q subscript 1 end subscript equals 6 q comma blank q subscript 2 end subscript equals 3 q
General
Maths-

Solution of differential equation d y minus sin space x sin y space d x equals 0 is

G i v e n comma space d y minus sin space x sin y d x equals 0
rightwards double arrow d y equals sin x space sin y space d x
rightwards double arrow fraction numerator d y over denominator d x end fraction fraction numerator 1 over denominator sin y end fraction equals sin x
rightwards double arrow fraction numerator d y over denominator d x space end fraction cos e c y equals sin x space
i n t e g r a t i n g space b o t h space s i d e s
rightwards double arrow log space tan y over 2 plus log space k equals negative cos space x
rightwards double arrow fraction numerator tan begin display style y over 2 end style over denominator C end fraction equals e to the power of negative cos x end exponent
rightwards double arrow e to the power of cos x space tan y over 2 end exponent equals C

Solution of differential equation d y minus sin space x sin y space d x equals 0 is

Maths-General
G i v e n comma space d y minus sin space x sin y d x equals 0
rightwards double arrow d y equals sin x space sin y space d x
rightwards double arrow fraction numerator d y over denominator d x end fraction fraction numerator 1 over denominator sin y end fraction equals sin x
rightwards double arrow fraction numerator d y over denominator d x space end fraction cos e c y equals sin x space
i n t e g r a t i n g space b o t h space s i d e s
rightwards double arrow log space tan y over 2 plus log space k equals negative cos space x
rightwards double arrow fraction numerator tan begin display style y over 2 end style over denominator C end fraction equals e to the power of negative cos x end exponent
rightwards double arrow e to the power of cos x space tan y over 2 end exponent equals C
General
physics-

In the given figure, a hollow spherical capacitor is shown. The electric field will not be zero at

The electric field of a hollow spherical capacitor is localised in between inner and outer surface of the spherical conductor.
Therefore, at point r subscript 1 end subscript less than r less than r subscript 2 end subscript, the electric field will not be zero.

In the given figure, a hollow spherical capacitor is shown. The electric field will not be zero at

physics-General
The electric field of a hollow spherical capacitor is localised in between inner and outer surface of the spherical conductor.
Therefore, at point r subscript 1 end subscript less than r less than r subscript 2 end subscript, the electric field will not be zero.
General
physics-

In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

Final charge on capacitor is
q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
R C=time constant of the circuit.
Putting q subscript 0 end subscript equals C V subscript 0 end subscript
therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C

In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

physics-General
Final charge on capacitor is
q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
R C=time constant of the circuit.
Putting q subscript 0 end subscript equals C V subscript 0 end subscript
therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C
General
Maths-

The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

G i v e n comma space 2 x fraction numerator d y over denominator d x end fraction minus y equals 3
rightwards double arrow 2. fraction numerator d y over denominator d x end fraction equals fraction numerator 3 plus y over denominator x end fraction
rightwards double arrow 2. fraction numerator d y over denominator d x end fraction. fraction numerator 1 over denominator 3 plus y end fraction equals 1 over x
i n t e g r a t i n g space b o t h space s i d e s space w e space g e t comma
rightwards double arrow 2 log open parentheses 3 plus y close parentheses equals log open parentheses x close parentheses
rightwards double arrow log open parentheses 3 plus y close parentheses squared equals log open parentheses x close parentheses
rightwards double arrow open parentheses 3 plus y close parentheses squared equals x plus C
t h e r e f o r e space t h e space s o l u t i o n space r e p r e s e n t s space p a r a b o l a s.

The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

Maths-General
G i v e n comma space 2 x fraction numerator d y over denominator d x end fraction minus y equals 3
rightwards double arrow 2. fraction numerator d y over denominator d x end fraction equals fraction numerator 3 plus y over denominator x end fraction
rightwards double arrow 2. fraction numerator d y over denominator d x end fraction. fraction numerator 1 over denominator 3 plus y end fraction equals 1 over x
i n t e g r a t i n g space b o t h space s i d e s space w e space g e t comma
rightwards double arrow 2 log open parentheses 3 plus y close parentheses equals log open parentheses x close parentheses
rightwards double arrow log open parentheses 3 plus y close parentheses squared equals log open parentheses x close parentheses
rightwards double arrow open parentheses 3 plus y close parentheses squared equals x plus C
t h e r e f o r e space t h e space s o l u t i o n space r e p r e s e n t s space p a r a b o l a s.
General
physics-

The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

physics-General
General
physics-

In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

physics-General
General
physics-

The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

physics-General
General
physics-

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

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

physics-General
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
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