Maths-
General
Easy

Question

Assertion (A): If the system of equations 3 x minus 2 y plus z equals 0 comma lambda x minus 14 y plus 15 z equals 0 comma x plus 2 y minus 3 z equals 0 have non zero solution then lambda equals 5
Reason (R): If the system of equations A X equals 0 has a non zero solution then bold italic A is singular

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true and R is not correct explanation of A
  3. A is true but R is false
  4. A is false but R is true

The correct answer is: Both A and R are true and R is the correct explanation of A

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Related Questions to study

General
physics-

Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulseJ equals M v is impared to the body at one of its ends, what would be its angular velocity?

Let omega be the angular velocity of the rod. Applying ,
Angular impulse = change in angular momentum about centre of mass of the system

J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega
therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega
therefore omega equals fraction numerator v over denominator L end fraction

Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulseJ equals M v is impared to the body at one of its ends, what would be its angular velocity?

physics-General
Let omega be the angular velocity of the rod. Applying ,
Angular impulse = change in angular momentum about centre of mass of the system

J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega
therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega
therefore omega equals fraction numerator v over denominator L end fraction
General
physics-

A binary star consists of two stars A (mass 2.2 M subscript s end subscript) and B (mass 11 M subscript s end subscript), where M subscript s end subscript is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction (as omega will be same in both cases)
equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1
equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1 (as m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis
equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis
equals 6
therefore The correct answer is 6.

A binary star consists of two stars A (mass 2.2 M subscript s end subscript) and B (mass 11 M subscript s end subscript), where M subscript s end subscript is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

physics-General
fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction (as omega will be same in both cases)
equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1
equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1 (as m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis
equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis
equals 6
therefore The correct answer is 6.
General
physics-

A force of- F stack k with hat on top acts on O, the origin of the coordinate system. The torque about the point 1, -(a) is


tau equals r cross times F
tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis
equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket
equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket

A force of- F stack k with hat on top acts on O, the origin of the coordinate system. The torque about the point 1, -(a) is

physics-General

tau equals r cross times F
tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis
equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket
equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket
General
Maths-

The area of the region bounded by y=|x-1| and y=1 in sq. units is

The area of the region bounded by y=|x-1| and y=1 in sq. units is

Maths-General
General
Maths-

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

Maths-General
General
physics-

A solid sphere of radius R has moment of inertia Iabout its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent
or fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent
therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R

A solid sphere of radius R has moment of inertia Iabout its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to

physics-General
fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent
or fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent
therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R
General
maths-

The area of the region bounded by y=1 plus open vertical bar x close vertical bar and the x-axis is

The area of the region bounded by y=1 plus open vertical bar x close vertical bar and the x-axis is

maths-General
General
maths-

Area of the region bounded by yequals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank is

Area of the region bounded by yequals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank is

maths-General
General
Maths-

Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

STEP BY STEP SOLUTION
Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

Areablank equals fraction numerator 1 over denominator 2 end fraction.4.2
equals 4 blanksq. units

Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

Maths-General
STEP BY STEP SOLUTION
Area of the region bounded by y=open vertical bar x close vertical bar blankand y=2 is

Areablank equals fraction numerator 1 over denominator 2 end fraction.4.2
equals 4 blanksq. units
General
maths-

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

maths-General
General
physics-

A Tjoint is formed by two identical rods A and Beach of mass m and length L in the X Yplane as shown. Its moment of inertia about axis coinciding with A is

I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

A Tjoint is formed by two identical rods A and Beach of mass m and length L in the X Yplane as shown. Its moment of inertia about axis coinciding with A is

physics-General
I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction
General
Maths-

The area bounded by the parabolablank x to the power of 2 end exponent=4ay, x-axis and the straight-line y=2a is

The area bounded by the parabolablank x to the power of 2 end exponent=4ay, x-axis and the straight-line y=2a is

Maths-General
General
physics-

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

Moment of inertia of the system about the centre of plane is given by
I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
equals 1.004 plus 2.008 plus 3.012 plus 4.016
equals 10.04 blank k g minus m to the power of 2 end exponent

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

physics-General
Moment of inertia of the system about the centre of plane is given by
I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets
equals 1.004 plus 2.008 plus 3.012 plus 4.016
equals 10.04 blank k g minus m to the power of 2 end exponent
General
physics-

For the given uniform square lamina A B C D, whose centre is O

Let the each side of square lamina is d.
So, I subscript E F end subscript equals I subscript G H end subscript (due to symmetry)
And I subscript A C end subscript equals I subscript B D end subscript (due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript
rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript (i)
and I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript
rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript (ii)
From Eqs. (i) and (ii), we get
I subscript A C end subscript equals I subscript E F end subscript
therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction
equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses
So, I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

For the given uniform square lamina A B C D, whose centre is O

physics-General
Let the each side of square lamina is d.
So, I subscript E F end subscript equals I subscript G H end subscript (due to symmetry)
And I subscript A C end subscript equals I subscript B D end subscript (due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript
rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript (i)
and I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript
rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript (ii)
From Eqs. (i) and (ii), we get
I subscript A C end subscript equals I subscript E F end subscript
therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction
equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses
So, I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript
General
chemistry-

Which is correct about the change given below

Which is correct about the change given below

chemistry-General