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Physics-

General

Easy

Question

A beam of light strikes one mirror of a right angle mirror assembly at an angle of incidence 45⁰ as shown in the figure The right angle mirror assembly is rotated such that the angle of incidence becomes 60⁰ Which of the following statements is (are) correct about the emerging light beam

It will move through an angle of 15⁰ with respect to the original emerging beam
it will move through an angle of 30⁰ with respect to the original emerging beam
it will move through an angle of 45⁰ with respect to the original emerging beam
it will emerge parallel to the original emerging beam

The correct answer is: it will emerge parallel to the original emerging beam

Related Questions to study

A boy is walking under an inclined mirror at a constant velocity V m/s along the X axis as shown in figure If the mirror is inclined at an angle q with the horizontal then what is the velocity of the image?

A boy is walking under an inclined mirror at a constant velocity V m/s along the X axis as shown in figure If the mirror is inclined at an angle q with the horizontal then what is the velocity of the image?

physics-General

Let $A equals open square brackets table row 1 0 0 row 5 2 0 row cell negative 1 end cell 6 1 end table close square brackets$ , then the ad joint of A is

A equals open square brackets table row 1 0 0 row 5 2 0 row cell negative 1 end cell 6 1 end table close square brackets

rightwards double arrow

a d j left parenthesis A right parenthesis equals open square brackets table row 2 cell negative 5 end cell 32 row 0 1 cell negative 6 end cell row 0 0 2 end table close square brackets to the power of T end exponent equals open square brackets table row 2 0 0 row cell negative 5 end cell 1 0 row 32 cell negative 6 end cell 2 end table close square brackets

Let $A equals open square brackets table row 1 0 0 row 5 2 0 row cell negative 1 end cell 6 1 end table close square brackets$ , then the ad joint of A is

A equals open square brackets table row 1 0 0 row 5 2 0 row cell negative 1 end cell 6 1 end table close square brackets

rightwards double arrow

a d j left parenthesis A right parenthesis equals open square brackets table row 2 cell negative 5 end cell 32 row 0 1 cell negative 6 end cell row 0 0 2 end table close square brackets to the power of T end exponent equals open square brackets table row 2 0 0 row cell negative 5 end cell 1 0 row 32 cell negative 6 end cell 2 end table close square brackets

If there be a compound of the formula $C H subscript 3 C left parenthesis O H right parenthesis subscript 3 comma$ which one of the following compounds would be obtained form it without treatment with any reagent?

Presence of two or more

O H g p

. On a carbon atom makes it unstable and compound loses

H subscript 2 O

If there be a compound of the formula $C H subscript 3 C left parenthesis O H right parenthesis subscript 3 comma$ which one of the following compounds would be obtained form it without treatment with any reagent?

chemistry-General

Presence of two or more

O H g p

. On a carbon atom makes it unstable and compound loses

H subscript 2 O

parallel

An object ‘O’ is placed infront of a small plane mirrorM₁ and a large convex mirrorM₂ of focal length f The distance between ‘O’ and M₁ is x and the distance between M₁ and M₂ is y The image of ‘O’ formed by M₁ and M₂ coincide The magnitude of ‘f’ is

An object ‘O’ is placed infront of a small plane mirrorM₁ and a large convex mirrorM₂ of focal length f The distance between ‘O’ and M₁ is x and the distance between M₁ and M₂ is y The image of ‘O’ formed by M₁ and M₂ coincide The magnitude of ‘f’ is

physics-General

Consider a disc rotating in the horizontal plane with a constant angular speed $omega$ about its centre $O$ . The disc has a shaded region on one side of the diameter and an unshanded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$ . The velocity of projection is in the $y minus z$ plane and is same for both pebbles with respect to the disc. Assume that i) they land back on the disc before the disc has completed $fraction numerator 1 over denominator 8 end fraction$ rotation. ii) their range is less than half the disc radius, and iii) $omega$ remains constant throughout. Then

To reach the unshaded portion particle

P

needs to travel horizontal range greater than

R sin invisible function application 45 degree

left parenthesis 0.7 blank R right parenthesis

but its range is less than

fraction numerator R over denominator 2 end fraction

. So it will fall on shaded portion

Q

is near to origin, its velocity will be nearly along

Q R

so its will fall in unshaded portion

Consider a disc rotating in the horizontal plane with a constant angular speed $omega$ about its centre $O$ . The disc has a shaded region on one side of the diameter and an unshanded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$ . The velocity of projection is in the $y minus z$ plane and is same for both pebbles with respect to the disc. Assume that i) they land back on the disc before the disc has completed $fraction numerator 1 over denominator 8 end fraction$ rotation. ii) their range is less than half the disc radius, and iii) $omega$ remains constant throughout. Then

physics-General

To reach the unshaded portion particle

P

needs to travel horizontal range greater than

R sin invisible function application 45 degree

left parenthesis 0.7 blank R right parenthesis

but its range is less than

fraction numerator R over denominator 2 end fraction

. So it will fall on shaded portion

Q

is near to origin, its velocity will be nearly along

Q R

so its will fall in unshaded portion

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

Equating the moments about

R

6 cross times P R equals 4 cross times R Q

P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

physics-General

Equating the moments about

R

6 cross times P R equals 4 cross times R Q

P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

parallel

On reacting with neutral ferric chloride, phenol gives

Phenol reacts with neutral

F e C l subscript 3

solution to give violet colour complex which is soluble in water.

On reacting with neutral ferric chloride, phenol gives

chemistry-General

Phenol reacts with neutral

F e C l subscript 3

solution to give violet colour complex which is soluble in water.

An air bubble in glass (m =1.5) is situated at a distance 3 cm from a convex surface of diameter 10 cm as shown The distance from surface at which the image of bubble appears is

An air bubble in glass (m =1.5) is situated at a distance 3 cm from a convex surface of diameter 10 cm as shown The distance from surface at which the image of bubble appears is

physics-General

The fig shows a mixture of blue, green, red colours incident on a right angled prism The critical angles of the material of prism for red, green and blue colours are 46⁰ , 44⁰ , 43⁰ respectively, The arrangement will separate

The fig shows a mixture of blue, green, red colours incident on a right angled prism The critical angles of the material of prism for red, green and blue colours are 46⁰ , 44⁰ , 43⁰ respectively, The arrangement will separate

physics-General

parallel

Ethanol reacts with thionyl chloride to give ethyl chloride and:

C subscript 2 H subscript 5 O H plus S O C l subscript 2 ⟶ C subscript 2 H subscript 5 C l plus S O subscript 2 plus H C l

Ethanol reacts with thionyl chloride to give ethyl chloride and:

chemistry-General

C subscript 2 H subscript 5 O H plus S O C l subscript 2 ⟶ C subscript 2 H subscript 5 C l plus S O subscript 2 plus H C l

The minimum speed for a particle at the lowest point of a vertical circle of radius $r$ , to describe the circle is $v$ . If the radius of the circle is reduced to one-fourth its value, the corresponding minimum speed will be

v equals square root of 5 g R end root

R to the power of ´ end exponent equals fraction numerator R over denominator 4 end fraction

v to the power of ´ end exponent equals square root of 5 g R ´ end root equals square root of 5 g R divided by 4 end root equals fraction numerator 1 over denominator 2 end fraction square root of 5 g R end root equals fraction numerator 1 over denominator 2 end fraction v

The minimum speed for a particle at the lowest point of a vertical circle of radius $r$ , to describe the circle is $v$ . If the radius of the circle is reduced to one-fourth its value, the corresponding minimum speed will be

physics-General

v equals square root of 5 g R end root

R to the power of ´ end exponent equals fraction numerator R over denominator 4 end fraction

v to the power of ´ end exponent equals square root of 5 g R ´ end root equals square root of 5 g R divided by 4 end root equals fraction numerator 1 over denominator 2 end fraction square root of 5 g R end root equals fraction numerator 1 over denominator 2 end fraction v

Scientific aspect of fermentation was first studied by:

Buchner studied fermentation.

Scientific aspect of fermentation was first studied by:

chemistry-General

Buchner studied fermentation.

parallel

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure A small object is placed on the principal axis of the combination, at a distance of 30 cm in front of the mirror The magnification of the image is

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure A small object is placed on the principal axis of the combination, at a distance of 30 cm in front of the mirror The magnification of the image is

physics-General

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure The combination behaves like

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure The combination behaves like

physics-General

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure The focal length of the combination has the magnitude

The convex surface of a thin concav–convex lens of glass of refractive index 1.5 has a radius of curvature of 20 cm The concave surface has a radius of curvature of 60 cm The convex side is silvered and placed on a horizontal surface as shown in the figure The focal length of the combination has the magnitude

physics-General

parallel

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