Maths-
General
Easy

Question

The number of numbers between 1 and 1010 which contain the digit 1 is -

  1. 1010 – 910 – 1    
  2. 910    
  3. 1010 – 810    
  4. None of these.    

The correct answer is: 1010 – 910 – 1

Book A Free Demo

+91

Grade*

Related Questions to study

General
maths-

The number of rectangles in the adjoining figure is –

The number of rectangles in the adjoining figure is –

maths-General
General
chemistry-

Assertion :this equilibrium favours backward direction.
Reason :is stronger base than C H subscript 2 end subscript C O O to the power of minus end exponent

Assertion :this equilibrium favours backward direction.
Reason :is stronger base than C H subscript 2 end subscript C O O to the power of minus end exponent

chemistry-General
General
physics-

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v blank a n d blank 2 v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A blank,these two particles will again reach The point A ?

A s first collision one particle having speed 2v will rotate
240 degree open parentheses o r fraction numerator 4 pi over denominator 3 end fraction close parentheseswhile other particle having speed v blankwill rotate
120 degree open parentheses o r fraction numerator 2 pi over denominator 3 end fraction close parentheses. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point A.

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v blank a n d blank 2 v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A blank,these two particles will again reach The point A ?

physics-General
A s first collision one particle having speed 2v will rotate
240 degree open parentheses o r fraction numerator 4 pi over denominator 3 end fraction close parentheseswhile other particle having speed v blankwill rotate
120 degree open parentheses o r fraction numerator 2 pi over denominator 3 end fraction close parentheses. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point A.
General
physics-

The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are open parentheses 0 , 2 close parentheses. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are open parentheses 0 , 2 close parentheses. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

physics-General
General
physics-

The potential energy of a particle varies with distance x as shown in the graph.
The force acting on the particle is zero at

F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
therefore F equals 0 for point B and C

The potential energy of a particle varies with distance x as shown in the graph.
The force acting on the particle is zero at

physics-General
F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
therefore F equals 0 for point B and C
General
physics-

A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

Work done = Area covered in between force displacement curve and displacement axis
= Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J

A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

physics-General
Work done = Area covered in between force displacement curve and displacement axis
= Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J
General
physics-

A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

By applying law of conservation of energy
m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root

A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

physics-General
By applying law of conservation of energy
m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root
General
physics-

Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

The resultant of 5 N along O C and 5 N along O A is
R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
equals square root of 72N along O B
The resultant of square root of 72 N along O B and square root of 72 N along O G is
R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.

Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

physics-General
The resultant of 5 N along O C and 5 N along O A is
R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
equals square root of 72N along O B
The resultant of square root of 72 N along O B and square root of 72 N along O G is
R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.
General
physics-

A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

Range of the projectile on an inclined plane (down the plane) is,
R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction

A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

physics-General
Range of the projectile on an inclined plane (down the plane) is,
R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction
General
physics-

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
According to work energy theorem
K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

physics-General
Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
According to work energy theorem
K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]
General
physics-

A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction

A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

physics-General
From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction
General
physics-

An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
Horizontal distance A B equals v t subscript O B end subscript
equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
equals 3333.33 blank m equals 3.33 blank k m

An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

physics-General
From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
Horizontal distance A B equals v t subscript O B end subscript
equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
equals 3333.33 blank m equals 3.33 blank k m
General
Maths-

If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

Maths-General
General
physics-

A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be


v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s

A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be

physics-General

v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s
General
physics-

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

Vertical height equals h equals l cos invisible function application 30 degree
Loss of potential energy equals m g h

equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

physics-General
Vertical height equals h equals l cos invisible function application 30 degree
Loss of potential energy equals m g h

equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l