Physics-
General
Easy

Question

As shown in the figure a particle starts its motion from 0 to A. And then it moves from. A to B. stack A B with minus on top is an are find the Path length

  1. 2r    
  2. r plus fraction numerator pi over denominator 3 end fraction    
  3. r open parentheses 1 plus fraction numerator pi over denominator 3 end fraction close parentheses    
  4. fraction numerator pi over denominator 3 end fraction left parenthesis r plus 1 right parenthesis    

The correct answer is: r open parentheses 1 plus fraction numerator pi over denominator 3 end fraction close parentheses

Related Questions to study

General
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A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and displacement.

A particle moves from A to P and then it moves from P to B as shown in the figure. Find path length and displacement.

physics-General
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A particle moves from A to B and then it moves from B to C as shown in figure. Calculate the ratio between path length and displacement.

A particle moves from A to B and then it moves from B to C as shown in figure. Calculate the ratio between path length and displacement.

physics-General
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As shown in the figure a particle moves from. 0 to A, and then A to B. Find pathlength and displacement.

As shown in the figure a particle moves from. 0 to A, and then A to B. Find pathlength and displacement.

physics-General
parallel
General
Maths-

The vectors a with not stretchy bar on top comma b with not stretchy bar on top comma c with not stretchy bar on top are equal in length and pair wise make equal angles. If a with not stretchy bar on top equals l with not stretchy bar on top plus ȷ with not stretchy bar on top comma b with not stretchy bar on top equals ȷ with not stretchy bar on top plus k with not stretchy bar on top, then c with not stretchy bar on top equals

The given vectors are a with rightwards arrow on top comma b with rightwards arrow on top a n d space c with rightwards arrow on top.
a with rightwards arrow on top equals i with hat on top plus j with hat on top space
b with rightwards arrow on top equals j with hat on top plus k with hat on top
L e t space c with space rightwards arrow on top equals x i with hat on top plus y j with hat on top space plus z k with hat on top
All the three vectors are of equal lengths.
open vertical bar a with rightwards arrow on top open vertical bar equals close vertical bar b with rightwards arrow on top open vertical bar equals close vertical bar c with rightwards arrow on top close vertical bar        ...(1)
open vertical bar a with rightwards arrow on top close vertical bar equals square root of 1 plus 1 end root
space space space space space space equals square root of 2
So, the length of all the vectors will be √2
Let the angle between each pair from the given vectors be A.
Let's write the dot product of each pair.
a with rightwards arrow on top. b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos A
open parentheses i with hat on top plus j with hat on top close parentheses. open parentheses j with hat on top plus k with hat on top close parentheses equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
space 0 space plus space 1 space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
R e a r r a n g e
open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space equals space 1 space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
The next pair will be
b with rightwards arrow on top. c with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
open parentheses j with hat on top plus k with hat on top close parentheses times open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
y space plus space z space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space space space space space space space space... left parenthesis space u sin g space 1 right parenthesis
y space plus space z space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis space space space space space space space space space space

c with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses. open parentheses i with hat on top plus j with hat on top close parentheses equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
x space plus space y space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space... left parenthesis u sin g space 1 right parenthesis
x space plus space y space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis
Now we will solve the equation
y + z = 1       ...(3)
x + y = 1      ...(4)
Subtracting  (4) from (3)
z - x = 0
x = z
Now, we know that length of all vectors is √2
open vertical bar c with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root
square root of 2 space equals square root of x squared plus y squared plus space z squared end root
space S q u a r e space b o t h space t h e space s i d e s space a n d space r e a r r a n g e
x squared space plus space y squared space plus space z squared space equals space 2
w e space k n o w space x space equals space z
2 z squared space plus space y squared space equals space 2
We will rearrange equation (3) and substitute in the above equation.
y + z = 1
y = 1 - z
2 z squared plus y squared equals 2
2 z squared plus left parenthesis 1 space minus space z right parenthesis squared space equals space 2 2 z squared space plus space 1 space minus space 2 z space plus space z squared space equals space 2
3 z squared space minus 2 z space minus 1 space equals space 0
F a c t o r i s e
3 z squared space minus 3 z space plus space z space minus space 1 space equals space 0
3 z left parenthesis z space minus space 1 right parenthesis space plus space 1 left parenthesis z space minus space 1 right parenthesis space equals space 0
left parenthesis 3 z space plus space 1 right parenthesis left parenthesis z space minus space 1 right parenthesis space equals space 0
z space equals space minus 1 third space o r space z space equals space 1
If we see the options, we don't have the value of z component as fraction. So, we will just focus on the value of z = 1
y = 1 - z
y = 1 - 1
y = 0
x + y = 1
x + 0 = 1
x = 1
c with rightwards arrow on top equals i with hat on top plus 0 j with overparenthesis on top plus k with hat on top
The solution is (1,0,1).

The vectors a with not stretchy bar on top comma b with not stretchy bar on top comma c with not stretchy bar on top are equal in length and pair wise make equal angles. If a with not stretchy bar on top equals l with not stretchy bar on top plus ȷ with not stretchy bar on top comma b with not stretchy bar on top equals ȷ with not stretchy bar on top plus k with not stretchy bar on top, then c with not stretchy bar on top equals

Maths-General
The given vectors are a with rightwards arrow on top comma b with rightwards arrow on top a n d space c with rightwards arrow on top.
a with rightwards arrow on top equals i with hat on top plus j with hat on top space
b with rightwards arrow on top equals j with hat on top plus k with hat on top
L e t space c with space rightwards arrow on top equals x i with hat on top plus y j with hat on top space plus z k with hat on top
All the three vectors are of equal lengths.
open vertical bar a with rightwards arrow on top open vertical bar equals close vertical bar b with rightwards arrow on top open vertical bar equals close vertical bar c with rightwards arrow on top close vertical bar        ...(1)
open vertical bar a with rightwards arrow on top close vertical bar equals square root of 1 plus 1 end root
space space space space space space equals square root of 2
So, the length of all the vectors will be √2
Let the angle between each pair from the given vectors be A.
Let's write the dot product of each pair.
a with rightwards arrow on top. b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos A
open parentheses i with hat on top plus j with hat on top close parentheses. open parentheses j with hat on top plus k with hat on top close parentheses equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
space 0 space plus space 1 space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
R e a r r a n g e
open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space equals space 1 space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
The next pair will be
b with rightwards arrow on top. c with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
open parentheses j with hat on top plus k with hat on top close parentheses times open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
y space plus space z space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space space space space space space space space... left parenthesis space u sin g space 1 right parenthesis
y space plus space z space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis space space space space space space space space space space

c with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses. open parentheses i with hat on top plus j with hat on top close parentheses equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
x space plus space y space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space... left parenthesis u sin g space 1 right parenthesis
x space plus space y space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis
Now we will solve the equation
y + z = 1       ...(3)
x + y = 1      ...(4)
Subtracting  (4) from (3)
z - x = 0
x = z
Now, we know that length of all vectors is √2
open vertical bar c with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root
square root of 2 space equals square root of x squared plus y squared plus space z squared end root
space S q u a r e space b o t h space t h e space s i d e s space a n d space r e a r r a n g e
x squared space plus space y squared space plus space z squared space equals space 2
w e space k n o w space x space equals space z
2 z squared space plus space y squared space equals space 2
We will rearrange equation (3) and substitute in the above equation.
y + z = 1
y = 1 - z
2 z squared plus y squared equals 2
2 z squared plus left parenthesis 1 space minus space z right parenthesis squared space equals space 2 2 z squared space plus space 1 space minus space 2 z space plus space z squared space equals space 2
3 z squared space minus 2 z space minus 1 space equals space 0
F a c t o r i s e
3 z squared space minus 3 z space plus space z space minus space 1 space equals space 0
3 z left parenthesis z space minus space 1 right parenthesis space plus space 1 left parenthesis z space minus space 1 right parenthesis space equals space 0
left parenthesis 3 z space plus space 1 right parenthesis left parenthesis z space minus space 1 right parenthesis space equals space 0
z space equals space minus 1 third space o r space z space equals space 1
If we see the options, we don't have the value of z component as fraction. So, we will just focus on the value of z = 1
y = 1 - z
y = 1 - 1
y = 0
x + y = 1
x + 0 = 1
x = 1
c with rightwards arrow on top equals i with hat on top plus 0 j with overparenthesis on top plus k with hat on top
The solution is (1,0,1).
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As shown in figure a block of mass m is aftachod with a cart. If the coefficient of static friction between the surface of cart and block is mu blankthan what would be accelcration alpha of cart to prevent the falling of block ?

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physics-General
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Physics-

An impulsive force of 100 N acts on a body for 1 sec What is the change in its linear momentum ?

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Physics-General
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maths-

If the vectors a with not stretchy bar on top equals open parentheses c log subscript 2 space x close parentheses ı with not stretchy bar on top minus 6 ı with not stretchy bar on top plus 3 k with not stretchy bar on top and b with not stretchy bar on top equals open parentheses log subscript 2 space x close parentheses ı with not stretchy bar on top plus 2 ȷ with not stretchy bar on top plus open parentheses 2 log subscript 2 space x close parentheses k with not stretchy bar on top make  an abtuse angle for any x element of left parenthesis 0 comma straight infinity right parenthesis, then C belangs to

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maths-General
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physics

A geostationary satellite is orbiting the earth at a height of 5R above that of surface of the earth. R being the radius of the earth. The time period of another satellite in hours at a height of  2R from the surface of earth is hr.

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physicsGeneral
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The most common minerals of phosphorus are

Phosphorus minerals is called as hydroxy apatite and fluorapatite

The most common minerals of phosphorus are

chemistry-General
Phosphorus minerals is called as hydroxy apatite and fluorapatite
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physics

The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as (G = Gravitational constant)

The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as (G = Gravitational constant)

physicsGeneral
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physics-

A mass of 6 kg is suspended by a rope of lengih 2 m form the ceiling . A force of 50 N in the horizonlal direction is applied at the mid Point P of the rope as shown in figure. what is the angle the rope makes. with the vertical in equilibrium ? open parentheses g equals 10 text end text m s to the power of 2 end exponent close parentheses Neglect mass of the rope:

A mass of 6 kg is suspended by a rope of lengih 2 m form the ceiling . A force of 50 N in the horizonlal direction is applied at the mid Point P of the rope as shown in figure. what is the angle the rope makes. with the vertical in equilibrium ? open parentheses g equals 10 text end text m s to the power of 2 end exponent close parentheses Neglect mass of the rope:

physics-General
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Two bodies A and B cach of mass m are fixed together by a massless spring A force F acts on the mass B as shown in figure. At the instant shown, a body A has an accelcration a. what is the acceleration of B?

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A block of mass m=4 kg is placed over a rough inclined plane as shown in figure, The cocfficient of friction between the block and plane is s=0.6.A force F=10 N is applied on the block of an angle at 30° The contact force between the block and the plane is

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physics-General
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