General
Easy
Maths-

If n element of N comma and theperiod of fraction numerator c o s invisible function application n x over denominator s i n invisible function application open parentheses fraction numerator x over denominator n end fraction close parentheses end fraction is 4 pi, then n equals

Maths-General

  1. 2    
  2. 4    
  3. 1    
  4. 3    

    Answer:The correct answer is: 2text L.C.M. of  end text open parentheses fraction numerator 2 pi over denominator n end fraction comma 2 pi n close parentheses equals 4 pi

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    Related Questions to study

    General
    physics-

    The adjoining diagram shows the spectral energy density distribution E subscript lambda end subscript of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is

    fraction numerator A subscript T end subscript over denominator A subscript 2000 end subscript end fraction equals fraction numerator 16 over denominator 1 end fraction [Given]
    Area under e subscript lambda end subscript minus lambda curve represents the emissive power of body and emissive power proportional to T to the power of 4 end exponent
    rightwards double arrow fraction numerator A subscript T end subscript over denominator A subscript 2000 end subscript end fraction equals open parentheses fraction numerator T over denominator 2000 end fraction close parentheses to the power of 4 end exponent rightwards double arrow fraction numerator 16 over denominator 1 end fraction equals open parentheses fraction numerator T over denominator 2000 end fraction close parentheses to the power of 4 end exponent rightwards double arrow T equals 4000 K

    The adjoining diagram shows the spectral energy density distribution E subscript lambda end subscript of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is

    physics-General
    fraction numerator A subscript T end subscript over denominator A subscript 2000 end subscript end fraction equals fraction numerator 16 over denominator 1 end fraction [Given]
    Area under e subscript lambda end subscript minus lambda curve represents the emissive power of body and emissive power proportional to T to the power of 4 end exponent
    rightwards double arrow fraction numerator A subscript T end subscript over denominator A subscript 2000 end subscript end fraction equals open parentheses fraction numerator T over denominator 2000 end fraction close parentheses to the power of 4 end exponent rightwards double arrow fraction numerator 16 over denominator 1 end fraction equals open parentheses fraction numerator T over denominator 2000 end fraction close parentheses to the power of 4 end exponent rightwards double arrow T equals 4000 K
    General
    physics-

    Five identical rods are joined as shown in figure. Point A and C are maintained at temperature 120 ℃ and 20 ℃ respectively. The temperature of junction B will be

    If thermal resistance of each rod is considered R then, the given combination can be redrawn as follows

    open parentheses H e a t blank c u r r e n t close parentheses subscript A C end subscript equals open parentheses H e a t blank c u r r e n t close parentheses subscript A B end subscript
    fraction numerator left parenthesis 120 minus 20 right parenthesis over denominator 2 R end fraction equals fraction numerator left parenthesis 120 minus theta right parenthesis over denominator R end fraction rightwards double arrow theta equals 70 ℃

    Five identical rods are joined as shown in figure. Point A and C are maintained at temperature 120 ℃ and 20 ℃ respectively. The temperature of junction B will be

    physics-General
    If thermal resistance of each rod is considered R then, the given combination can be redrawn as follows

    open parentheses H e a t blank c u r r e n t close parentheses subscript A C end subscript equals open parentheses H e a t blank c u r r e n t close parentheses subscript A B end subscript
    fraction numerator left parenthesis 120 minus 20 right parenthesis over denominator 2 R end fraction equals fraction numerator left parenthesis 120 minus theta right parenthesis over denominator R end fraction rightwards double arrow theta equals 70 ℃
    General
    physics-

    One end of a thermally insulated rod is kept at a temperature T subscript 1 end subscript and other at T subscript 2 end subscript. The rod is composed of two sections of lengths l subscript 1 end subscript and l subscript 2 end subscript and thermal conductivities K subscript 1 end subscript and K subscript 2 end subscript respectively. The temperature at the interface of the two sections is

    Let temperature at the interface is T.
    For part AB,

    fraction numerator Q subscript 1 end subscript over denominator t end fraction proportional to fraction numerator open parentheses T subscript 1 end subscript minus T close parentheses K subscript 1 end subscript over denominator l subscript 1 end subscript end fraction
    For part B C comma
    fraction numerator Q subscript 2 end subscript over denominator t end fraction proportional to fraction numerator open parentheses T minus T subscript 2 end subscript close parentheses K subscript 2 end subscript over denominator l subscript 2 end subscript end fraction
    At equilibrium, fraction numerator Q subscript 1 end subscript over denominator t end fraction equals fraction numerator Q subscript 2 end subscript over denominator t end fraction
    therefore fraction numerator open parentheses T subscript 1 end subscript minus T close parentheses K subscript 1 end subscript over denominator l subscript 1 end subscript end fraction equals fraction numerator open parentheses T minus T subscript 2 end subscript close parentheses K subscript 2 end subscript over denominator l subscript 2 end subscript end fraction
    rightwards double arrow T equals fraction numerator T subscript 1 end subscript K subscript 1 end subscript l subscript 2 end subscript plus T subscript 2 end subscript K subscript 2 end subscript l subscript 1 end subscript over denominator K subscript 1 end subscript l subscript 2 end subscript plus K subscript 2 end subscript l subscript 1 end subscript end fraction

    One end of a thermally insulated rod is kept at a temperature T subscript 1 end subscript and other at T subscript 2 end subscript. The rod is composed of two sections of lengths l subscript 1 end subscript and l subscript 2 end subscript and thermal conductivities K subscript 1 end subscript and K subscript 2 end subscript respectively. The temperature at the interface of the two sections is

    physics-General
    Let temperature at the interface is T.
    For part AB,

    fraction numerator Q subscript 1 end subscript over denominator t end fraction proportional to fraction numerator open parentheses T subscript 1 end subscript minus T close parentheses K subscript 1 end subscript over denominator l subscript 1 end subscript end fraction
    For part B C comma
    fraction numerator Q subscript 2 end subscript over denominator t end fraction proportional to fraction numerator open parentheses T minus T subscript 2 end subscript close parentheses K subscript 2 end subscript over denominator l subscript 2 end subscript end fraction
    At equilibrium, fraction numerator Q subscript 1 end subscript over denominator t end fraction equals fraction numerator Q subscript 2 end subscript over denominator t end fraction
    therefore fraction numerator open parentheses T subscript 1 end subscript minus T close parentheses K subscript 1 end subscript over denominator l subscript 1 end subscript end fraction equals fraction numerator open parentheses T minus T subscript 2 end subscript close parentheses K subscript 2 end subscript over denominator l subscript 2 end subscript end fraction
    rightwards double arrow T equals fraction numerator T subscript 1 end subscript K subscript 1 end subscript l subscript 2 end subscript plus T subscript 2 end subscript K subscript 2 end subscript l subscript 1 end subscript over denominator K subscript 1 end subscript l subscript 2 end subscript plus K subscript 2 end subscript l subscript 1 end subscript end fraction
    General
    physics-

    The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along B D. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L divided by C is

    Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2 blank m i n equals m C left parenthesis 90 minus 80 right parenthesis
    In 4 blank m i n comma blankheat supplied equals 2 m C open parentheses 90 minus 80 close parentheses
    therefore 2 m blank C open parentheses 90 minus 80 close parentheses equals m L rightwards double arrow fraction numerator L over denominator C end fraction equals 20

    The figure given below shows the cooling curve of pure wax material after heating. It cools from A to B and solidifies along B D. If L and C are respective values of latent heat and the specific heat of the liquid wax, the ratio L divided by C is

    physics-General
    Let the quantity of heat supplied per minute be Q. Then quantity of heat supplied in 2 blank m i n equals m C left parenthesis 90 minus 80 right parenthesis
    In 4 blank m i n comma blankheat supplied equals 2 m C open parentheses 90 minus 80 close parentheses
    therefore 2 m blank C open parentheses 90 minus 80 close parentheses equals m L rightwards double arrow fraction numerator L over denominator C end fraction equals 20
    General
    maths-

    If s i n invisible function application x equals 1 divided by 2, then s i n invisible function application 3 x equals_______________

    If s i n invisible function application x equals 1 divided by 2, then s i n invisible function application 3 x equals_______________

    maths-General
    General
    physics-

    Which one of the following is v subscript m end subscript minus T graph for perfectly black body? v subscript m end subscriptis the frequency of radiation with maximum intensity, T is the absolute temperature.

    Intensity is directly proportional to energy.

    Which one of the following is v subscript m end subscript minus T graph for perfectly black body? v subscript m end subscriptis the frequency of radiation with maximum intensity, T is the absolute temperature.

    physics-General
    Intensity is directly proportional to energy.
    General
    physics-

    The graph signifies

    The graph signifies

    physics-General
    General
    physics-

    Which curve shows the rise of temperature with the amount of heat supplied, for a piece of ice?

    Initially on heating temperature rises from negative 73 ℃ (200K) to 0degree(273K). Then ice melts and temperature does not rise. After the whole ice has melted, temperature begins to rise until it reaches 100℃ (373K). Then it becomes constant and after that it changes to vapours.

    Which curve shows the rise of temperature with the amount of heat supplied, for a piece of ice?

    physics-General
    Initially on heating temperature rises from negative 73 ℃ (200K) to 0degree(273K). Then ice melts and temperature does not rise. After the whole ice has melted, temperature begins to rise until it reaches 100℃ (373K). Then it becomes constant and after that it changes to vapours.
    General
    physics-

    Three rods made of same material and having same cross-section are joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

    Let the temperature of function be theta, then
    H equals H subscript 1 end subscript plus H subscript 2 end subscript
    rightwards double arrow fraction numerator K A left parenthesis theta minus 0 right parenthesis over denominator L end fraction equals fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction plus fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction

    Or theta equals 90 minus theta plus 90 minus theta
    Or theta equals 180 minus 2 theta
    Or 3 theta equals 180
    Or theta equals 60 ℃

    Three rods made of same material and having same cross-section are joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

    physics-General
    Let the temperature of function be theta, then
    H equals H subscript 1 end subscript plus H subscript 2 end subscript
    rightwards double arrow fraction numerator K A left parenthesis theta minus 0 right parenthesis over denominator L end fraction equals fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction plus fraction numerator K A left parenthesis 90 minus theta right parenthesis over denominator L end fraction

    Or theta equals 90 minus theta plus 90 minus theta
    Or theta equals 180 minus 2 theta
    Or 3 theta equals 180
    Or theta equals 60 ℃
    General
    physics-

    The figure shows a glass tube (linear co-efficient of expansion is alpha) completely filled with a liquid of volume expansion co-efficient gamma. On heating length of the liquid column does not change. Choose the correct relation between gamma and alpha

    When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus gamma must be less than 3 alpha
    Now we can write V equals V subscript 0 end subscript left parenthesis 1 plus gamma increment T right parenthesis
    Since V equals A l subscript 0 end subscript equals open square brackets A subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses close square brackets l subscript 0 end subscript equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses
    Hence V subscript 0 end subscript open parentheses 1 plus gamma increment T close parentheses equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses rightwards double arrow gamma equals 2 alpha

    The figure shows a glass tube (linear co-efficient of expansion is alpha) completely filled with a liquid of volume expansion co-efficient gamma. On heating length of the liquid column does not change. Choose the correct relation between gamma and alpha

    physics-General
    When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus gamma must be less than 3 alpha
    Now we can write V equals V subscript 0 end subscript left parenthesis 1 plus gamma increment T right parenthesis
    Since V equals A l subscript 0 end subscript equals open square brackets A subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses close square brackets l subscript 0 end subscript equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses
    Hence V subscript 0 end subscript open parentheses 1 plus gamma increment T close parentheses equals V subscript 0 end subscript open parentheses 1 plus 2 alpha increment T close parentheses rightwards double arrow gamma equals 2 alpha
    General
    physics-

    The spectrum of a black body at two temperatures 27 ℃ and 327 ℃ is shown in the figure. Let A subscript 1 end subscript and A subscript 2 end subscript be the areas under the two curves respectively. The value of fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction is

    Area under given curve represents emissive power and emissive power proportional to T to the power of 4 end exponent rightwards double arrow A proportional to T to the power of 4 end exponent
    rightwards double arrow fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction equals fraction numerator T subscript 2 end subscript superscript 4 end superscript over denominator T subscript 1 end subscript superscript 4 end superscript end fraction equals fraction numerator open parentheses 273 plus 327 close parentheses to the power of 4 end exponent over denominator open parentheses 273 plus 27 close parentheses to the power of 4 end exponent end fraction equals open parentheses fraction numerator 600 over denominator 300 end fraction close parentheses to the power of 4 end exponent equals fraction numerator 16 over denominator 1 end fraction

    The spectrum of a black body at two temperatures 27 ℃ and 327 ℃ is shown in the figure. Let A subscript 1 end subscript and A subscript 2 end subscript be the areas under the two curves respectively. The value of fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction is

    physics-General
    Area under given curve represents emissive power and emissive power proportional to T to the power of 4 end exponent rightwards double arrow A proportional to T to the power of 4 end exponent
    rightwards double arrow fraction numerator A subscript 2 end subscript over denominator A subscript 1 end subscript end fraction equals fraction numerator T subscript 2 end subscript superscript 4 end superscript over denominator T subscript 1 end subscript superscript 4 end superscript end fraction equals fraction numerator open parentheses 273 plus 327 close parentheses to the power of 4 end exponent over denominator open parentheses 273 plus 27 close parentheses to the power of 4 end exponent end fraction equals open parentheses fraction numerator 600 over denominator 300 end fraction close parentheses to the power of 4 end exponent equals fraction numerator 16 over denominator 1 end fraction
    General
    physics-

    In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures t subscript 1 end subscript and t subscript 2 end subscript. The liquid columns in the two arms have heights l subscript 1 end subscript and l subscript 2 end subscript respectively. The coefficient of volume expansion of the liquid is equal to

    Suppose, height of liquid in each arm before rising the temperature is l.

    With temperature rise height of liquid in each arm increases i. e. blank l subscript 1 end subscript greater than l blank a n d blank l subscript 2 end subscript greater than l
    Also l equals fraction numerator l subscript 1 end subscript over denominator 1 plus gamma t subscript 1 end subscript end fraction equals fraction numerator l subscript 2 end subscript over denominator 1 plus gamma t subscript 2 end subscript end fraction
    rightwards double arrow l subscript 1 end subscript plus gamma l subscript 1 end subscript t subscript 2 end subscript equals l subscript 2 end subscript plus gamma l subscript 2 end subscript t subscript 1 end subscript rightwards double arrow gamma equals fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator l subscript 2 end subscript t subscript 1 end subscript minus l subscript 1 end subscript t subscript 2 end subscript end fraction

    In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures t subscript 1 end subscript and t subscript 2 end subscript. The liquid columns in the two arms have heights l subscript 1 end subscript and l subscript 2 end subscript respectively. The coefficient of volume expansion of the liquid is equal to

    physics-General
    Suppose, height of liquid in each arm before rising the temperature is l.

    With temperature rise height of liquid in each arm increases i. e. blank l subscript 1 end subscript greater than l blank a n d blank l subscript 2 end subscript greater than l
    Also l equals fraction numerator l subscript 1 end subscript over denominator 1 plus gamma t subscript 1 end subscript end fraction equals fraction numerator l subscript 2 end subscript over denominator 1 plus gamma t subscript 2 end subscript end fraction
    rightwards double arrow l subscript 1 end subscript plus gamma l subscript 1 end subscript t subscript 2 end subscript equals l subscript 2 end subscript plus gamma l subscript 2 end subscript t subscript 1 end subscript rightwards double arrow gamma equals fraction numerator l subscript 1 end subscript minus l subscript 2 end subscript over denominator l subscript 2 end subscript t subscript 1 end subscript minus l subscript 1 end subscript t subscript 2 end subscript end fraction
    General
    physics-

    Six identical metallic rods are joined together in a pattern as shown in the figure. Points A and D are maintained at temperature 60 ℃ and 240 ℃. The temperature of the junction B will be

    R equals fraction numerator l over denominator K A end fraction
    fraction numerator T subscript A end subscript minus T subscript B end subscript over denominator R end fraction equals fraction numerator T subscript B end subscript minus T subscript C end subscript over denominator R end fraction equals fraction numerator T subscript C end subscript minus T subscript D end subscript over denominator R end fraction
    60 minus T subscript B end subscript equals T subscript B end subscript minus T subscript C end subscript(i)

    60 minus T subscript B end subscript equals T subscript C end subscript minus 240(ii)
    Solving (i) and (ii)
    T subscript B end subscript equals 120 ℃

    Six identical metallic rods are joined together in a pattern as shown in the figure. Points A and D are maintained at temperature 60 ℃ and 240 ℃. The temperature of the junction B will be

    physics-General
    R equals fraction numerator l over denominator K A end fraction
    fraction numerator T subscript A end subscript minus T subscript B end subscript over denominator R end fraction equals fraction numerator T subscript B end subscript minus T subscript C end subscript over denominator R end fraction equals fraction numerator T subscript C end subscript minus T subscript D end subscript over denominator R end fraction
    60 minus T subscript B end subscript equals T subscript B end subscript minus T subscript C end subscript(i)

    60 minus T subscript B end subscript equals T subscript C end subscript minus 240(ii)
    Solving (i) and (ii)
    T subscript B end subscript equals 120 ℃
    General
    physics-

    A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. when bimetallic strip is placed in a cold bath

    The metal X has a higher coefficient of expansion compared to that for metal Y so, on placing bimetallic strip in a cold bath, X will shrink more than Y. Hence, the strip will bend towards the left.

    A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. when bimetallic strip is placed in a cold bath

    physics-General
    The metal X has a higher coefficient of expansion compared to that for metal Y so, on placing bimetallic strip in a cold bath, X will shrink more than Y. Hence, the strip will bend towards the left.
    General
    physics-

    The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

    Let the temperature of common interface be T ℃. Rate of heat flow
    H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
    therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
    And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    In steady state, the rate of heat flow should be same in whole system i e comma
    H subscript 1 end subscript equals blank H subscript 2 end subscript
    rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
    rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
    rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
    Hence, heat flow from composite slab is
    H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    = fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
    Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
    By comparing Eqs. (ii) and (iii), we get
    rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction

    The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

    physics-General
    Let the temperature of common interface be T ℃. Rate of heat flow
    H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
    therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
    And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    In steady state, the rate of heat flow should be same in whole system i e comma
    H subscript 1 end subscript equals blank H subscript 2 end subscript
    rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
    rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
    rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
    Hence, heat flow from composite slab is
    H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    = fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
    Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
    By comparing Eqs. (ii) and (iii), we get
    rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction