Physics
General
Easy

Question

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

  1. 10ms-2
  2. 15ms-2
  3. 7.5ms-2
  4. 30ms-2

The correct answer is: 7.5ms-2

Book A Free Demo

+91

Grade*

Related Questions to study

General
physics

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.

physicsGeneral
General
physics-

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
General
biology

In angiosperms, female gametophyte is represented by

In angiosperms, female gametophyte is represented by

biologyGeneral
General
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 ]

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
General
Maths-

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

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

Maths-General
General
physics

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

physicsGeneral
General
physics-

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
or 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
or 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
General
physics

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

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

physicsGeneral
General
physics-

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-

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-

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 is 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 is 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 is
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 is
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
General
physics-

A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T

A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

physics-General
Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T