A particle is projected vertically upwards with velocity .30 ms-Turito

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Easy

Physics

A particle is projected vertically upwards with velocity .30 ms^-1 Find the ratio of average speed and instantaneous velocity after 6s.[g=10 ms^-1 ]

PhysicsGeneral

3
$\frac{1}{2}$
4
2

Answer:The correct answer is: $1 half$

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$y equals f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus vertical line x vertical line end fraction comma x element of R comma y element of R text is end text$

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A particle moves 4 m in the south direction. Then it moves 3 m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity

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In the adjacent figure is shown a closed path $P.$ A long straight conductor carrying a current $I$ passes through $O$ and perpendicular to the plane of the paper. Then which of the following holds good?

A long straight conductor carrying current

I

passes through

O comma

then by symmetry, all points of the circular path are equivalent and hence the magnitude of magnetic field should be same at these points.
The circulation of magnetic field along the circle is

not stretchy contour integral B. d l equals mu subscript 0 end subscript I

(using Ampere’s law)

In the adjacent figure is shown a closed path $P.$ A long straight conductor carrying a current $I$ passes through $O$ and perpendicular to the plane of the paper. Then which of the following holds good?

physics-General

A long straight conductor carrying current

I

passes through

O comma

then by symmetry, all points of the circular path are equivalent and hence the magnitude of magnetic field should be same at these points.
The circulation of magnetic field along the circle is

not stretchy contour integral B. d l equals mu subscript 0 end subscript I

(using Ampere’s law)

General

physics-

A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

The resultant magnetic field at the centre of the coil is

Magnetic field a

B

B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction

Magnetic field at

A

B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction

The resultant magnetic field at the centre

open vertical bar B subscript A end subscript close vertical bar equals open vertical bar B subscript B end subscript close vertical bar

So, magnetic field become is zero.

A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

The resultant magnetic field at the centre of the coil is

physics-General

Magnetic field a

B

B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction

Magnetic field at

A

B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction

The resultant magnetic field at the centre

open vertical bar B subscript A end subscript close vertical bar equals open vertical bar B subscript B end subscript close vertical bar

So, magnetic field become is zero.

General

physics-

The figure shows the cross-section of a long cylindrical conductor of radius a carrying a uniformly distributed current i. The magnetic field due to current at P is

The current enclosed with in the circle

fraction numerator i over denominator pi a to the power of 2 end exponent end fraction blank comma blank pi r to the power of 2 end exponent equals fraction numerator i over denominator a to the power of 2 end exponent end fraction r to the power of 2 end exponent

Ampere’s law

not stretchy contour integral B. d l equals mu subscript 0 end subscript i ´

gives

B blank.2 pi r equals fraction numerator mu subscript 0 end subscript i r to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction

B equals blank fraction numerator mu subscript 0 end subscript i r over denominator 2 pi a to the power of 2 end exponent end fraction

The figure shows the cross-section of a long cylindrical conductor of radius a carrying a uniformly distributed current i. The magnetic field due to current at P is

physics-General

The current enclosed with in the circle

fraction numerator i over denominator pi a to the power of 2 end exponent end fraction blank comma blank pi r to the power of 2 end exponent equals fraction numerator i over denominator a to the power of 2 end exponent end fraction r to the power of 2 end exponent

Ampere’s law

not stretchy contour integral B. d l equals mu subscript 0 end subscript i ´

gives

B blank.2 pi r equals fraction numerator mu subscript 0 end subscript i r to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction

B equals blank fraction numerator mu subscript 0 end subscript i r over denominator 2 pi a to the power of 2 end exponent end fraction

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In angiosperms, female gametophyte is represented by

biologyGeneral

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A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the $X minus Y$ plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

900553

If we take a small strip of

d r

at distance

r

from centre, then number of turns in this strip would be,

d N equals open parentheses fraction numerator N over denominator b minus a end fraction close parentheses d r

Magnetic field due to this element at the centre of the coil will be

d B equals fraction numerator mu subscript 0 end subscript open parentheses d N close parentheses I over denominator 2 r end fraction equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction fraction numerator d r over denominator r end fraction

B equals not stretchy integral subscript r equals a end subscript superscript r equals b end superscript d B equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction ln invisible function application open parentheses fraction numerator b over denominator a end fraction close parentheses

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the $X minus Y$ plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

900553

physics-General

If we take a small strip of

d r

at distance

r

from centre, then number of turns in this strip would be,

d N equals open parentheses fraction numerator N over denominator b minus a end fraction close parentheses d r

Magnetic field due to this element at the centre of the coil will be

d B equals fraction numerator mu subscript 0 end subscript open parentheses d N close parentheses I over denominator 2 r end fraction equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction fraction numerator d r over denominator r end fraction

B equals not stretchy integral subscript r equals a end subscript superscript r equals b end superscript d B equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction ln invisible function application open parentheses fraction numerator b over denominator a end fraction close parentheses

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The motion of a particle along a straight line is described by the function .X=(3T-2)² Calculate the acceleration after 10 s.

physicsGeneral

General

physics

Rohit completes a semi circular path of radius R in 10 seconds. Calculate average speed and average velocity in ms^-1.

physicsGeneral

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If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient $eta comma$ the Poisson’s ratio $sigma$ will be

Young’s modulus,

Y equals fraction numerator 3 ɳ over denominator 2 end fraction

We know that

Y equals 2 ɳ open parentheses 1 plus sigma close parentheses

therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses

⟹

blank sigma equals negative fraction numerator 1 over denominator 4 end fraction

If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient $eta comma$ the Poisson’s ratio $sigma$ will be

physics-General

Young’s modulus,

Y equals fraction numerator 3 ɳ over denominator 2 end fraction

We know that

Y equals 2 ɳ open parentheses 1 plus sigma close parentheses

therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses

⟹

blank sigma equals negative fraction numerator 1 over denominator 4 end fraction

General

physics

Given figure shows a graph at acceleration $not stretchy rightwards arrow$ time for a rectilinear motion. Find average acceleration in first 10 seconds

physicsGeneral

General

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Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

increment L equals fraction numerator F L over denominator A Y end fraction

Because, wires of the same material are stretched by the same load. So, F and Y will be constant.

therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction

increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent

therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent

therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent

therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent

We can see that, L=100 cm and r = 1mm will elongate most.

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

physics-General

increment L equals fraction numerator F L over denominator A Y end fraction

Because, wires of the same material are stretched by the same load. So, F and Y will be constant.

therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction

increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent

therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent

therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent

therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction

equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent

We can see that, L=100 cm and r = 1mm will elongate most.

General

physics-

Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P,Q and R as shown in the figure. With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in case of

P

and

R comma

current divides equally in two halves because of equal resistances in the two halves. Due to equal currents in two halves the magnetic field at the centre will be zero.

Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P,Q and R as shown in the figure. With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in case of

physics-General

P

and

R comma

current divides equally in two halves because of equal resistances in the two halves. Due to equal currents in two halves the magnetic field at the centre will be zero.

General

physics-

A current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
The magnetic field at the centre of the loop is $fraction numerator mu subscript 0 end subscript i over denominator R end fraction$ times.
$left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree$

(i) Magnetic field at the centre due to the curved portion

D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses

According to right hand screw rule, the magnetic field will be into the plane of paper.

(ii) Magnetic field at

M

due to

A B

is zero.
(iii) Magnetic field at the centre due to the curved portion

B C

fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses.

According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at

M

due to

D C

is zero.
Hence, the resultant magnetic field at

M

equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction

A current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
The magnetic field at the centre of the loop is $fraction numerator mu subscript 0 end subscript i over denominator R end fraction$ times.
$left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree$

physics-General

(i) Magnetic field at the centre due to the curved portion

D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses

According to right hand screw rule, the magnetic field will be into the plane of paper.

(ii) Magnetic field at

M

due to

A B

is zero.
(iii) Magnetic field at the centre due to the curved portion

B C

fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses.

According to
right hand screw rule, the magnetic field will be into the plane of paper.
(iv) Magnetic field at

M

due to

D C

is zero.
Hence, the resultant magnetic field at

M

equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction

General

physics-

A part of a long wire carrying a current $i$ is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

The magnitude of the magnetic field at point

O

due to straight part of wire is

B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction

B subscript 1 end subscript

is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre

O

due to the current loop of radius

r

B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

B subscript 2 end subscript

is also perpendicular to the page, directed upwards (right hand screw rule).

B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses

equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses

A part of a long wire carrying a current $i$ is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

physics-General

The magnitude of the magnetic field at point

O

due to straight part of wire is

B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction

B subscript 1 end subscript

is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre

O

due to the current loop of radius

r

B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

B subscript 2 end subscript

is also perpendicular to the page, directed upwards (right hand screw rule).

B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses

equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses

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A particle is projected vertically upwards with velocity .30 ms-1 Find the ratio of average speed and instantaneous velocity after 6s.[g=10 ms-1 ]

342

Answer:The correct answer is:

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A particle moves 4 m in the south direction. Then it moves 3 m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity

A particle moves 4 m in the south direction. Then it moves 3 m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity

In the adjacent figure is shown a closed path A long straight conductor carrying a current passes through and perpendicular to the plane of the paper. Then which of the following holds good?

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A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagramThe resultant magnetic field at the centre of the coil is

A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagramThe resultant magnetic field at the centre of the coil is

The figure shows the cross-section of a long cylindrical conductor of radius a carrying a uniformly distributed current i. The magnetic field due to current at P is

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In angiosperms, female gametophyte is represented by

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A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the plane and a steady current flows through the wire. The Z-component of the magnetic field at the centre of the spiral is 900553

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the plane and a steady current flows through the wire. The Z-component of the magnetic field at the centre of the spiral is 900553

The motion of a particle along a straight line is described by the function .X=(3T-2)2 Calculate the acceleration after 10 s.

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If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient the Poisson’s ratio will be

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Given figure shows a graph at acceleration time for a rectilinear motion. Find average acceleration in first 10 seconds

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Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P,Q and R as shown in the figure. With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in case of

Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P,Q and R as shown in the figure. With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in case of

A current is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.The magnetic field at the centre of the loop is times.

A current is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.The magnetic field at the centre of the loop is times.

A part of a long wire carrying a current is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

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A particle is projected vertically upwards with velocity .30 ms^-1 Find the ratio of average speed and instantaneous velocity after 6s.[g=10 ms^-1 ]

3
$\frac{1}{2}$
4
2

Answer:The correct answer is: $1 half$

In the adjacent figure is shown a closed path $P.$ A long straight conductor carrying a current $I$ passes through $O$ and perpendicular to the plane of the paper. Then which of the following holds good?

In the adjacent figure is shown a closed path $P.$ A long straight conductor carrying a current $I$ passes through $O$ and perpendicular to the plane of the paper. Then which of the following holds good?

A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

The resultant magnetic field at the centre of the coil is

A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

The resultant magnetic field at the centre of the coil is

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the $X minus Y$ plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

900553

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the $X minus Y$ plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

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The motion of a particle along a straight line is described by the function .X=(3T-2)² Calculate the acceleration after 10 s.

The motion of a particle along a straight line is described by the function .X=(3T-2)² Calculate the acceleration after 10 s.

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Rohit completes a semi circular path of radius R in 10 seconds. Calculate average speed and average velocity in ms^-1.

If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient $eta comma$ the Poisson’s ratio $sigma$ will be

If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient $eta comma$ the Poisson’s ratio $sigma$ will be

Given figure shows a graph at acceleration $not stretchy rightwards arrow$ time for a rectilinear motion. Find average acceleration in first 10 seconds

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A part of a long wire carrying a current $i$ is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

A part of a long wire carrying a current $i$ is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is