Maths-

#### The equation of the plane containing the line where al + bm + cn is equal to

Maths-General

- -1
- 2
- 0
- 1

#### Answer:The correct answer is: 0 Since these two lines are intersecting so shortest distance between the lines will be 0.

Hence (c) is the correct answer.

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#### A point *P* moves in counter-clockwise direction on a circular path as shown in the figure. The movement of *P* is such that it sweeps out length where is in metre and *t *is in second. The radius of the path is 20 m. The acceleration of *P* when *t* =2s is nearly

#### A point *P* moves in counter-clockwise direction on a circular path as shown in the figure. The movement of *P* is such that it sweeps out length where is in metre and *t *is in second. The radius of the path is 20 m. The acceleration of *P* when *t* =2s is nearly

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#### A piece of wire is bent in the shape of a parabola $y=k{x}^{2}(y$-axis vertical) with a bead of mass $m$ 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 $x$-axis with a constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$-axis is

$ma\mathrm{cos}\theta =mg\mathrm{cos}(90-\theta )$
$\Rightarrow \frac{a}{g}=\mathrm{tan}\theta \Rightarrow \frac{a}{g}=\frac{dy}{dx}$
$\Rightarrow \frac{d}{dx}{\left(kx\right)}^{2}=\frac{a}{g}\Rightarrow x=\frac{a}{2gk}$

#### A piece of wire is bent in the shape of a parabola $y=k{x}^{2}(y$-axis vertical) with a bead of mass $m$ 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 $x$-axis with a constant acceleration $a$. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$-axis is

physics-General

$ma\mathrm{cos}\theta =mg\mathrm{cos}(90-\theta )$
$\Rightarrow \frac{a}{g}=\mathrm{tan}\theta \Rightarrow \frac{a}{g}=\frac{dy}{dx}$
$\Rightarrow \frac{d}{dx}{\left(kx\right)}^{2}=\frac{a}{g}\Rightarrow x=\frac{a}{2gk}$

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

physics-General

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 small body of mass slides down from the top of a hemisphere of radius . The surface of block and hemisphere are frictionless. The height at which the body lose contact with the surface of the sphere is

#### A small body of mass slides down from the top of a hemisphere of radius . The surface of block and hemisphere are frictionless. The height at which the body lose contact with the surface of the sphere is

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maths-

#### The shortest distance between the two straight line$\frac{x-4/3}{2}=\frac{y+6/5}{3}=\frac{z-3/2}{4}$ and $\frac{5y+6}{8}=\frac{2z-3}{9}=\frac{3x-4}{5}$ is

#### The shortest distance between the two straight line$\frac{x-4/3}{2}=\frac{y+6/5}{3}=\frac{z-3/2}{4}$ and $\frac{5y+6}{8}=\frac{2z-3}{9}=\frac{3x-4}{5}$ is

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maths-

#### The value of ${\int}_{0}^{100}\u200a\left\{\sqrt{x}\right\}dx$ (where {x} is the fractional part of x) is

#### The value of ${\int}_{0}^{100}\u200a\left\{\sqrt{x}\right\}dx$ (where {x} is the fractional part of x) is

maths-General

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 x-axis as 0.1 + 1.0 + 1.0 = 0
but normal to the plane will be perpendicular to x-axis.
Hence (c) is the correct answer.

#### The value of ${\int}_{0}^{1}\u200a|\mathrm{sin}2\pi x|\mid dx$ is equal to

maths-General

Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0
but normal to the plane will be perpendicular to x-axis.
Hence (c) is the correct answer.

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

maths-General

Since these two lines are intersecting so shortest distance between the lines will be 0.
Hence (c) is the correct answer.

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 $\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}$.
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

maths-General

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 $\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}$.
Hence (c) is the correct choice.

maths-

maths-General

physics-

#### Average torque on a projectile of mass , initial speed and angles of projection , between initial and final position and as shown in figure about the point of projection is

Time of flight.

Horizontal range,

Change in angular momentum,

about point of projection

Horizontal range,

Change in angular momentum,

about point of projection

#### Average torque on a projectile of mass , initial speed and angles of projection , between initial and final position and as shown in figure about the point of projection is

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Time of flight.

Horizontal range,

Change in angular momentum,

about point of projection

Horizontal range,

Change in angular momentum,

about point of projection

maths-

(where a, b are integers) =

The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x

= 1, = 2, = –3

\ Required point is (0, 5, 7).

Hence (c) is the correct answer.

_{1}, y_{1}, z_{1}), then= 1, = 2, = –3

\ Required point is (0, 5, 7).

Hence (c) is the correct answer.

(where a, b are integers) =

maths-General

The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x

= 1, = 2, = –3

\ Required point is (0, 5, 7).

Hence (c) is the correct answer.

_{1}, y_{1}, z_{1}), then= 1, = 2, = –3

\ Required point is (0, 5, 7).

Hence (c) is the correct answer.

maths-

The foot of the perpendicular on the line drown from the origin is C if the line cuts the x-axis and y-axis 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 x-axis and y-axis at A and B respectively then BC : CA is

maths-General

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

=

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

physics-General

physics-

#### Three balls are dropped from the top of a building with equal speed at different angles. When the balls strike ground their velocities are and respectively, then

All the balls are projected from the same height, therefore their velocities will be equal.

#### Three balls are dropped from the top of a building with equal speed at different angles. When the balls strike ground their velocities are and respectively, then

physics-General

All the balls are projected from the same height, therefore their velocities will be equal.