Chemistry-
General
Easy

Question

Compounds (A) and (B) are – 

  1. N a C l O subscript 3 end subscript comma N a C l O    
  2. N a O C l subscript 2 end subscript comma N a O C l    
  3. N a C l O subscript 4 end subscript comma N a C l O subscript 3 end subscript    
  4. N a O C l comma N a C l O subscript 3 end subscript    

The correct answer is: N a O C l comma N a C l O subscript 3 end subscript

Book A Free Demo

+91

Grade*

Related Questions to study

General
Maths-

2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

Maths-General
General
maths-

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

maths-General
General
Maths-

If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

Maths-General
General
Maths-

If the quadratic equation a x squared plus 2 c x plus b equals 0 and a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis have a common root then a plus 4 b plus 4 c is equal to

If the quadratic equation a x squared plus 2 c x plus b equals 0 and a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis have a common root then a plus 4 b plus 4 c is equal to

Maths-General
General
physics-

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14m s to the power of negative 1 end exponent to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14m s to the power of negative 1 end exponent to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

physics-General
At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent
General
Maths-

In a Δabc if b+c=3a then cot invisible function application straight B over 2 times cot invisible function application straight C over 2 has the value equal to –

In a Δabc if b+c=3a then cot invisible function application straight B over 2 times cot invisible function application straight C over 2 has the value equal to –

Maths-General
General
Maths-

In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

Maths-General
General
Maths-

In a triangle ABC, a: b: c = 4: 5: 6. Then 3A + B equals to :

In a triangle ABC, a: b: c = 4: 5: 6. Then 3A + B equals to :

Maths-General
General
physics-

A bullet of mass m is fired with a velocity of 50 m s to the power of negative 1 end exponent at an angle theta with the horizontal. At the highest point of its trajectory, it collides had on with a bob of massless string of length l equals 10 divided by 3m and gets embedded in the bob. After the collision, the string moves to an angle of 120degree. What is the angle theta ?

Velocity of bullet at highest point of its trajectory = 50 cos invisible function application theta in horizontal direction.
As bullet of mass m collides with pendulum bob of mass 3m and two stick together, their common velocity
v to the power of ´ end exponent equals fraction numerator m subscript 1 end subscript 50 cos invisible function application theta over denominator m plus 3 n end fraction equals fraction numerator 25 over denominator 2 end fraction cos invisible function application theta m s to the power of negative 1 end exponent
As now under this velocity v ´ pendulum bob goes up to an angel 120degree, hence
fraction numerator v to the power of ´ 2 end exponent over denominator 2 g end fraction equals h equals l open parentheses 1 minus cos invisible function application 120 degree close parentheses equals fraction numerator 10 over denominator 3 end fraction open square brackets 1 minus open parentheses f minus fraction numerator 1 over denominator 2 end fraction close parentheses close square brackets equals 5
rightwards double arrow blank v to the power of ´ 2 end exponent equals 2 cross times 10 cross times 5 equals 100 or v to the power of ´ end exponent equals 10
Comparing two answer of v ´, we get
fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses

A bullet of mass m is fired with a velocity of 50 m s to the power of negative 1 end exponent at an angle theta with the horizontal. At the highest point of its trajectory, it collides had on with a bob of massless string of length l equals 10 divided by 3m and gets embedded in the bob. After the collision, the string moves to an angle of 120degree. What is the angle theta ?

physics-General
Velocity of bullet at highest point of its trajectory = 50 cos invisible function application theta in horizontal direction.
As bullet of mass m collides with pendulum bob of mass 3m and two stick together, their common velocity
v to the power of ´ end exponent equals fraction numerator m subscript 1 end subscript 50 cos invisible function application theta over denominator m plus 3 n end fraction equals fraction numerator 25 over denominator 2 end fraction cos invisible function application theta m s to the power of negative 1 end exponent
As now under this velocity v ´ pendulum bob goes up to an angel 120degree, hence
fraction numerator v to the power of ´ 2 end exponent over denominator 2 g end fraction equals h equals l open parentheses 1 minus cos invisible function application 120 degree close parentheses equals fraction numerator 10 over denominator 3 end fraction open square brackets 1 minus open parentheses f minus fraction numerator 1 over denominator 2 end fraction close parentheses close square brackets equals 5
rightwards double arrow blank v to the power of ´ 2 end exponent equals 2 cross times 10 cross times 5 equals 100 or v to the power of ´ end exponent equals 10
Comparing two answer of v ´, we get
fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses
General
physics-

A spherical hollow is made in a lead sphere of radius R such that its surface touches the outside surface of lead sphere and passes through the centre. What is the shift in the centre of lead sphere as a result of this hollowing?

Let centre of mass of lead sphere after hollowing be at point O subscript 2 end subscript comma where O O subscript 2 end subscript equals x
Mass of spherical hollow m equals fraction numerator fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator R over denominator 2 end fraction close parentheses to the power of 2 end exponent M over denominator open parentheses fraction numerator 4 over denominator 2 end fraction pi R to the power of 3 end exponent close parentheses end fraction equals fraction numerator M over denominator 8 end fraction and
x equals O O subscript 1 end subscript equals fraction numerator R over denominator 2 end fraction

therefore blank x equals fraction numerator M cross times 0 minus open parentheses fraction numerator M over denominator 8 end fraction close parentheses cross times fraction numerator R over denominator 2 end fraction over denominator M minus fraction numerator M over denominator 8 end fraction end fraction equals fraction numerator fraction numerator M R over denominator 16 end fraction over denominator fraction numerator 7 M over denominator 8 end fraction end fraction equals negative fraction numerator R over denominator 14 end fraction
therefore shift equals fraction numerator R over denominator 14 end fraction

A spherical hollow is made in a lead sphere of radius R such that its surface touches the outside surface of lead sphere and passes through the centre. What is the shift in the centre of lead sphere as a result of this hollowing?

physics-General
Let centre of mass of lead sphere after hollowing be at point O subscript 2 end subscript comma where O O subscript 2 end subscript equals x
Mass of spherical hollow m equals fraction numerator fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator R over denominator 2 end fraction close parentheses to the power of 2 end exponent M over denominator open parentheses fraction numerator 4 over denominator 2 end fraction pi R to the power of 3 end exponent close parentheses end fraction equals fraction numerator M over denominator 8 end fraction and
x equals O O subscript 1 end subscript equals fraction numerator R over denominator 2 end fraction

therefore blank x equals fraction numerator M cross times 0 minus open parentheses fraction numerator M over denominator 8 end fraction close parentheses cross times fraction numerator R over denominator 2 end fraction over denominator M minus fraction numerator M over denominator 8 end fraction end fraction equals fraction numerator fraction numerator M R over denominator 16 end fraction over denominator fraction numerator 7 M over denominator 8 end fraction end fraction equals negative fraction numerator R over denominator 14 end fraction
therefore shift equals fraction numerator R over denominator 14 end fraction
General
chemistry-

On heating a mixture of N H subscript 4 end subscript C lans K N O subscript 2 end subscript, we get –

On heating a mixture of N H subscript 4 end subscript C lans K N O subscript 2 end subscript, we get –

chemistry-General
General
physics-

Three identical blocks A comma B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m s to the power of negative 1 end exponent. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is

For collision between blocks A and B,
e equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator u subscript A end subscript minus u subscript B end subscript end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 minus 0 end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 end fraction
therefore v subscript B end subscript minus v subscript A end subscript equals 10 e equals 10 cross times 0.5 equals 5 blank horizontal ellipsis. left parenthesis i right parenthesis
from principle of momentum conservation,
m subscript A end subscript u subscript A end subscript plus m subscript B end subscript u subscript B end subscript equals m subscript A end subscript v subscript A end subscript plus m subscript B end subscript v subscript B end subscript
Or m cross times 10 plus 0 equals m v subscript A end subscript plus m v subscript B end subscript
therefore blank v subscript A end subscript plus v subscript B end subscript equals 10 blank horizontal ellipsis. left parenthesis i i right parenthesis
Adding Eqs. (i) and (ii), we get
v subscript B end subscript equals 7.5 blank m s to the power of negative 1 end exponent blank horizontal ellipsis left parenthesis i i i right parenthesis
Similarly for collision between B and C
v subscript C end subscript minus v subscript B end subscript equals 7.5 e equals 7.5 cross times 0.5 equals 3.75
therefore blank v subscript C end subscript minus v subscript B end subscript equals 3.75 m s to the power of negative 1 end exponent …(iv)
Adding Eqs. (iii) and (iv) we get
2 v subscript C end subscript equals 11.25
therefore blank v subscript C end subscript equals fraction numerator 11.25 over denominator 2 end fraction equals 5.6 m s to the power of negative 1 end exponent

Three identical blocks A comma B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m s to the power of negative 1 end exponent. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is

physics-General
For collision between blocks A and B,
e equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator u subscript A end subscript minus u subscript B end subscript end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 minus 0 end fraction equals fraction numerator v subscript B end subscript minus v subscript A end subscript over denominator 10 end fraction
therefore v subscript B end subscript minus v subscript A end subscript equals 10 e equals 10 cross times 0.5 equals 5 blank horizontal ellipsis. left parenthesis i right parenthesis
from principle of momentum conservation,
m subscript A end subscript u subscript A end subscript plus m subscript B end subscript u subscript B end subscript equals m subscript A end subscript v subscript A end subscript plus m subscript B end subscript v subscript B end subscript
Or m cross times 10 plus 0 equals m v subscript A end subscript plus m v subscript B end subscript
therefore blank v subscript A end subscript plus v subscript B end subscript equals 10 blank horizontal ellipsis. left parenthesis i i right parenthesis
Adding Eqs. (i) and (ii), we get
v subscript B end subscript equals 7.5 blank m s to the power of negative 1 end exponent blank horizontal ellipsis left parenthesis i i i right parenthesis
Similarly for collision between B and C
v subscript C end subscript minus v subscript B end subscript equals 7.5 e equals 7.5 cross times 0.5 equals 3.75
therefore blank v subscript C end subscript minus v subscript B end subscript equals 3.75 m s to the power of negative 1 end exponent …(iv)
Adding Eqs. (iii) and (iv) we get
2 v subscript C end subscript equals 11.25
therefore blank v subscript C end subscript equals fraction numerator 11.25 over denominator 2 end fraction equals 5.6 m s to the power of negative 1 end exponent
General
physics-

Two identical masses A and B are hanging stationary by a light pulley (shown in the figure). A shell C moving upwards with velocity v collides with the block B and gets stick to it. Then

When C collides with B then due to impulsive force, combined mass (B plus C) starts to move upward. Consequently the string becomes slack

Two identical masses A and B are hanging stationary by a light pulley (shown in the figure). A shell C moving upwards with velocity v collides with the block B and gets stick to it. Then

physics-General
When C collides with B then due to impulsive force, combined mass (B plus C) starts to move upward. Consequently the string becomes slack
General
physics-

In the given figure four identical spheres of equal mass m are suspended by wires of equal length l subscript 0 end subscript, so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is

When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l subscript 0 end subscript equals kinetic energy at bottom
Or m g l subscript 0 end subscript equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
Or v equals square root of 2 g l subscript 0 end subscript end root
Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to square root of 2 g l subscript 0 end subscript end root

In the given figure four identical spheres of equal mass m are suspended by wires of equal length l subscript 0 end subscript, so that all spheres are almost touching to each other. If the sphere 1 is released from the horizontal position and all collisions are elastic, the velocity of sphere 4 just after collision is

physics-General
When the sphere 1 is released from horizontal position, then from energy conservation, potential energy at height l subscript 0 end subscript equals kinetic energy at bottom
Or m g l subscript 0 end subscript equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
Or v equals square root of 2 g l subscript 0 end subscript end root
Since, all collisions are elastic, so velocity of sphere 1 is transferred to sphere 2, then from 2 to 3 and finally from 3 to 4. Hence, just after collision, the sphere 4 attains a velocity equal to square root of 2 g l subscript 0 end subscript end root
General
physics-

Two negatively charges particles having charges e subscript 1 end subscript and e subscript 2 end subscript and masses m subscript 1 end subscript and m subscript 2 end subscript respectively are projected one after another into a region with equal initial velocity. The electric field E is along the y-axis, while the direction of projection makes an angle a with the y-axis. If the ranges of the two particles along the x-axis are equal then one can conclude that

Here, effected gravitational acceleration is
g to the power of ´ end exponent equals fraction numerator m g minus q E over denominator m end fraction
therefore blank R equals fraction numerator v subscript 0 end subscript superscript 2 end superscript sin invisible function application 2 alpha over denominator g ´ end fraction
It means, g ’ for both particles are same
This is possible when
m subscript 1 end subscript equals m subscript 2 end subscript and e subscript 1 end subscript equals e subscript 2 end subscript

Two negatively charges particles having charges e subscript 1 end subscript and e subscript 2 end subscript and masses m subscript 1 end subscript and m subscript 2 end subscript respectively are projected one after another into a region with equal initial velocity. The electric field E is along the y-axis, while the direction of projection makes an angle a with the y-axis. If the ranges of the two particles along the x-axis are equal then one can conclude that

physics-General
Here, effected gravitational acceleration is
g to the power of ´ end exponent equals fraction numerator m g minus q E over denominator m end fraction
therefore blank R equals fraction numerator v subscript 0 end subscript superscript 2 end superscript sin invisible function application 2 alpha over denominator g ´ end fraction
It means, g ’ for both particles are same
This is possible when
m subscript 1 end subscript equals m subscript 2 end subscript and e subscript 1 end subscript equals e subscript 2 end subscript