The function 2x 3 – 9x 2 + 12x – 3 is increasing when x belongs to the interval

The point on the ellipse 16x² + 9y² = 400 where ordinate decrease at the same rate at which the abscissa increases is

For such questions, we should know how to find the derivates. We should know the concept of abscissa and ordinate.

The point on the ellipse 16x² + 9y² = 400 where ordinate decrease at the same rate at which the abscissa increases is

For such questions, we should know how to find the derivates. We should know the concept of abscissa and ordinate.

The displacement of a particle in time ‘t’ is given by S = t³ – t² – 8t – 18. The acceleration of the particle when its velocity vanishes is

For such questions, we should know relation between displacement, velocity and time. We should also know how to find first and second order derivate.

The displacement of a particle in time ‘t’ is given by S = t³ – t² – 8t – 18. The acceleration of the particle when its velocity vanishes is

For such questions, we should know relation between displacement, velocity and time. We should also know how to find first and second order derivate.

If $S subscript n end subscript equals stretchy sum from k equals 1 to n of a subscript k end subscript$ and $stack l i m with n rightwards arrow infinity below a subscript n end subscript equals a comma$ then $stack l i m with n rightwards arrow infinity below fraction numerator S subscript n plus 1 end subscript minus S subscript n end subscript over denominator square root of stretchy sum from k equals 1 to n of k end root end fraction$ is equal to

$stack l i m with n rightwards arrow infinity below open square brackets fraction numerator 1 over denominator n to the power of 3 end exponent plus 1 end fraction plus fraction numerator 4 over denominator n to the power of 3 end exponent plus 1 end fraction plus fraction numerator 9 over denominator n to the power of 3 end exponent plus 1 end fraction plus horizontal ellipsis horizontal ellipsis. plus fraction numerator n to the power of 2 end exponent over denominator n to the power of 3 end exponent plus 1 end fraction close square brackets equals$

$stack l i m with x rightwards arrow 0 below fraction numerator square root of 1 plus x end root minus square root of 1 minus x end root over denominator sin to the power of negative 1 end exponent invisible function application x end fraction equals$

The figure shows a transparent slab of length 1 m placed in air whose refractive index in x direction varies as $mu equals 1 plus x to the power of 2 end exponent left parenthesis 0 less than x less than 1 right parenthesis$ . The optical path length of ray R will be

A parallel beam of light 500 nm is incident at an angle 30° with the normal to the slit plane in a young's double slit experiment. The intensity due to each slit is Io. Point O is equidistant from $S subscript 1 end subscript$ and $S subscript 2 end subscript$ . The distance between slits is 1 mm.

In a Young's Double slit experiment, first maxima is observed at a fixed point P on the screen. Now the screen is continuously moved away from the plane of slits. The ratio of intensity at point P to the intensity at point O (centre of the screen)

An YDSE is carried out in a liquid of refractive index $mu equals 1.3$ and a thin film of air is formed in front of the lower slit as shown in the figure. If a maxima of third order is formed at the origin O, find the thickness of the air film. The wavelength of light in air is $lambda subscript 0 end subscript equals 0.78 mu m$ and D/d=1000.

Two coherent radio point sources that are separated by 2.0 m are radiating in phase with a wavelength of 0.25 m. If a detector moves in a large circle around their midpoint, at how many points will the detector show a maximum signal ?

Two coherent narrow slits emitting light of wavelength $lambda$ in the same phase are placed parallel to each other at a small separation of $2 lambda$ . The light is collected on a screen S which is placed at a distance $D left parenthesis much greater-than lambda right parenthesis$ from the slit $S subscript 1 end subscript$ as shown in figure. Find the finite distance x such that the intensity of P is equal to intensity at O.

Consider the arrangement shown in figure. By some mechanism, the separation between the slits $S subscript 3 end subscript$ and $S subscript 4 end subscript$ can be changed. The intensity is measured at the point P which is at the common perpendicular bisector of $S subscript 1 end subscript S subscript 2 end subscript$ and $S subscript 3 end subscript S subscript 4 end subscript$ . When $z equals fraction numerator D lambda over denominator 2 blank d end fraction$ , the intensity measured at P is I. Find this intensity when z is equal to

1) $fraction numerator D lambda over denominator d end fraction$ ,
2) $fraction numerator 3 D lambda over denominator 2 blank d end fraction$ , and
3) $fraction numerator 2 D lambda over denominator d end fraction$ .

$stack l i m with x rightwards arrow pi divided by 2 below fraction numerator a to the power of cot invisible function application x end exponent minus a to the power of cos invisible function application x end exponent over denominator c o t invisible function application x minus c o s invisible function application x end fraction equals$

In Young's double slit experiment $S subscript 1 end subscript$ and $S subscript 2 end subscript$ are two slits. Films of thicknesses $t subscript 1 end subscript$ and $t subscript 2 end subscript$ and refractive indices $m subscript 1 end subscript$ and $m subscript 2 end subscript$ are placed in front of $S subscript 1 end subscript$ and $S subscript 2 end subscript$ respectively. If $m subscript 1 end subscript t subscript 1 end subscript equals m subscript 2 end subscript t subscript 2 end subscript$ , then the central maximum will :