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Physics-

General

Easy

Question

Tangential acceleration of a particle moving in a circle of radius 1 m varies with time t as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of 30° with radial acceleration is

4 sec
4/3 sec
2^2/3 sec
$\sqrt{2} s e c$

The correct answer is: 2^2/3 sec

Related Questions to study

A truck starting from rest moves with an acceleration of 5 m/s² for 1 sec and then moves with constant velocity. The velocity w.r.t ground v/s time graph for block in truck is (Assume that block does not fall off the truck)

A truck starting from rest moves with an acceleration of 5 m/s² for 1 sec and then moves with constant velocity. The velocity w.r.t ground v/s time graph for block in truck is (Assume that block does not fall off the truck)

physics-General

If position time graph of a particle is sine curve as shown, what will be its velocity-time graph

If position time graph of a particle is sine curve as shown, what will be its velocity-time graph

physics-General

The value of $not stretchy integral subscript negative 2 end subscript superscript 3 end superscript vertical line 1 minus x to the power of 2 end exponent vertical line d x$ is‐

The value of $not stretchy integral subscript negative 2 end subscript superscript 3 end superscript vertical line 1 minus x to the power of 2 end exponent vertical line d x$ is‐

parallel

$not stretchy integral subscript negative pi end subscript superscript pi end superscript fraction numerator 2 x left parenthesis 1 plus blank s i n blank x right parenthesis over denominator 1 plus c o s to the power of 2 end exponent x end fraction equals$

$not stretchy integral subscript negative pi end subscript superscript pi end superscript fraction numerator 2 x left parenthesis 1 plus blank s i n blank x right parenthesis over denominator 1 plus c o s to the power of 2 end exponent x end fraction equals$

If $I subscript n end subscript equals not stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript t a n to the power of n end exponent blank x$ $d x$ then the value of $n left parenthesis I subscript n minus 1 end subscript plus I subscript n plus 1 end subscript right parenthesis$ is‐

If $I subscript n end subscript equals not stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript t a n to the power of n end exponent blank x$ $d x$ then the value of $n left parenthesis I subscript n minus 1 end subscript plus I subscript n plus 1 end subscript right parenthesis$ is‐

Statement‐I:: $not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript left parenthesis square root of x plus 1 right parenthesis d x less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis comma$ $n element of N.$
Statement‐II:: If $f left parenthesis x right parenthesis$ is continuous and increasing in $left square bracket 0 , 1 right square bracket$ , then $not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript f left parenthesis$ xXlx $less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis$ , where $n element of N$

Statement‐I:: $not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript left parenthesis square root of x plus 1 right parenthesis d x less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis comma$ $n element of N.$
Statement‐II:: If $f left parenthesis x right parenthesis$ is continuous and increasing in $left square bracket 0 , 1 right square bracket$ , then $not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript f left parenthesis$ xXlx $less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis$ , where $n element of N$

parallel

$not stretchy integral subscript 2 end subscript superscript 3 end superscript fraction numerator square root of x over denominator square root of left parenthesis 5 minus x right parenthesis end root plus square root of x end fraction d x equals$

$not stretchy integral subscript 2 end subscript superscript 3 end superscript fraction numerator square root of x over denominator square root of left parenthesis 5 minus x right parenthesis end root plus square root of x end fraction d x equals$

$stack lim with n rightwards arrow infinity below invisible function application fraction numerator pi over denominator n end fraction left square bracket blank s i n blank fraction numerator pi over denominator n end fraction plus blank s i n blank fraction numerator 2 pi over denominator n end fraction plus horizontal ellipsis plus blank s i n blank fraction numerator left parenthesis n minus 1 right parenthesis pi over denominator n end fraction right square bracket$ is equals to:

$stack lim with n rightwards arrow infinity below invisible function application fraction numerator pi over denominator n end fraction left square bracket blank s i n blank fraction numerator pi over denominator n end fraction plus blank s i n blank fraction numerator 2 pi over denominator n end fraction plus horizontal ellipsis plus blank s i n blank fraction numerator left parenthesis n minus 1 right parenthesis pi over denominator n end fraction right square bracket$ is equals to:

$not stretchy integral subscript e to the power of e to the power of e end exponent end exponent end subscript superscript e to the power of e to the power of e to the power of e end exponent end exponent end exponent end superscript fraction numerator d x over denominator x blank l n blank x times blank l n blank left parenthesis blank l n blank x right parenthesis times blank l n blank left parenthesis blank l n blank left parenthesis blank l n blank x right parenthesis right parenthesis end fraction$ equals

$not stretchy integral subscript e to the power of e to the power of e end exponent end exponent end subscript superscript e to the power of e to the power of e to the power of e end exponent end exponent end exponent end superscript fraction numerator d x over denominator x blank l n blank x times blank l n blank left parenthesis blank l n blank x right parenthesis times blank l n blank left parenthesis blank l n blank left parenthesis blank l n blank x right parenthesis right parenthesis end fraction$ equals

parallel

If $integral subscript 0 superscript pi divided by 3 end superscript fraction numerator blank c o s blank x over denominator 3 plus 4 blank s i n blank x end fraction d x equals k blank l o g blank left parenthesis fraction numerator 3 plus 2 square root of 3 over denominator 3 end fraction right parenthesis$ then k is‐

If $integral subscript 0 superscript pi divided by 3 end superscript fraction numerator blank c o s blank x over denominator 3 plus 4 blank s i n blank x end fraction d x equals k blank l o g blank left parenthesis fraction numerator 3 plus 2 square root of 3 over denominator 3 end fraction right parenthesis$ then k is‐

In the figure shown in YDSE, a parallel beam of light is incident on the slit from a medium of refractive index n₁ The wavelength of light in this medium is l₁ A transparent slab of thickness ‘t’ and refractive index n₃ is put infront of one slit. The medium between the screen and the plane of the slits is n₂ The phase difference between the light waves reaching point ‘O’ (symmetrical, relative to the slits) is :

In the figure shown in YDSE, a parallel beam of light is incident on the slit from a medium of refractive index n₁ The wavelength of light in this medium is l₁ A transparent slab of thickness ‘t’ and refractive index n₃ is put infront of one slit. The medium between the screen and the plane of the slits is n₂ The phase difference between the light waves reaching point ‘O’ (symmetrical, relative to the slits) is :

physics-General

In YDSE, the source placed symmetrically with respect to the slit is now moved parallel to the plane of the slits so that it is closer to the upper slit, as shown. Then,

In YDSE, the source placed symmetrically with respect to the slit is now moved parallel to the plane of the slits so that it is closer to the upper slit, as shown. Then,

physics-General

parallel

A parallel beam of light 500nm 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 S₁ and S₂ . The distance between slits is 1mm.

A parallel beam of light 500nm 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 S₁ and S₂ . The distance between slits is 1mm.

physics-General

In the figure shown if a parallel beam of white light is incident on the plane of the slits then the distance of the white spot on the screen from O is [Assume d << D, $lambda$ << d]

In the figure shown if a parallel beam of white light is incident on the plane of the slits then the distance of the white spot on the screen from O is [Assume d << D, $lambda$ << d]

physics-General

Two slits are separated by 0.3 mm. A beam of 500 nm light strikes the slits producing an interference pattern. The number of maxima observed in the angular range – 30° < $theta$ < 30°.

Two slits are separated by 0.3 mm. A beam of 500 nm light strikes the slits producing an interference pattern. The number of maxima observed in the angular range – 30° < $theta$ < 30°.

physics-General

parallel

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