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Maths-

General

Easy

Question

If $stack a with minus on top$ and $stack b with minus on top$ are two unit vectors and $ϕ$ is the angle between them, then $fraction numerator 1 over denominator 2 end fraction vertical line stack a with minus on top minus stack b with minus on top vertical line$ is equal to

0
$π / 2$
$| s i n (ϕ / 2) |$
$| c o s (ϕ / 2) |$

The correct answer is: $vertical line s i n invisible function application left parenthesis ϕ divided by 2 right parenthesis vertical line$

$vertical line stack a with minus on top minus stack b with minus on top vertical line to the power of 2 end exponent equals vertical line minus vertical line to the power of 2 end exponent plus vertical line stack b with minus on top vertical line to the power of 2 end exponent minus 2 vertical line stack a with minus on top vertical line vertical line stack b with rightwards arrow on top vertical line c o s invisible function application ϕ$
$equals 1 plus 1 minus 2 left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis c o s invisible function application ϕ$
$equals 2 left parenthesis 1 minus c o s invisible function application ϕ right parenthesis$
$equals 2 open parentheses 2 s i n to the power of 2 end exponent invisible function application ϕ divided by 2 close parentheses$
$equals 4 s i n to the power of 2 end exponent invisible function application ϕ divided by 2$
$equals vertical line stack a with minus on top minus stack b with minus on top vertical line equals 2 s i n invisible function application ϕ divided by 2$
$equals fraction numerator open vertical bar table row cell minus end cell row cell a minus b end cell end table close vertical bar over denominator 2 end fraction equals s i n invisible function application ϕ divided by 2$

Related Questions to study

Statement-1:The area of the triangle formed by the points A(20,22) ;B(21,24) and C(22,23) is the same as the area of the triangle formed by the point P(0,0) ;Q(1,2) and R(2,1) . Because
Statement-2: The area of the triangle is invariant w.r.t translation of the coordinate axes.

Statement-1:The area of the triangle formed by the points A(20,22) ;B(21,24) and C(22,23) is the same as the area of the triangle formed by the point P(0,0) ;Q(1,2) and R(2,1) . Because
Statement-2: The area of the triangle is invariant w.r.t translation of the coordinate axes.

$stack a with minus on top equals stack i with minus on top plus stack j with minus on top plus stack k with minus on top comma stack b with minus on top equals 2 stack i with minus on top plus 3 stack j with minus on top plus 5 stack k with minus on top$ and $stack c with minus on top$ is a unit vector. The maximum value of the scalar triple product $left square bracket stack a b c with bar on top right square bracket$ is

$stack a with minus on top equals stack i with minus on top plus stack j with minus on top plus stack k with minus on top comma stack b with minus on top equals 2 stack i with minus on top plus 3 stack j with minus on top plus 5 stack k with minus on top$ and $stack c with minus on top$ is a unit vector. The maximum value of the scalar triple product $left square bracket stack a b c with bar on top right square bracket$ is

In metals the time of relaxation of electrons

In metals the time of relaxation of electrons

Physics-General

parallel

$P left parenthesis stack p with minus on top right parenthesis text end text$ and $Q left parenthesis stack q with minus on top right parenthesis$ are the position vectors of two fixed points and $R left parenthesis stack r with minus on top right parenthesis$ is the position vector of a variable point. If R moves such that $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis cross times left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$ =0 then the locus of R is

$P left parenthesis stack p with minus on top right parenthesis text end text$ and $Q left parenthesis stack q with minus on top right parenthesis$ are the position vectors of two fixed points and $R left parenthesis stack r with minus on top right parenthesis$ is the position vector of a variable point. If R moves such that $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis cross times left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$ =0 then the locus of R is

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

Given : $cos invisible function application xcos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x close parentheses equals fraction numerator 1 over denominator 4 end fraction comma x element of open square brackets 0 , 6 pi close square brackets$
Sum of all solutions is

Given : $cos invisible function application xcos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x close parentheses equals fraction numerator 1 over denominator 4 end fraction comma x element of open square brackets 0 , 6 pi close square brackets$
Sum of all solutions is

parallel

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If $fraction numerator 3 plus i 2 s i n invisible function application theta over denominator 1 minus i 2 s i n invisible function application theta end fraction$ is real , then

This section contains 3 multiple choice questions numbered 14 to 16 .Each question has 4 choices (A), (B),(c) , and (D), out of which ONE or MORE may be correct
If $fraction numerator 3 plus i 2 s i n invisible function application theta over denominator 1 minus i 2 s i n invisible function application theta end fraction$ is real , then

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Physics-General

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

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Physics-General

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Figure shows four charges q₁, q₂, q₃ and q₄ fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q₁, q₂, q₃ and q₄ is

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A Gaussian surface S enclosed two charges q₁ = q, q₂ = -q and q₃ = q. The net flux passing through S is

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Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration is zero, if Q is equal to

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