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Physics-

General

Easy

Question

Frequency ranges for micro waves are :

$3 \times 10^{9} to 3 \times 10^{4} H z$
$3 \times 10^{13} to 3 \times 10^{9} H z$
$3 \times 10^{14} to 3 \times 10^{9} H z$
$3 \times 10^{11} to 3 \times 10^{9} H z$

The correct answer is: $3 cross times 10 to the power of 13 end exponent text to end text 3 cross times 10 to the power of 9 end exponent H z$

Related Questions to study

The higher frequency TV broad casting bands range is

The higher frequency TV broad casting bands range is

physics-General

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure, then tension in the string is

T sin invisible function application theta equals M omega to the power of 2 end exponent R

T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta

(ii)
From (i) and (ii)

T equals M omega to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure, then tension in the string is

Physics-General

T sin invisible function application theta equals M omega to the power of 2 end exponent R

T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta

(ii)
From (i) and (ii)

T equals M omega to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

$cos invisible function application left parenthesis alpha minus beta right parenthesis equals 1 text and end text cos invisible function application left parenthesis alpha plus beta right parenthesis equals 1 over c$ where $alpha comma beta element of left square bracket negative pi comma pi right square bracket$ Pairs $alpha comma beta$ which satisfy both the equations is/are

$cos invisible function application left parenthesis alpha minus beta right parenthesis equals 1 text and end text cos invisible function application left parenthesis alpha plus beta right parenthesis equals 1 over c$ where $alpha comma beta element of left square bracket negative pi comma pi right square bracket$ Pairs $alpha comma beta$ which satisfy both the equations is/are

parallel

The number of integral values of k for which the equation $7 cos invisible function application x plus 5 sin invisible function application x equals 2 k plus 1$ has a solution is

The number of integral values of k for which the equation $7 cos invisible function application x plus 5 sin invisible function application x equals 2 k plus 1$ has a solution is

The number of distinct real roots of $open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row cell sin invisible function application x end cell cell cos invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell sin invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell cos invisible function application x end cell cell sin invisible function application x end cell end table close vertical bar equals 0$ in the interval $negative pi over 4 less or equal than x less or equal than pi over 4$ is

The number of distinct real roots of $open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row cell sin invisible function application x end cell cell cos invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell sin invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell cos invisible function application x end cell cell sin invisible function application x end cell end table close vertical bar equals 0$ in the interval $negative pi over 4 less or equal than x less or equal than pi over 4$ is

If $0 less or equal than x less than 2 pi$ , then the number of real values of x, which satisfy the equation $cos invisible function application x plus cos invisible function application 2 x plus cos invisible function application 3 x plus cos invisible function application 4 x equals 0$ is :

If $0 less or equal than x less than 2 pi$ , then the number of real values of x, which satisfy the equation $cos invisible function application x plus cos invisible function application 2 x plus cos invisible function application 3 x plus cos invisible function application 4 x equals 0$ is :

parallel

The number of solutions of the equation $cos squared invisible function application open parentheses x plus pi over 6 close parentheses plus cos squared invisible function application x minus 2 cos invisible function application open parentheses x plus pi over 6 close parentheses cos invisible function application open parentheses pi over 6 close parentheses equals sin squared invisible function application pi over 6$ in the interval $left parenthesis negative pi divided by 2 comma pi divided by 2 right parenthesis$ is

The number of solutions of the equation $cos squared invisible function application open parentheses x plus pi over 6 close parentheses plus cos squared invisible function application x minus 2 cos invisible function application open parentheses x plus pi over 6 close parentheses cos invisible function application open parentheses pi over 6 close parentheses equals sin squared invisible function application pi over 6$ in the interval $left parenthesis negative pi divided by 2 comma pi divided by 2 right parenthesis$ is

The possible values of $theta element of left parenthesis 0 comma pi right parenthesis$ such that $sin invisible function application theta plus sin invisible function application 4 theta plus sin invisible function application 7 theta equals 0$ are :

The possible values of $theta element of left parenthesis 0 comma pi right parenthesis$ such that $sin invisible function application theta plus sin invisible function application 4 theta plus sin invisible function application 7 theta equals 0$ are :

The number of values of x in the interval [0, 3 $straight pi$ ] satisfying the equation , $2 sin squared invisible function application x plus 5 sin invisible function application x minus 3 equals 0$ is

The number of values of x in the interval [0, 3 $straight pi$ ] satisfying the equation , $2 sin squared invisible function application x plus 5 sin invisible function application x minus 3 equals 0$ is

parallel

The number of solutions of tan x + sec x = 2 cos x in [0, 2 $straight pi$ ], is

The number of solutions of tan x + sec x = 2 cos x in [0, 2 $straight pi$ ], is

Over $left square bracket negative pi comma pi right square bracket$ the solution of $2 sin squared invisible function application open parentheses x plus pi over 4 close parentheses plus square root of 3 cos invisible function application 2 x greater or equal than 0$ is

Over $left square bracket negative pi comma pi right square bracket$ the solution of $2 sin squared invisible function application open parentheses x plus pi over 4 close parentheses plus square root of 3 cos invisible function application 2 x greater or equal than 0$ is

Solution to inequality cos 2x + 5 cos x + 3 ≥ 0 over $left square bracket negative pi comma pi right square bracket$ is

Solution to inequality cos 2x + 5 cos x + 3 ≥ 0 over $left square bracket negative pi comma pi right square bracket$ is

parallel

Let $f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction$ The sum of all the solutions of f (x) = 0 in [0, 100p] is

table row cell table row blank cell f left parenthesis x right parenthesis equals fraction numerator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent over denominator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses plus 1 minus c o t invisible function application x plus c o t invisible function application x end fraction end cell row blank cell text  defined for  end text R minus n pi comma n element of I end cell row blank cell f left parenthesis x right parenthesis equals fraction numerator left parenthesis c o t invisible function application x right parenthesis to the power of 4 end exponent over denominator 1 plus c o t to the power of 2 end exponent invisible function application x end fraction end cell row blank cell f left parenthesis x right parenthesis equals 0 rightwards double arrow c o t invisible function application x equals 0 rightwards double arrow x equals left parenthesis 2 n minus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell row blank cell therefore x equals fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction comma fraction numerator 5 pi over denominator 2 end fraction comma horizontal ellipsis horizontal ellipsis comma fraction numerator 199 pi over denominator 2 end fraction end cell row blank cell s u m equals fraction numerator pi over denominator 2 end fraction left square bracket 1 plus 3 plus 5 plus horizontal ellipsis horizontal ellipsis plus 199 right square bracket end cell end table end cell row cell text end text text ( end text text 100 end text text end text text s end text text o end text text l end text text u end text text t end text text i end text text o end text text n end text text s end text text ) end text text end text end cell row cell equals fraction numerator pi over denominator 2 end fraction times fraction numerator 100 over denominator 2 end fraction times 200 equals 5000 pi end cell end table

Let $f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction$ The sum of all the solutions of f (x) = 0 in [0, 100p] is

table row cell table row blank cell f left parenthesis x right parenthesis equals fraction numerator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent over denominator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses plus 1 minus c o t invisible function application x plus c o t invisible function application x end fraction end cell row blank cell text  defined for  end text R minus n pi comma n element of I end cell row blank cell f left parenthesis x right parenthesis equals fraction numerator left parenthesis c o t invisible function application x right parenthesis to the power of 4 end exponent over denominator 1 plus c o t to the power of 2 end exponent invisible function application x end fraction end cell row blank cell f left parenthesis x right parenthesis equals 0 rightwards double arrow c o t invisible function application x equals 0 rightwards double arrow x equals left parenthesis 2 n minus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell row blank cell therefore x equals fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction comma fraction numerator 5 pi over denominator 2 end fraction comma horizontal ellipsis horizontal ellipsis comma fraction numerator 199 pi over denominator 2 end fraction end cell row blank cell s u m equals fraction numerator pi over denominator 2 end fraction left square bracket 1 plus 3 plus 5 plus horizontal ellipsis horizontal ellipsis plus 199 right square bracket end cell end table end cell row cell text end text text ( end text text 100 end text text end text text s end text text o end text text l end text text u end text text t end text text i end text text o end text text n end text text s end text text ) end text text end text end cell row cell equals fraction numerator pi over denominator 2 end fraction times fraction numerator 100 over denominator 2 end fraction times 200 equals 5000 pi end cell end table

The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n $element of$ I

We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x

x equals fraction numerator n pi over denominator 3 end fraction comma n element of I

The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n $element of$ I

We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x

x equals fraction numerator n pi over denominator 3 end fraction comma n element of I

The sum of all the solution of cotθ= sin 2θ(θ ≠ n $straight pi$ , n integer), 0 , $less or equal than theta less or equal than pi$ is

For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0

table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table

Then the sum of roots is 3p/2

The sum of all the solution of cotθ= sin 2θ(θ ≠ n $straight pi$ , n integer), 0 , $less or equal than theta less or equal than pi$ is

For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0

table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table

Then the sum of roots is 3p/2

parallel

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