Maths-
General
Easy

Question

If sin space 2 theta minus cos space 3 theta equals 0 and theta is acute then theta

  1. 36 to the power of ring operator
  2. 54 to the power of ring operator
  3. 18 to the power of ring operator
  4. 30 to the power of ring operator

The correct answer is: 18 to the power of ring operator

Related Questions to study

General
maths-

If sin space open parentheses x plus 28 to the power of ring operator close parentheses equals cos space open parentheses 3 x minus 78 to the power of ring operator close parentheses then x=

If sin space open parentheses x plus 28 to the power of ring operator close parentheses equals cos space open parentheses 3 x minus 78 to the power of ring operator close parentheses then x=

maths-General
General
physics-

The adjoining figure shows a force of 40 N pulling a body of mass 5 kg in a direction  above the horizontal. The body is in rest on a smooth horizontal surface. Assuming acceleration of free-fall is 10 . Which of the following statements I and II is/are correct?

I) The weight of the 5 kg mass acts vertically downwards

II) The net vertical force acting on the body is 30 N.

The adjoining figure shows a force of 40 N pulling a body of mass 5 kg in a direction  above the horizontal. The body is in rest on a smooth horizontal surface. Assuming acceleration of free-fall is 10 . Which of the following statements I and II is/are correct?

I) The weight of the 5 kg mass acts vertically downwards

II) The net vertical force acting on the body is 30 N.

physics-General
General
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A sphere of mass m is kept in equilibrium with the help of several springs as shown in the figure. Measurement shows that one of the springs applies a force   on the sphere. With what acceleration the sphere will move immediately after this particular spring is cut?

A sphere of mass m is kept in equilibrium with the help of several springs as shown in the figure. Measurement shows that one of the springs applies a force   on the sphere. With what acceleration the sphere will move immediately after this particular spring is cut?

physics-General
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If you want to pile up sand onto a circular area of radius R. The greatest height of the sand pile that can be erected without spilling the sand onto the surrounding area, if  is the coefficient of friction between sand particle is

If you want to pile up sand onto a circular area of radius R. The greatest height of the sand pile that can be erected without spilling the sand onto the surrounding area, if  is the coefficient of friction between sand particle is

physics-General
General
physics-

Two forces are simultaneously applied on an object. What third force would make the net force to point to the left (–x direction) ?

Two forces are simultaneously applied on an object. What third force would make the net force to point to the left (–x direction) ?

physics-General
General
physics-

A car is going at a speed of 6 m/s when it encounters a 15 m slope of angle  . The friction coefficient between the road and tyre is 0.5. The driver applies the brakes. The minimum speed of car with which it can reach the bottom is open parentheses straight g equals 10 straight m divided by straight s squared close parentheses

A car is going at a speed of 6 m/s when it encounters a 15 m slope of angle  . The friction coefficient between the road and tyre is 0.5. The driver applies the brakes. The minimum speed of car with which it can reach the bottom is open parentheses straight g equals 10 straight m divided by straight s squared close parentheses

physics-General
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General
physics-

A block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in figure. The coefficient of friction between the block and the plane is µ = 0.5. What minimum force F should be applied perpendicular to the plane on the block, so that block does not slip on the plane (g=10m/s to the power of 2 end exponent )

A block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in figure. The coefficient of friction between the block and the plane is µ = 0.5. What minimum force F should be applied perpendicular to the plane on the block, so that block does not slip on the plane (g=10m/s to the power of 2 end exponent )

physics-General
General
physics-

A block of mass of 10 kg lies on a rough inclined plane of inclination theta equals s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses with the horizontal when a force of 30N is applied on the block parallel to and upward the plane, the total force exerted by the plane on the block is nearly along (coefficient of friction is µ = fraction numerator 3 over denominator 4 end fraction ) ( g = 10 m/S subscript equals end subscript superscript 2 end superscript )

A block of mass of 10 kg lies on a rough inclined plane of inclination theta equals s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses with the horizontal when a force of 30N is applied on the block parallel to and upward the plane, the total force exerted by the plane on the block is nearly along (coefficient of friction is µ = fraction numerator 3 over denominator 4 end fraction ) ( g = 10 m/S subscript equals end subscript superscript 2 end superscript )

physics-General
General
physics-

A block of mass 3 kg is at rest on a rough inclined plane as shown in the figure. The magnitude of net force exerted by the surface on the block will be (g=10 m/S subscript equals end subscript superscript 2 end superscript )

A block of mass 3 kg is at rest on a rough inclined plane as shown in the figure. The magnitude of net force exerted by the surface on the block will be (g=10 m/S subscript equals end subscript superscript 2 end superscript )

physics-General
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General
Maths-

If a with minus on top equals i with minus on top plus j with minus on top plus k with minus on top text  and  end text b with minus on top equals stack i with minus on top minus j with minus on top with blank on top then the vectors Error converting from MathML to accessible text. and Error converting from MathML to accessible text.

The vectors a with rightwards arrow on top a n d b with rightwards arrow on top are given as follows:
table attributes columnalign left end attributes row cell a with rightwards arrow on top equals i with hat on top plus j with hat on top plus k with hat on top end cell row cell b with rightwards arrow on top equals i with hat on top minus j with hat on top end cell end table
Let the other three vectors be denoted as P with rightwards arrow on top comma stack Q space with rightwards arrow on top a n d space R with rightwards arrow on top
P with rightwards arrow on top equals open parentheses a with rightwards arrow on top.1 with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. j with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. k with overparenthesis on top close parentheses k with hat on top
Q with rightwards arrow on top equals open parentheses b with rightwards arrow on top. i with hat on top close parentheses i with hat on top plus open parentheses b with rightwards arrow on top. j with hat on top close parentheses j with hat on top plus open parentheses b with rightwards arrow on top. k with hat on top close parentheses k with hat on top
R with rightwards arrow on top equals i with hat on top plus j with hat on top minus 2 k with overparenthesis on top
We have to find the relation between the vectors.
Now, we will first find the components of the vectors
a with rightwards arrow on top. i with overparenthesis on top equals left parenthesis i with hat on top plus i with hat on top plus k with hat on top right parenthesis open parentheses i with hat on top close parentheses equals 1
stack a. with rightwards arrow on top j with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space 1
a with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis k with hat on top equals 1
b with rightwards arrow on top. i with overparenthesis on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top close parentheses equals 1
b with rightwards arrow on top. j with hat on top space equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space minus 1
b with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis k with hat on top right parenthesis space equals space 0
So, the vectors become.
P with rightwards arrow on top equals i with hat on top plus j with hat on top plus k with hat on top
Q with rightwards arrow on top equals i with hat on top minus j with hat on top
Now, we will take their dot product to check if they are mutually perpendicular or not.
P with bar on top. Q with bar on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses. open parentheses i with hat on top minus j with hat on top close parentheses
space space space space space space space space equals space 1 space minus space 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c l u a r.
Q with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses
space space space space space space space space equals space 1 minus 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c u l a r

P with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses left parenthesis i with hat on top plus j with hat on top minus 2 k with hat on top right parenthesis
space space space space space space space space equals space 1 space plus space 1 space minus space 2
space space space space space space space space equals space 0

space space space space space space space space
So, the given vectors are perpendicular.
If we see all the vectors are perpendicular to each other.
The given vectors are mutually perpendicular.

If a with minus on top equals i with minus on top plus j with minus on top plus k with minus on top text  and  end text b with minus on top equals stack i with minus on top minus j with minus on top with blank on top then the vectors Error converting from MathML to accessible text. and Error converting from MathML to accessible text.

Maths-General
The vectors a with rightwards arrow on top a n d b with rightwards arrow on top are given as follows:
table attributes columnalign left end attributes row cell a with rightwards arrow on top equals i with hat on top plus j with hat on top plus k with hat on top end cell row cell b with rightwards arrow on top equals i with hat on top minus j with hat on top end cell end table
Let the other three vectors be denoted as P with rightwards arrow on top comma stack Q space with rightwards arrow on top a n d space R with rightwards arrow on top
P with rightwards arrow on top equals open parentheses a with rightwards arrow on top.1 with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. j with overparenthesis on top close parentheses i with overparenthesis on top plus open parentheses a with rightwards arrow on top. k with overparenthesis on top close parentheses k with hat on top
Q with rightwards arrow on top equals open parentheses b with rightwards arrow on top. i with hat on top close parentheses i with hat on top plus open parentheses b with rightwards arrow on top. j with hat on top close parentheses j with hat on top plus open parentheses b with rightwards arrow on top. k with hat on top close parentheses k with hat on top
R with rightwards arrow on top equals i with hat on top plus j with hat on top minus 2 k with overparenthesis on top
We have to find the relation between the vectors.
Now, we will first find the components of the vectors
a with rightwards arrow on top. i with overparenthesis on top equals left parenthesis i with hat on top plus i with hat on top plus k with hat on top right parenthesis open parentheses i with hat on top close parentheses equals 1
stack a. with rightwards arrow on top j with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space 1
a with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top plus j with hat on top plus k with hat on top right parenthesis k with hat on top equals 1
b with rightwards arrow on top. i with overparenthesis on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top close parentheses equals 1
b with rightwards arrow on top. j with hat on top space equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis j with hat on top right parenthesis space equals space minus 1
b with rightwards arrow on top. k with hat on top equals left parenthesis i with hat on top minus j with hat on top right parenthesis left parenthesis k with hat on top right parenthesis space equals space 0
So, the vectors become.
P with rightwards arrow on top equals i with hat on top plus j with hat on top plus k with hat on top
Q with rightwards arrow on top equals i with hat on top minus j with hat on top
Now, we will take their dot product to check if they are mutually perpendicular or not.
P with bar on top. Q with bar on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses. open parentheses i with hat on top minus j with hat on top close parentheses
space space space space space space space space equals space 1 space minus space 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c l u a r.
Q with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top minus j with hat on top close parentheses open parentheses i with hat on top plus j with hat on top minus 2 k with hat on top close parentheses
space space space space space space space space equals space 1 minus 1
space space space space space space space space space equals space 0
S o comma space t h e space g i v e n space v e c t o r s space a r e space p e r p e n d i c u l a r

P with rightwards arrow on top. R with rightwards arrow on top equals open parentheses i with hat on top plus j with hat on top plus k with hat on top close parentheses left parenthesis i with hat on top plus j with hat on top minus 2 k with hat on top right parenthesis
space space space space space space space space equals space 1 space plus space 1 space minus space 2
space space space space space space space space equals space 0

space space space space space space space space
So, the given vectors are perpendicular.
If we see all the vectors are perpendicular to each other.
The given vectors are mutually perpendicular.
General
physics-

In the shown situation, which of the following is/are possible ?

If F1 ¹ F2 , then system will move with acceleration so spring force ¹0
If F1 = 40 N & F2 = 60 N then a equals fraction numerator F subscript 2 end subscript minus F subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 20 over denominator 100 end fraction equals fraction numerator 1 over denominator 5 end fraction m divided by s to the power of 2 end exponentand spring force F1 + m1 a = 40 plus fraction numerator 1 over denominator 5 end fraction (60) = 52 N
If F1 = 60 N & F2 = 40 N then spring force = 52 N

In the shown situation, which of the following is/are possible ?

physics-General
If F1 ¹ F2 , then system will move with acceleration so spring force ¹0
If F1 = 40 N & F2 = 60 N then a equals fraction numerator F subscript 2 end subscript minus F subscript 1 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 20 over denominator 100 end fraction equals fraction numerator 1 over denominator 5 end fraction m divided by s to the power of 2 end exponentand spring force F1 + m1 a = 40 plus fraction numerator 1 over denominator 5 end fraction (60) = 52 N
If F1 = 60 N & F2 = 40 N then spring force = 52 N
General
physics-

A block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in figure. The coefficient of friction between the block and the plane is µ = 0.5. What minimum force F should be applied perpendicular to the plane on the block, so that block does not slip on the plane (g=10m/s to the power of 2 end exponent )

A block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in figure. The coefficient of friction between the block and the plane is µ = 0.5. What minimum force F should be applied perpendicular to the plane on the block, so that block does not slip on the plane (g=10m/s to the power of 2 end exponent )

physics-General
parallel
General
physics-

A block of mass of 10 kg lies on a rough inclined plane of inclination theta equals s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses with the horizontal when a force of 30N is applied on the block parallel to and upward the plane, the total force exerted by the plane on the block is nearly along (coefficient of friction is µ = fraction numerator 3 over denominator 4 end fraction ) ( g = 10 m/S subscript equals end subscript superscript 2 end superscript )

A block of mass of 10 kg lies on a rough inclined plane of inclination theta equals s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses with the horizontal when a force of 30N is applied on the block parallel to and upward the plane, the total force exerted by the plane on the block is nearly along (coefficient of friction is µ = fraction numerator 3 over denominator 4 end fraction ) ( g = 10 m/S subscript equals end subscript superscript 2 end superscript )

physics-General
General
physics-

A block of mass 3 kg is at rest on a rough inclined plane as shown in the figure. The magnitude of net force exerted by the surface on the block will be (g=10 m/S subscript equals end subscript superscript 2 end superscript )

A block of mass 3 kg is at rest on a rough inclined plane as shown in the figure. The magnitude of net force exerted by the surface on the block will be (g=10 m/S subscript equals end subscript superscript 2 end superscript )

physics-General
General
physics-

As shown in figure, the left block rests on a table at distance null = Normal reaction between A & B N subscript 2 end subscript = Normal reaction between B & C Which of the following statement(s) is/are correct ?

As shown in figure, the left block rests on a table at distance null = Normal reaction between A & B N subscript 2 end subscript = Normal reaction between B & C Which of the following statement(s) is/are correct ?

physics-General
parallel

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