Physics-
General
Easy

Question

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 )

  1. OA    
  2. OB    
  3. OC    
  4. OD    

The correct answer is: OB

Related Questions to study

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
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
parallel
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
parallel
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
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physics-

As shown in figure, the left block rests on a table at distance lambda  from the edge while the right block is kept at the same level so that thread is unstretched and does not sag and then released. What will happen first ?


As shown in figure, the left block rests on a table at distance lambda  from the edge while the right block is kept at the same level so that thread is unstretched and does not sag and then released. What will happen first ?


physics-General
General
physics-

If the string is pulled down with a force of 120 N as shown in the figure, then the acceleration of 8 kg block would

If the string is pulled down with a force of 120 N as shown in the figure, then the acceleration of 8 kg block would

physics-General
parallel
General
physics-

With what acceleration ‘a’ shown the elevator descends so that the block of mass M exerts a force of fraction numerator M g over denominator 10 end fraction blankon the weighing machine? [g = acceleration due to gravity]

With what acceleration ‘a’ shown the elevator descends so that the block of mass M exerts a force of fraction numerator M g over denominator 10 end fraction blankon the weighing machine? [g = acceleration due to gravity]

physics-General
General
physics-

An astronaut accidentally gets separated out of his small spaceship accelerating in inter-stellar space at a constant acceleration of 10 m/S to the power of 2 end exponent . What is the acceleration of the astronaut at the instant he is outside the spaceship ?

When the astronaut is outside the spaceship, the net external force (except negligible gravitational force due to spaceship) is zero as he is isolated from all interactions.

An astronaut accidentally gets separated out of his small spaceship accelerating in inter-stellar space at a constant acceleration of 10 m/S to the power of 2 end exponent . What is the acceleration of the astronaut at the instant he is outside the spaceship ?

physics-General
When the astronaut is outside the spaceship, the net external force (except negligible gravitational force due to spaceship) is zero as he is isolated from all interactions.
General
physics-

The two diodes A and B are biased as shown, then

The two diodes A and B are biased as shown, then

physics-General
parallel
General
physics-

The value of current in the following diagram will be:

The value of current in the following diagram will be:

physics-General
General
physics-

The value of current in the adjoining diagram will be:

The value of current in the adjoining diagram will be:

physics-General
General
physics-

A semiconductor P-N junction is to be forward biased with a battery of e.m.f. 1.5 Volt. If a potential difference of 0.5V appears on the junction which does not depend on current and on passing 10mA current through the junction there occurs huge Joule loss, then to use the junction at 5mA current, the resistance required to be connected in its series will be:

A semiconductor P-N junction is to be forward biased with a battery of e.m.f. 1.5 Volt. If a potential difference of 0.5V appears on the junction which does not depend on current and on passing 10mA current through the junction there occurs huge Joule loss, then to use the junction at 5mA current, the resistance required to be connected in its series will be:

physics-General
parallel

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