Physics
General
Easy

Question

What does the speedometer measure kept in motorbike ?

  1. Average Velocity
  2. Average speed
  3. instantaneous speed
  4. instantaneous Velocity

The correct answer is: instantaneous speed

Related Questions to study

General
physics-

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle A O C equals 60 degree. If B subscript 1 and B subscript 2 are the magnitudes of the magnetic fields at O due to the currents in  ABC and ADC respectively, then ratio fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction is

From Biot-Savart law the magnetic field at the centre is directly proportional to the length of current carrying segment.
therefore fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction equals fraction numerator l e n g t h blank o f blank A B C over denominator l e n g t h blank o f blank A D C end fraction
equals fraction numerator a n g l e blank s u b t e n d e d blank b y blank A B C over denominator a n g l e blank s u b t e n d e d blank b y blank A D C end fraction
equals fraction numerator left parenthesis 360 degree minus 60 degree right parenthesis over denominator 60 degree end fraction equals fraction numerator 300 over denominator 60 end fraction equals fraction numerator 5 over denominator 1 end fraction

A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle A O C equals 60 degree. If B subscript 1 and B subscript 2 are the magnitudes of the magnetic fields at O due to the currents in  ABC and ADC respectively, then ratio fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction is

physics-General
From Biot-Savart law the magnetic field at the centre is directly proportional to the length of current carrying segment.
therefore fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction equals fraction numerator l e n g t h blank o f blank A B C over denominator l e n g t h blank o f blank A D C end fraction
equals fraction numerator a n g l e blank s u b t e n d e d blank b y blank A B C over denominator a n g l e blank s u b t e n d e d blank b y blank A D C end fraction
equals fraction numerator left parenthesis 360 degree minus 60 degree right parenthesis over denominator 60 degree end fraction equals fraction numerator 300 over denominator 60 end fraction equals fraction numerator 5 over denominator 1 end fraction
General
physics-

Current I is flowing in conductor shaped as shown in the figure. The radius of the curved part is r and the length of straight portion is very large. The value of the magnetic field at the centre O will be

B subscript A end subscript equals 0

B subscript B end subscript equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis I over denominator r end fraction blank circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction
B subscript C end subscript equals fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction circled times
So, net magnetic field at the centre
equals B subscript A end subscript plus B subscript B end subscript plus blank B subscript C end subscript
equals 0 plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction plus fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses

Current I is flowing in conductor shaped as shown in the figure. The radius of the curved part is r and the length of straight portion is very large. The value of the magnetic field at the centre O will be

physics-General
B subscript A end subscript equals 0

B subscript B end subscript equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis I over denominator r end fraction blank circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction
B subscript C end subscript equals fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction circled times
So, net magnetic field at the centre
equals B subscript A end subscript plus B subscript B end subscript plus blank B subscript C end subscript
equals 0 plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction plus fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses
General
physics

A body is moving in x direction with constant acceleration a Find the difference of the displacement covered by it in nth second and (n-1) th second.

A body is moving in x direction with constant acceleration a Find the difference of the displacement covered by it in nth second and (n-1) th second.

physicsGeneral
parallel
General
physics

A body starts its motion with zero velocity and its acceleration is 3m/s2. Find the distance travelled by it in fifth second.

A body starts its motion with zero velocity and its acceleration is 3m/s2. Find the distance travelled by it in fifth second.

physicsGeneral
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

In 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
In 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

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

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

physicsGeneral
parallel
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
parallel
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

w e space h a v e comma space
f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator x over denominator 1 plus x end fraction comma space i f space x greater or equal than 0 end cell row cell fraction numerator x over denominator 1 minus x end fraction comma space i f space x less or equal than 0 end cell end table close
C a s e space i comma space w h e n space x greater or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 plus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 plus x subscript 2 end fraction rightwards double arrow x subscript 1 plus x subscript 1 x subscript 2 equals x subscript 2 plus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 plus x end fraction rightwards double arrow y plus x y equals x rightwards double arrow x left parenthesis 1 minus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 minus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 minus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 minus y end fraction over denominator 1 plus fraction numerator y over denominator 1 minus y end fraction end fraction equals fraction numerator y over denominator 1 minus y plus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

C a s e space i comma space w h e n space x less or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 minus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 minus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 minus x subscript 2 end fraction rightwards double arrow x subscript 1 minus x subscript 1 x subscript 2 equals x subscript 2 minus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 minus x end fraction rightwards double arrow y minus x y equals x rightwards double arrow x left parenthesis 1 plus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 plus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 plus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 plus y end fraction over denominator 1 minus fraction numerator y over denominator 1 plus y end fraction end fraction equals fraction numerator y over denominator 1 plus y minus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

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
w e space h a v e comma space
f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator x over denominator 1 plus x end fraction comma space i f space x greater or equal than 0 end cell row cell fraction numerator x over denominator 1 minus x end fraction comma space i f space x less or equal than 0 end cell end table close
C a s e space i comma space w h e n space x greater or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 plus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 plus x subscript 2 end fraction rightwards double arrow x subscript 1 plus x subscript 1 x subscript 2 equals x subscript 2 plus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 plus x end fraction rightwards double arrow y plus x y equals x rightwards double arrow x left parenthesis 1 minus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 minus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 minus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 minus y end fraction over denominator 1 plus fraction numerator y over denominator 1 minus y end fraction end fraction equals fraction numerator y over denominator 1 minus y plus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

C a s e space i comma space w h e n space x less or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 minus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 minus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 minus x subscript 2 end fraction rightwards double arrow x subscript 1 minus x subscript 1 x subscript 2 equals x subscript 2 minus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 minus x end fraction rightwards double arrow y minus x y equals x rightwards double arrow x left parenthesis 1 plus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 plus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 plus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 plus y end fraction over denominator 1 minus fraction numerator y over denominator 1 plus y end fraction end fraction equals fraction numerator y over denominator 1 plus y minus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.
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
parallel
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
parallel

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