Physics
General
Easy
Question
The resolving limit of healthy eye is about
 or



The correct answer is: or
Resolving limit of eye is one minute (1').
Related Questions to study
physics
A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm. What type and of what focal length of lens he should use to see clearly the object placed 60 cm away
For viewing far objects, concave lenses are used and for concave lens
u = wants to see ; v = can see
so from .
u = wants to see ; v = can see
so from .
A person is suffering from myopic defect. He is able to see clear objects placed at 15 cm. What type and of what focal length of lens he should use to see clearly the object placed 60 cm away
physicsGeneral
For viewing far objects, concave lenses are used and for concave lens
u = wants to see ; v = can see
so from .
u = wants to see ; v = can see
so from .
physics
A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true
In minimum deviation position refracted ray inside the prism is parallel to the base of the prism
A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true
physicsGeneral
In minimum deviation position refracted ray inside the prism is parallel to the base of the prism
physics
In the given figure, what is the angle of prism
Angle of prism is the angle between incident and emergent surfaces.
In the given figure, what is the angle of prism
physicsGeneral
Angle of prism is the angle between incident and emergent surfaces.
physics
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and material are now added to P as shown in the figure. The ray will suffer
As the prisms Q and R are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and material are now added to P as shown in the figure. The ray will suffer
physicsGeneral
As the prisms Q and R are of the same material and have identical shape they combine to form a slab with parallel faces. Such a slab does not cause any deviation.
Physics
A bob of mass is suspended by a massless string of length . The horizontal velocity at position is just sufficient to make it reach the point . The angle at which the speed of the bob is half of that at , satisfies
A bob of mass is suspended by a massless string of length . The horizontal velocity at position is just sufficient to make it reach the point . The angle at which the speed of the bob is half of that at , satisfies
PhysicsGeneral
maths
The foot of the perpendicular on the line drown from the origin is C if the line cuts the xaxis and yaxis at A and B respectively then BC : CA is
Let direction ratios of the required line be <a, b, c>
Therefore a  2 b  2 c = 0
And 2 b + c = 0
Þ c =  2 b
a  2 b + 4b = 0 Þ a =  2 b
Therefore direction ratios of the required line are < 2b, b,  2b> = <2,  1, 2>
direction cosines of the required line
=
Therefore a  2 b  2 c = 0
And 2 b + c = 0
Þ c =  2 b
a  2 b + 4b = 0 Þ a =  2 b
Therefore direction ratios of the required line are < 2b, b,  2b> = <2,  1, 2>
direction cosines of the required line
=
The foot of the perpendicular on the line drown from the origin is C if the line cuts the xaxis and yaxis at A and B respectively then BC : CA is
mathsGeneral
Let direction ratios of the required line be <a, b, c>
Therefore a  2 b  2 c = 0
And 2 b + c = 0
Þ c =  2 b
a  2 b + 4b = 0 Þ a =  2 b
Therefore direction ratios of the required line are < 2b, b,  2b> = <2,  1, 2>
direction cosines of the required line
=
Therefore a  2 b  2 c = 0
And 2 b + c = 0
Þ c =  2 b
a  2 b + 4b = 0 Þ a =  2 b
Therefore direction ratios of the required line are < 2b, b,  2b> = <2,  1, 2>
direction cosines of the required line
=
maths
(where a, b are integers) =
The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x_{1}, y_{1}, z_{1}), then
= 1, = 2, = –3
\ Required point is (0, 5, 7).
Hence (c) is the correct answer.
= 1, = 2, = –3
\ Required point is (0, 5, 7).
Hence (c) is the correct answer.
(where a, b are integers) =
mathsGeneral
The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x_{1}, y_{1}, z_{1}), then
= 1, = 2, = –3
\ Required point is (0, 5, 7).
Hence (c) is the correct answer.
= 1, = 2, = –3
\ Required point is (0, 5, 7).
Hence (c) is the correct answer.
maths
mathsGeneral
maths
${\int}_{\pi /4}^{\pi /4}\u200a\frac{{e}^{x}(x\mathrm{sin}x)}{{e}^{2x}1}dx$ is equal to
Let direction cosines of straight line be l, m, n
\ 4l + m + n = 0
l – 2m + n = 0
Þ $\frac{l}{3}=\frac{m}{3}=\frac{n}{9}$ Þ $\frac{l}{1}=\frac{m}{+1}=\frac{n}{3}$
\ Equation of straight line is $\frac{x2}{1}=\frac{y+1}{1}=\frac{z+1}{3}$.
Hence (c) is the correct choice.
${\int}_{\pi /4}^{\pi /4}\u200a\frac{{e}^{x}(x\mathrm{sin}x)}{{e}^{2x}1}dx$ is equal to
mathsGeneral
Let direction cosines of straight line be l, m, n
\ 4l + m + n = 0
l – 2m + n = 0
Þ $\frac{l}{3}=\frac{m}{3}=\frac{n}{9}$ Þ $\frac{l}{1}=\frac{m}{+1}=\frac{n}{3}$
\ Equation of straight line is $\frac{x2}{1}=\frac{y+1}{1}=\frac{z+1}{3}$.
Hence (c) is the correct choice.
maths
Let $f:R\to R,f\left(x\right)=\left\{\begin{array}{c}x[x\left]\right,\left[x\right]\\ x[x+1\left]\right,\left[x\right]\end{array}\right.$$\begin{array}{r}\text{is odd}\\ 1\text{is even where [.]}\end{array}$ denotes greatest integer function, then ${\int}_{2}^{4}\u200af\left(x\right)dx$ is equal to
Since these two lines are intersecting so shortest distance between the lines will be 0.
Hence (c) is the correct answer.
Let $f:R\to R,f\left(x\right)=\left\{\begin{array}{c}x[x\left]\right,\left[x\right]\\ x[x+1\left]\right,\left[x\right]\end{array}\right.$$\begin{array}{r}\text{is odd}\\ 1\text{is even where [.]}\end{array}$ denotes greatest integer function, then ${\int}_{2}^{4}\u200af\left(x\right)dx$ is equal to
mathsGeneral
Since these two lines are intersecting so shortest distance between the lines will be 0.
Hence (c) is the correct answer.
maths
The value of ${\int}_{0}^{1}\u200a\mathrm{sin}2\pi x\mid dx$ is equal to
Given plane y + z + 1 = 0 is parallel to xaxis as 0.1 + 1.0 + 1.0 = 0
but normal to the plane will be perpendicular to xaxis.
Hence (c) is the correct answer.
The value of ${\int}_{0}^{1}\u200a\mathrm{sin}2\pi x\mid dx$ is equal to
mathsGeneral
Given plane y + z + 1 = 0 is parallel to xaxis as 0.1 + 1.0 + 1.0 = 0
but normal to the plane will be perpendicular to xaxis.
Hence (c) is the correct answer.
Maths
The value of (where {x} is the fractional part of x) is
The value of (where {x} is the fractional part of x) is
MathsGeneral
Maths
The shortest distance between the two straight line and is
The given equations are
and
The shortest distance between the two cartesian lines
=
and
The shortest distance between the two cartesian lines
=
The shortest distance between the two straight line and is
MathsGeneral
The given equations are
and
The shortest distance between the two cartesian lines
=
and
The shortest distance between the two cartesian lines
=
physics
A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $\theta $ at which the speed of the bob is half of that at A, satisfies
Velocity of the bob at the point A
$v=\sqrt{5gL}$(i)
${\left(\frac{v}{2}\right)}^{2}={v}^{2}2gh\left(ii\right)$
$h=L(1\mathrm{cos}\theta )(iii)$
$SolvingEqs.\left(i\right),\left(ii\right)and\left(iii\right),weget$
$\mathrm{cos}\theta =\frac{7}{8}$
$or\theta ={cos}^{1}\left(\frac{7}{8}\right)=151\xb0$
A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $\theta $ at which the speed of the bob is half of that at A, satisfies
physicsGeneral
Velocity of the bob at the point A
$v=\sqrt{5gL}$(i)
${\left(\frac{v}{2}\right)}^{2}={v}^{2}2gh\left(ii\right)$
$h=L(1\mathrm{cos}\theta )(iii)$
$SolvingEqs.\left(i\right),\left(ii\right)and\left(iii\right),weget$
$\mathrm{cos}\theta =\frac{7}{8}$
$or\theta ={cos}^{1}\left(\frac{7}{8}\right)=151\xb0$
Physics
A piece of wire is bent in the shape of a parabola axis vertical) with a bead of mass on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the axis with a constant acceleration . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the axis is
A piece of wire is bent in the shape of a parabola axis vertical) with a bead of mass on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the axis with a constant acceleration . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the axis is
PhysicsGeneral