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94051.

Find the synonym of words written in capitals.Summit A) rear B) bottom C) peak D) front

Answer»

Correct option is C) peak

94052.

What is Bump Maps? Explain in brief.

Answer»

Bump maps are grayscale textures you map to objects to create the illusion of surface relief on an otherwise flat object.

94053.

Displacement maps cannot cast or receive shadows. (a) True(b) False

Answer»

Correct option: False

94054.

What makes high pass sharpening a unique method?

Answer»

The difference between high pass sharpening and most other methods of sharpening is that high pass sharpening does not actually adjust or change any pixels in your original image. Also, because high pass exists as a separate layer, you can adjust the layer’s Opacity and Blending Modes to control the strength of sharpening over the entire image.

94055.

Find the synonymPeak A) bottom B) summit C) rear D) front

Answer»

Correct option is B) summit

94056.

Explain how to create a texture background.

Answer»

1. In the Environment section of the camera’s Attribute Editor (View > Camera Attribute Editor), click the Create button to the right of the Image Plane attribute. Maya creates an image plane and connects it to the camera. 

2. Set the Type attribute for the image plane to Texture. 

3. Click the texture button for the Texture attribute. The Create Render Node window appears. 

4.Select the texture you want to use as a background (for example, an Environment Texture). Maya creates the texture and connects it to the image plane.

94057.

What is Displacement Maps? Explain in brief.

Answer»

Displacement maps are grayscale textures you map to objects to create true surface relief (elevations and depressions) on an otherwise flat object.

94058.

Explain how to set the position of Image Plane.

Answer»
  • Use the Placement attributes to position an image plane relative to the camera. 
  • Use the Placement Extras attributes to control which portion of an image file is visible on the image plane.
94059.

Find the synonymBump A) jolt B) piece C) lump D) handle

Answer»

Correct option is A) jolt

94060.

What does IOR stands for?

Answer»

IOR stands for Index of Refraction

94061.

A partical revolves in clockwise direction (as seen from point A) in a circle C of radius 1 cm and completes one revolution in 2 sec. The axis of the circle and the principal axis of the mirror M coincides, call it AB. The radius of curvature of the mirror of 20cm. Then, the direction of revolution (as seen from A) of the image of the partical and its speed is A. Clockwise, `1.57 cm//s`B. Clockwise, `3.14 cm//s`C. Anticlockwise, `1.57 cm//s`D. Anticlockwise, `3.14 cm//s`

Answer» Correct Answer - A
By mirror formula : `(1)/(v)+(1)/(-10)=(1)/(10)`
`rArr v=+5 cm :. m=+(1)/(2)`
the image revolves in circle of radius `(1)/(2) cm`.
Image of a radius is erect `rArr` image will revolve in the same direction as the particle. The image will complete one revolution in the same the `2s`
velocity of image `v`
`=omega r=(2pi)/(2)xx(1)/(2)=(pi)/(2) cm//s=1.57 cm//s`.
94062.

Explain how to create a Bump map in Maya.

Answer»

Bump maps are grayscale textures you map to objects to create the illusion of surface relief (elevations and depressions) on an otherwise flat object. 

With bump maps, depressions and elevations look real because they don’t alter the geometry of the surface the way Displacement maps do. Bump maps just change the direction of the surface’s normal based on the bump map’s Alpha Gain value. 

Use bump maps to create very shallow reliefs. For example, you can make objects look like they are embossed, have shallow rolling hills, and so on. 

Because bump maps are not true surface relief, they:

  • cannot cast or receive shadows 
  • cannot be seen if you silhouette the mapped object 
  • take less time to render than displacement maps
94063.

In which side Channel Box editor in Maya uses interface? (a) Right (b) Left (c) Bottom (d) None of these

Answer»

Correct option: (a) Right

94064.

Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in the figure. Ball 1 moves with a velocity `v_(1)` towards ball 2. All collisions are assumed to be elastic. If `Mltm`, the number of collisions between the balls will be. A. 1B. 2C. 3D. 4

Answer» Correct Answer - B
After collision between 2 `&` 3 moves, 2 moves forward so only 2 collision.
94065.

What is IOR?

Answer»

The IOR parameter (Index of Refraction) defines the material's Fresnel reflectivity and is by default the angular function used. Effectively the IOR will define the balance between reflections on surfaces facing the viewer and on surface edges.

94066.

A bird in air is diving vertically over a tank with speed 6 cm//s Base of the tank is silvered A fish in the tank is rising upward along the same line with speed 4 cm//s[Take:mu_(water)=4//3] `{:(,"Column-I",,"Column-II"),("(A)","Speed of the image of fish as seen by the bird directly",,"(P)12"),("(B)",underset("from the mirror as seen by the bird")" Speed of the image of fish formed after reflection",,"(Q) 4"),("(C)","Speed of image of bird relative to the fish looking upwards",,"(R) 9"),("(D)",underset("downwards in the mirror")" Speed of image of bird relative to the fish looking",,"(S) 3"):}`

Answer» Correct Answer - [(A)R;(B)S;(C)P;(D)Q]
(C) Velocity of bird in water `= 6XX4/3=8 downarrow w.r.t fish = 8+4 =12 downarrow`
(D) Velocity of bird in water after reflection from mirror =`8 uparrow w.r.t fish = 8-4 =4 uparrow`
(A) Velocity of fish in air `= 4xx 3/4 = 3 uparrow`
Velocity of fish w.r.t `= 3+6=9 uparrow`
(B) Velocity of image of fish after reflection from mirror in air `=4xx3/4 = 3 uparrow`
w.r.t bird `=-3+6=3 uparrow`
94067.

If a prism having refractive index `sqrt 2` has angle of minimum deviation equal to the angle of refraction of the prism, then the angle of refraction of the prism is:A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`

Answer» Correct Answer - D
Put `A=delta_(min)` and `mu= sqrt(2)`
The relation `mu=(sin ((A+delta_(min))/(2)))/(sin((A)/(2)))`
`sqrt(2)=(sinA)/(sin A//2) rArr sqrt(2)=2 cos A//2`
`rArr cos A//2 =(1)/(sqrt(2)) rArr A//2 =45^(@)`
`rArr A=90^(@)`.
94068.

Explain Interface overview in Maya with suitable example.

Answer»

There are a lot of items displayed in the Maya user interface which are as follows: 

1. Menu Bar: The Main Menu bar appears at the top of the Maya interface directly below the Maya title bar and displays the chosen menu set. Each menu set corresponds to a module within Maya: Animation, Polygons, Surfaces, Rendering, and Dynamics. Modules are a method for grouping related features and tools. 

2. Status Line: The Status Line, located directly below the Main Menu bar, contains a variety of items, most of which are used while modeling or working with objects within Maya. Many of the Status Line items are represented by a graphical icon. The icons save space in the Maya interface and allow for quick access to tools used most often. 

3. Shelf: The Shelf is located directly below the Status line. The Maya Shelf is useful for storing tools and items that you use frequently or have customized for your own use. You can keep the tools and items you use most frequently in a location that provides handy access. Maya has some of the Shelf items pre-configured for your use.

4. Workspace: The Maya workspace is where you conduct most of your work within Maya. The workspace is the central window where your objects and most editor panels appear.

  • The panel is labeled persp at the bottom to indicate that you are viewing the Maya scene from a perspective camera view. 
  • The panel has its own menu bar at the top left corner of the panel. These menus allow you to access tools and functions related to that specific panel. 
  • The grid is displayed with two heavy lines intersecting at the center of the Maya scene. This central location is called the origin. The origin is the center of Maya’s 3D world, and with all object’s directional values measured from this location.

5. Channel Box: The Channel Box is the primary, fastest, and most streamlined tool for editing object attributes. It lets you quickly change attribute values, set keys on keyable attributes, lock or unlock attributes, and create expressions on attributes. 

6. Quick Layout Buttons: The quick layout buttons display just below the Tool Box and let you switch between useful panel layouts 

7. Toolbox: The Tool Box displays on the left side of the Maya interface by It contains the most commonly-used tools for working in Maya like., Move, Rotate, Scale and Last used tool 

8. Command Line: The command line lets you type single MEL or Python commands without having to open the Script editor. You can switch between MEL and Python mode by clicking the MEL/Python button. The result from the command appears in the output 

9. Time and Range Slider: The Time Slider controls the playback range, keys, and breakdowns within the playback range

94069.

How to apply Texture mapping Explain?

Answer»

To apply a texture to an object, you map the texture to an attribute on the object's material. 

The attribute to which the texture is connected determines how the texture is used and how it affects the final output. 

To map a texture to a material using the Attribute Editor 

1. Select your material in the Hyper shade. 

In the material Attribute Editor, click the map button beside the attribute that you want to connect a texture to. The Create Render Node window appears. 

2. Select a texture from the Create Render Node window. 

If you are selecting a 2D texture, right-click the texture and select a mapping method (Create texture (create normally), Create as projection, Create as stencil). If you do not select a mapping method, the default method (Create texture) is used. 

If you are selecting the File texture, map to the image file by clicking the browse icon beside the Image Name attribute.

94070.

How will you define Q factor

Answer»

The quality factor or Q factor is a dimensionless parameter that describes how under-damped an oscillator or resonator is, or equivalently, characterizes a resonator's bandwidth relative to its center frequency.

94071.

According to a FIR in police station, car A was at rest waiting for a red light at a crossing when it was hit by similar car B from rear side. Both cars had then hard brekes on, and from their skid marks it is surmiesed that they skidded together about 8m in the original direction of travel before coming to rest. Assuming a stopping force of about 0.4 times the combined weights of the cars. the speed of car B just before the collision is approximately `[g=10 m//s^(2)]`A. 8m/sB. 4m/sC. 16 m/sD. 32 m/s

Answer» Correct Answer - C
`S=8,v=0,a=-(0.4)g=-4`
`rArru=sqrt(2xx4xx8)=8=` velocity after collision and velocity after collision and velocity before collision be `v_(0)`
2mu=`mv_(0)`
`rArr=v_(0)=16 m//s`
94072.

Explain the use of specular maps.

Answer»

Some surfaces are shinier than others (for example a wet fish has a shinier surface than a dry leaf). By mapping a texture to the Specular attribute of an object’s material, you create a specular map which lets you describe how shine appears on objects (by controlling highlight).Shiny objects reflect light directly; matte objects diffuse light. Specular highlights show the places on the object where the light sources are reflected at consistent angles; reflections on an object show, among other things, light bounced from surrounding objects. 

Specular highlights depend directly on the view (camera), not the position of the light, like diffuse shading does. 

The size of a specular highlight on a surface makes the surface look either flat or shiny. 

Note: Only materials with specular attributes (Anisotropic, Blinn, Phong, and PhongE) have surface highlights. The specular highlight is the white shiny glow on the material.

94073.

What is a use of specular maps?

Answer»

Mapping a texture to the Specular attribute of an object’s material, you create a specular map which lets you describe how shine appears on objects (by controlling highlight).

Common surface material Specular Shading attributes Some surfaces are shinier than others (for example a wet fish has a shinier surface than a dry leaf). Depending on how shiny a surface is, it reflects light in different ways.

Shiny objects reflect light directly; matte objects diffuse light. Specular highlights show the places on the object where the light sources are reflected at consistent angles; reflections on an object show, among other things, light bounced from surrounding objects.

Specular highlights depend directly on the view (camera), not the position of the light, like diffuse shading does.

The size of a specular highlight on a surface makes the surface look either flat or shiny

94074.

An impulse `J` is applied on a ring of mass `m` along a line passing through its centre `O`. The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is A. `(J)/(m)`B. `(J)/(2m)`C. `(J)/(4m)`D. `(J)/(3m)`

Answer» Correct Answer - A
`E=|Z|l=|4+3j|2=sqrt(4^(2)+3^(2))xx2=10V,phi = "arg " Z = tan^(-1).(3)/(4)=37^(@)`
94075.

In the figure, an object is placed at distance `25 cm` from the surface of a convex mirror, and a plane mirror is set so that image formed by the two mirrors lie adjacent to each other in the same plane. The plane mirror is placed at `20 cm` from the object. What is the radius of curvature of the convex mirror ? .

Answer» Correct Answer - R = 75 cm
Image due to plane mirror will form at a distance of `20 cm` left of the mirror.
Since image formed by two mirrors lie adjacent to each other.
For convex mirror, image position is `15 cm` towards left
`u=-25 cm rArr v=+15 cm`
using `(1)/(v)+(1)/(u)=(1)/(f)=(2)/(R)`
`(1)/(15)-(1)/(25)=(2)/(R) R =75 cm`.
94076.

A solid hemisphere and a solid cone have a common base and are made of same material. The centre of mass of the common structure coincides with the centre of the common base. If R is the radius of hemisphere and h is height of the cone, then A. `(h)/(R )=sqrt(3)`B. `(h)/(R )=(1)/sqrt(3)`C. `(h)/(R )=3`D. `(h)/(R )=(1)/(3)`

Answer» Correct Answer - C
We have `(1)/(2)Mv^(2)=(1)/(2)kl^(2)`
`v=sqrt((k)/(M))l`
Maximum momentum `=Mv=Msqrt((k)/(M))l=sqrt(Mk)l`
94077.

A uniform solid right circular cone of base radius `R` is joined to a uniform solid hemisphere of radius `R` and of the same density, as shown. The centre of mass of the composite solid lies at the centre of base of the cone. The height of the cone is A. 1.5RB. `sqrt(3)R`C. 3RD. `2sqrt(3)R`

Answer» Correct Answer - B
`((1)/(3)piR^(2)h)pxx(h)/(4)=((2)/(3)piR^(3))p(3R)/(8)`
`h=sqrt(3)R`
94078.

What is sexing

Answer»

Through sexing, biologists and agricultural workers determine the sex of livestock and other animals they work with. 

94079.

An isolated nucleus which was initially at rest, disintegrates into two nuclei due to internal nuclear forces and no`gamma` rays are produced. If the ratio of their kinetic energy is found to beA. Ratio of their de-broglie wavelength is `(sqrt(64))/(sqrt(27))` respectivelyB. Ratio of their speed is 64/27 respectivelyC. Ratio of their nuclear radius is 5/4 respectivelyD. None of these

Answer» `P_(1)=P_(2)=P`
`m_(1)v_(1)=m_(2)v_(2)`
`((P^(2))/(2m_(1)))//((P^(2))/(2m_(2)))=64/27`
`(m_(2))/(m_(1))-64/27 =(v_(1))/(v_(2))`
`(lambda_(1))/(lambda_(2))=(h//P_(1))/(h//P_(2))=1:1`
`(R_(1))/(R_(2))=((A_(1))/(A_(2)))^(1//3)=3/4`
94080.

Evaluate \(\rm \int_0^4 \frac{1}{1+ \sqrt x}dx\)1. 2 - 2log 32. 4 - log 33. 4 - 2log 34. None of the above

Answer» Correct Answer - Option 3 : 4 - 2log 3

Concept:

\(\rm \int \frac{1}{x} dx = \log x + c\)

Calculation:

I = \(\rm \int_0^4 \frac{1}{1+ \sqrt x}dx\)

Let 1 + \(\rm \sqrt x\) = t       .... (1)

Differentiating with respect to x, we get

\(\rm \Rightarrow (0+\frac{1}{2\sqrt x})dx = dt\)

\(\rm \Rightarrow dx = {2\sqrt x}dt\)

From equation (1), we get 

\(\rm \sqrt x\) = t - 1

∴ dx = 2(t - 1)dt

x04
t13

 

Now,

I = \(\rm \int_1^3 \frac{2(t-1)}{t}dt \)

\(\rm 2\int_1^3 \left(1-\frac{1}{t} \right )dt \)

\(\rm 2\left[t - \log t \right ]_1^3\)

= 2 [(3 - log 3) - (1 - log 1)]

= 2(2 - log 3)

= 4 - 2log 3

94081.

What is the equation of the tangent to the curve y = sin x at (0, 0)?1. x + y = 02. x - y = 03. 2x + y = 04. 2x - y = 0

Answer» Correct Answer - Option 2 : x - y = 0

Concept:

Steps to find the equation of the tangent to the curve:

1) Find the first derivative of f(x).

2) Use the point-slope formula to find the equation for the tangent line.

Point-slope is the general form:  

y - y₁=m(x - x₁), Where m = slope of tangent = \(\rm \frac {dy}{dx}\)

Calculation:

Here, y = sin x

\(\rm \frac {dy}{dx}\) = cos x

\(\rm \frac{dy}{dx}|_{x=0}\)= cos 0 = 1

So, Slope = 1

So, Equation of tangent to the curve is (y - 0) = 1(x - 0)

⇒ y = x

⇒ x - y = 0

Hence, option (2) is correct. 

94082.

If radius of circle is increasing at rate 0.5 cm/sec what is the rate of increase of its circumference ? 1. 0. 7 cm/sec 2. 0. 5π  cm/sec 3. 2π  cm/sec 4. π cm/sec

Answer» Correct Answer - Option 4 : π cm/sec 

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

Circumference of circle = 2πr

 

Calculation:

Here, \(\rm \frac {dr}{dt}\) = 0.5 cm/sec

Circumference of circle, C = 2πr

Now taking derivatives on both sides, we get 

\(\rm \frac {dC}{dt}\) = 2π \(\rm \frac {dr}{dt}\)

= 2π (0.5)

= π cm/sec

Hence, option (4) is correct. 

94083.

For the given curve: y = 2x – x2, when x increases at the rate of 3 units/sec, then how the slope of curve changes?1. Increasing, at 6 units/sec2. decreasing, at 6 units/sec3. Increasing, at 3 units/sec4. decreasing, at 3 units/sec

Answer» Correct Answer - Option 2 : decreasing, at 6 units/sec

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

 

Calculation:

Given that, y = 2x – x2 and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

94084.

In a curve y = t2 + 2t and x = t3, find the slope \(\rm dy\over dx\) at t = 51. \(\rm \frac{4}{5}\)2. \(\rm \frac{4}{3}\)3. \(\rm \frac{12}{5}\)4. \(\rm \frac{4}{25}\)

Answer» Correct Answer - Option 4 : \(\rm \frac{4}{25}\)

Concept:

Parametric Form:

If f(x) and g(x) are the functions in x, then 

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

 

Calculation:

Given y = t2 + 2t

\(\rm {dy\over dt}\) = 2t + 2

Also x = t3

\(\rm dx\over dt\) = 3t2 

Now \(\rm dy\over dx\) = \(\rm \frac{dy\over dt}{dx\over dt}\) 

\(\rm dy\over dx\) = \(\rm \frac{2t +2}{3t^2}\)

At t = 5,

\(\rm {dy\over dx}\) = \(\rm \frac{2(5) +2}{3(5)^2}\) = \(\rm \frac{12}{75}\) = \(\rm \frac{4}{25}\)

94085.

Let the slope of the curve y = sin-1 (cos x) be tan θ, then the value of θ in the interval (0, π) is1. π/62. π/43. 3π/44. -π/4

Answer» Correct Answer - Option 3 : 3π/4

Concept:

Formula:

  • cos-1 (cos x) = x
  • sin-1 (sin x) = x
  • Slope of the curve = tan θ = \(\rm \frac {dy}{dx}\)

 

Calculation:

Given: Slope = tan θ

y = sin-1 (cos x)

⇒ y = sin-1 (sin (π/2 -x))

⇒ y = π/2 – x                                        (∵ sin-1 (sin x) = x)

Differentiating both sides with respect to x, we get

\(\rm \frac {dy}{dx}\) = -1

As we know, Slope = \(\rm \frac {dy}{dx}\)

⇒ tan θ  = -1

∴ θ = 3π/4

94086.

The tangent to the curve y2 = 16x and touches the curve at a point (1, 4). Find the distance of origin from the tangent1. \(\rm {1\over\sqrt5}\)2. \(\rm {2\over\sqrt5}\)3. \(\rm {3\over\sqrt5}\)4. \(\rm {4\over\sqrt5}\)

Answer» Correct Answer - Option 2 : \(\rm {2\over\sqrt5}\)

Concept:

The slope of the line tangent to the curve;

\(\rm {y_2-y_1\over x_2-x_1}=\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)_{at(x_1, y_1)}\) = m

Where (x1, y1) are the point on which tangent touches the curve and (x2, y2) is any other point on the tangent

Equation of the line with slope m and an on line points (x1, y1) is:

\(\rm (y-y_1) = m(x-x_1)\)

Distance of a point (p, q) from a line ax + by + c = 0:

D = \(\rm \left|{ap+bq+c\over\sqrt{a^2+b^2}}\right|\)

Calculation:

Given curve y2 = 16x

Differentiating w.r.t x on both sides

⇒ \(\rm \frac{\mathrm{d} }{\mathrm{d} x}y^2=\frac{\mathrm{d} }{\mathrm{d} x}16x\)

⇒ 2y \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}\) = 16

The point on which tangent touches the curve is (1, 4)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = {8\over 4}\) = 2

⇒ Slope of tangent m = 2

Now equation of the tangent: 

\(\rm (y-y_1) = m(x-x_1)\)

⇒ y - 4 = 2(x - 1) 

⇒ y - 2x - 2 = 0​

So, distance of the origin (0, 0) from normal is;

D = \(\rm \left|{ap+bq+c\over\sqrt{a^2+b^2}}\right|\)

⇒ D = \(\rm \left|{1\times0+(-2)\times0+(-2)\over\sqrt{1^2+(-2)^2}}\right|\) 

⇒ D = \(\boldsymbol{\rm \left|{-2\over\sqrt5}\right| = {2\over\sqrt5}}\)

94087.

Two plane parallel conducting plates `1.5xx10^(-2)m` apart are held horizontal one above the other in air. The upper plate is maintained at a positive potential of`1.5kV` while the other plate is earthed. Calculate the number of electrons which must be attached to a small oil drop of mass `4.8xx10^(-15)kg` between the plates of move the drop with constant speed. Neglect the density and viscosity of air.

Answer» Correct Answer - 3
`F=qE=-"ne" V/d=mg`
`n=(mgd)/(eV)=(4.8xx10^(-15)xx10xx1.5xx10^(-2))/(1.6xx10^(-19)xx1500)=3`
94088.

If the electric field in a region is given as `E=y^(2)hati+2xyhatj` and the potential is assumed `4` Volts at the origin, find the potential, in Volts, at the point `(2,1,9)`. All values are in SI units.

Answer» Correct Answer - 2
`V_((0,0,0))-V_((2,1,9))=int_((0,0,0))^((2,1,9)) y^(2)dx+2xydy=int_((0,0,0))^((2,1,9)) d(xy^(2))`
`=xy_(("0,0,0"))^(2^(("2,1,9")))=2`
94089.

A rod of length L and cross section area A has variable density according to the relation `rho (x)=rho_(0)+kx` for `0 le x le L/2` and `rho(x)=2x^(2)` for `L/2 le x le L` where `rho_(0)` and k are constants. Find the mass of the rod.

Answer» Correct Answer - `((7L^(3))/12+rho_(0)L/2+(kL^(2))/8)A`
`M_(1)=underset(0)overset(L//2)(int)A(rho_(0)+kx)dx=(rho_(0) L/2+(kL^(2))/8)A`
`M_(2)=underset(L//2)overset(L)(int)A(2x^(2)dx)=2/3[L^(3)-L^(3)/8]=(14L^(3))/24 A`
`M_("total")=M_(1)+M_(2)=((14L^(3))/24+rho_(0)L/2+(kL^(2))/8)A`
94090.

An earth satellite is revolving in a circular orbit of radius a withh a velocity `v_(0)`. A gun in the satellite is directly aimed toward earth. A bullet is fired from the gun with muzzle velocity `(v_(0))/2`. Find the ratio of distance of farthest and closest approach of bullet from centre of earth. (Assume that mass of the satellite is very-very large with respect to the mass of the bullet)

Answer» Correct Answer - 3
`v_(0)=sqrt((GM)/a)`
`mv_(0)a=mvr`
`1/2m(v_(0)^(2)+(v_(0)^(2))/4)-(GMm)/a=1/2mv^(2)-(GMm)/r`
Solving `(r_(max))/(r_(min))=(2a)/(2a//3)=3`
94091.

A sphere of radius `R` has a variable charge density `rho(r)=rho_(0)r^(n-1)(rleR)`. If ratio of "potential at centre" and "potential at surface" be `3//2` then find the value of `n`

Answer» Correct Answer - 1
Method 1: If `rho(r)=rho_(0)(="constant")` then we know that required `(3/2(KQ)/R)/((KQ)/R)=3/2`
Hence `n=1`
Method 2: You may use integration method also to solve this problem.
94092.

The electric field in a region is radially outwards with magnitude `E=alphar//epsilon_(0)`. IN a sphere of radius `R` centered at the origin, calculate the value of charge in coulombs `alpha=5/(pi)` if `V//m^(2)` and `R=(3/10)^(1//3)m`.

Answer» Correct Answer - 6
`q=epsilon_(0)ointE.ds=epsilon_(0)(alphar)/(epsilon_(0))4pir^(2)=4pir^(3)alpha=6`.
94093.

The coordinates of focus of a parabola which touches `x=1, y=1, x+2y=2 and y=2x+ 2` areA. `((-(4)/(3),-(13)/(3)))`B. `(-(22)/(31),-(7)/(3))`C. `(-(1)/(20),(17)/(20))`D. none of these

Answer» Correct Answer - C
Let (a,b) be the foucs of the parabola then mirror image of this focus on the above tangents lie on directrix and directrix passes through `(0,0)`.
then we get focus `(-(1)/(20),(17)/(20))`
94094.

If `alpha,beta` are the roots of the equation `x^(2)+4x+p=0,` where `=Sigma_(r-0)^(n) ""^(n)C_(r)(1+rt)/((1+nt)^(r ))(-1)^(r )`, then `|alpha-beta|` is equal to:

Answer» Correct Answer - D
`p=underset(r=0)overset(n)Sigma.^(n)(1+rt)/(1+nt)^(r) (-1)^(r)=`
`underset(r=0)overset(n)Sigma.^(n)C_(r)((-1)/(1+nt))^(r)+ntunderset(r=0)overset(n)Sigma.^(n-1)C_(r-1)((-1)/(1+nt))^(r)`
`=(1-(1)/(1+nt))^(n)-(nt)/(1+nt)underset(r=0)overset(n)sum.^(n-1)C_(r-1)((-1)/(1+nt))^(r-1)`
`=((nt)/(1+nt))^(n)-(nt)/(1+nt)(1-(1)/(1+nt))^(n-1)=0`
`implies x^(2)+4 x=0 implies x=0,-4implies|alpha-beta|=4`
94095.

Consider a hyperbola H : `x^(2)-y^(2)` =k and a parabola `P:y=x^(2)` then identify the correct statements(S)A. If point of intrsections of P and H are concyclic then `k lt 2`B. If P and H touch each other then `k = 1//4`C. If `k=-1//3` and `m_(1)`, are the slopes of common tangents to P and H then `(3m_(1).^(2)+8)(3m_(2).^(2)+8)=112`D. If P,H do not touch but intersect at exactly two points then `k lt 0`

Answer» Correct Answer - A::B::C::D
`y-y^(2)=k`
for concyclic `D ge 0` ge
for common tangent
`y=mx pm sqrt(km^(2)-k)` and `y = mx-(1)/(4)m^(2)`
comparing `(3m_(1).^(2)+8)(3m_(2).^(2)+8)=112`
94096.

Let two paraboles have a common axis where focus of each being exterior to the other and lt `l_(1),l_(2)` be their latus rectums then the locus of the mid points of the intercepts between the parabolas made on the lines parallel to the common axis is aA. Straight line if `l_(1)=l_(2)AAl_(1),l_(2)in R^(+)`B. Parabola if `l_(1)nel_(2)AAl_(1),l_(2)inR^(+)`C. Parabola`AAl_(1),l_(2) in R^(+)`D. Hyperbola if `l_(1)=2l_(2)AAl_(1),l_(2)in R^(+)`

Answer» Correct Answer - A::B
`p_(1):y^(2)=l_(1)x`
`P_(2):y^(2)=-l_(2)(x-b)`
point on `P_(2)(y^(2)/(l_(1)),y)` and `P_(2) (b-y^(2)/(l_(2)),y)`
find locus of midpoint of above points
94097.

Let AB be a chord of the parabola `x^(2) = 4y`. A circle drawn with AB as diameter passes through the vertex C of the parabola. If area of `DeltaABC = 20 sq`. units then the coordinates of A can beA. `(6,9)`B. `(-6,9)`C. `(-8,16)`D. `(8,16)`

Answer» Correct Answer - C::D
`A(2t_(1),t_(1).^(2))B(2t_(2),t_(2).^(2))`
`t_(1)t_(2) = - 4`
Area `|(2t_(1),t_(1).^(2),1),(2t_(2),t_(2).^(2),1),(0,0,1)|=20`
solving `t_(1)=pm4,pm1`
94098.

`x^(2)+y^(2)-4x-4y+7=0` and `x^(2)+y^(2)4x+4y+7=0` are equation of excircle opposite to B and C respectively in a `DeltaABC`, thenA. `A = (0,0)`B. `AB=AC`C. `DeltaABC"can be equilateral"`D. `DeltaABC "can not be equilateral"`

Answer» Correct Answer - A::B::D
`O_(1)=(2,2),r_(1)=1`
`O_(2)=(-2,-2),r_(2)=1`
`impliesA=(0,0)`
94099.

A number is chosen at random from the number 10to99. By seeing the number a man will laugh if product of the digits is 12. If he choose three number with replacement then the probability that he will laugh at least once isA. `1-((3)/(5))^(3)`B. `(2(45^(2)+43^(2)-45xx43))/(45^(3)`C. `(98)/(125)`D. `1-((43)/(45))^(3)`

Answer» Correct Answer - D
There can be four such numbers i.e 43,34,62,26.
Whose product of digit is 12
implies Probability that the man will laugh by seeing the chosen numbers `=(4)/(90)=(2)/(45)`
implies required probability
`=1-(1-(2)/(45))^(3)=1-((43)/45)^(3)`
94100.

If the circle `C_(1)` touches x-axis and the line `y=xtantheta(tanthetagt0)` in first quadrant and circle`C_(2)` touches the `y=xtantheta` at the same point at which `C_(1)` touches it such that ratio of radius of `C_(1)` and `C_(2)` is 2:1, then `tan(theta)/(2)=sqrt(a-B)/(c)` where a,b,c,epsilon N and `HCF(b,c)=1`A. `a=13`B. `b=3`C. `c=2`D. `a=17`

Answer» Correct Answer - B::C::D
Let `m=tantheta` ,then
`angleAOO_(1)=(theta)/(2),angleO_(2)OO_(1)=45^(@)`
`impliesangleAOO_(2)=45^(@)-(theta)/(2)`
`tan,(theta)/(2)=(r_(1))/(OA)`
`tan (45^(@)-(theta)/(2))=(r^(2))/(OA)`
`(1)/(tan.(theta)/(2))((1-tan.(theta)/(2)))/((1+tan.(theta)/(2)))=(r_(2))/(r_(1))=(1)/(2)`
`impliestan(theta)/(2)=(-3+sqrt(17))/(2)`