1. For a glass plate transitivity and reflectivity are specified as 0.86 and 0.08 respectively, the absorptivity of the glass plate is
(a) 0.86
(b) 0.08
(c) 1.00
(d) 0.06
(1 Mark, 1988)

Ans: d
Explanation:
We know: \alpha +\rho +\tau =1
=> \alpha +0.08 + 0.86 =1
=> Absorptivity (\alpha) = 1 – 0.08 – 0.86 = 0.06
2. A diffuse radiation surface has
(a) Radiation intensity independent of angle
(b) Emissive power independent of angle
(c) Emissive power independent of wave length
(d) Radiation intensity independent of both angle and wavelength
(1 Mark, 1991)

Ans: a
3. The radiative heat transfer rate per unit area (W/m^{2}) between two plane parallel grey surfaces (emissivity = 0.9) maintained at 400 K and 300 K is:
(a) 992
(b) 812
(c) 464
(d) 567
(2 Mark, 1993)

Ans: b
Explanation:
Q=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1 }{ { \varepsilon }_{ 1 } } +\frac { 1 }{ { \varepsilon }_{ 2 } } 1 }=> Q=\frac { 5.67\times { 10 }^{ 8 }\left( { 400 }^{ 4 }{ 300 }^{ 4 } \right) }{ \frac { 1 }{ 0.9 } +\frac { 1 }{ 0.9 } 1 } = 812 W/m^{2}
4. For the circular tube of equal length and diameter shown below, the view factor F_{13} is 0.17. the view factor F_{12} in this case will be
(a) 0.17
(b) 0.21
(c) 0.79
(d) 0.83
(2 Mark, 2001)

Ans: d
Explanation:
We know: F_{11 }+ F_{12} + F_{13} = 1
=> 0 + F_{12} +0.17 = 1
=> F_{12 }= 0.83
5. What is the value of the view factor for two inclined flat plates having common edge of equal width, and with an angle of 20 degrees?
(a) 0.83
(b) 1.17
(c) 0.66
(d) 1.34
(2 Mark, 2002)
6. A plate having 10 cm^{2} area each side is hanging in the middle of a room of 100 m^{2} total surface area. The plate temperature and emissivity are respectively 800 K and 0.6. The temperature and emissivity values for the surfaces of the room are 300K and 0.3 respectively. Boltzmann’s constant σ = 5.67×10^{8} W/m^{2}K^{4}. The total heat loss from the two surfaces of the plate is
(a) 13.66 W
(b) 27.32 W
(c) 27.87 W
(d) 13.66 MW
(1 Mark, 2003)

Ans: b
Explanation:
A plate having 10 cm^{2 }area each side.
Total area of the plate from which heat is released = 2 * 10 = 20 cm^{2}
Heat exchange between any two bodies:
{ Q }=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1{ \epsilon }_{ 1 } }{ { A }_{ 1 }{ \epsilon }_{ 1 } } +\frac { 1 }{ { A }_{ 1 }{ F }_{ 12 } } +\frac { 1{ \epsilon }_{ 2 } }{ { A }_{ 2 }{ \epsilon }_{ 2 } } }=> { Q }=\frac { { A }_{ 1 }\sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1{ \epsilon }_{ 1 } }{ { \epsilon }_{ 1 } } +\frac { 1 }{ { F }_{ 12 } } +\left( \frac { { A }_{ 1 } }{ { A }_{ 2 } } \right) \frac { 1{ \epsilon }_{ 2 } }{ { \epsilon }_{ 2 } } }
=> { Q }=\frac { \left( 20\times { 10 }^{ 4 } \right) \times \left( 5.67\times { 10 }^{ 8 } \right) \times \left( 800^{ 4 }300^{ 4 } \right) }{ \frac { 10.6 }{ 0.6 } +\frac { 1 }{ 1 } +\left( \frac { \left( 20\times { 10 }^{ 4 } \right) }{ 100 } \right) \frac { 10.3 }{ 0.3 } } = 27.32 W
7. A solid cylinder (surface 2) is located at the center of a hollow sphere (surface 1). The diameter of the sphere is 1m, while the cylinder has a diameter and length of 0.5 m each. The radiation configuration factor F_{11} is
(a) 0.375
(b) 0.625
(c) 0.75
(d) 1
(2 Mark, 2005)

Ans: b
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1}F_{12} = A_{2 }[As F_{21 }= 1]
=> F_{12 }= \frac { { A }_{ 2 } }{ { A }_{ 1 } } =\frac { 2\pi { R }_{ i }L+\left( 2\times \pi \times { R }_{ i }^{ 2 } \right) }{ 4\pi { R }_{ o }^{ 2 } } =\frac { 2\pi \times 0.25\times 0.5+\left( 2\times \pi \times { 0.25 }^{ 2 } \right) }{ 4\pi \times 0.5^{ 2 } } =0.375
Now, F_{11 }+ F_{12} = 1
=> F_{11 }= 1 – F_{12}
=> F_{11 }= 1 – 0.375 = 0.625
8. The following figure was generated from experimental data relating spectral black body emissive power to wave length at three temperatures T_{1}, T_{2} and T_{3} T_{1} > T_{2} > T_{3. }The conclusion is that the measurements are
(a) Correct because the maxima in E_{bλ} show that correct trend
(b) Correct because Planck’s law is satisfied
(c) Wrong because the Stefan Boltzmann law is not satisfied
(d) Wrong because Wien’s displacement law is not satisfied
(1 Mark, 2005)

Ans: d
Explanation:
According to Wein’s displacement law:
T\alpha \frac { 1 }{ { \lambda }_{ max } }i.e. The peak should move towards the left as the temperature increases.
But in the question thee peak moved towards the right as the temperature increases.
So, option “d” is correct.
10. A hollow enclosure is formed between two infinitely long concentric cylinders of radii 1m and 2m, respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface2) and the outer surface of the smaller cylinder (surface1). The radiating surfaces are diffuse and the medium in the enclosure is nonparticipating. The fraction of the thermal radiation leaving the larger surface and striking itself is
(a) 0.25
(b) 0.5
(c) 0.75
(d) 1
(2 Mark, 2008)

Ans: b
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1}F_{21} = A_{2 }[As F_{12 }= 1]
=> F_{21 }= \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 2\pi r_{ 1 }L }{ 2\pi r_{ 2 }L } =\frac { 1 }{ 2 }
Now, F_{22 }+ F_{21} = 1
=> F_{22 }= 1 – F_{21 }= 1 0.5 = 0.5
Common Data Questions: 11 & 12.
Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a twosurface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10^{8 }W/m^{2}K^{4}.
11. The irradiation (in kW/m^{2}) for the upper plate (plate 1) is
(a) 2.5
(b) 3.6
(c) 17.0
(d) 19.5
(2 Mark, 2009)

Ans: d
Explanation:
Irradiation on the plate 1 = Energy emitted by plate 2 + Reflected portion of emitted energy by plate 1 on plate 2
=> Irradiation on the plate 1 = \epsilon \sigma { T }_{ 2 }^{ 4 }+\left( 1\alpha \right) \sigma { T }_{ 1 }
=> Irradiation on the plate 1 = 0.7\times \left( 5.67\times { 10 }^{ 8 } \right) { \times }{ 500 }^{ 4 }+\left( 10.7 \right) \times \left( 5.67\times { 10 }^{ 8 } \right) { \times }{ 1000 }^{ 4 } = 19.5 kW/m^{2}
12. If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/m^{2}) between plate 1 and plate 2 is
(a) 17.0
(b) 19.0
(c) 23.0
(d) 31.7
(2 Mark, 2009)

Ans: d
Explanation:
Q=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1 }{ { \varepsilon }_{ 1 } } +\frac { 1 }{ { \varepsilon }_{ 2 } } 1 }=> Q=\frac { 5.67\times { 10 }^{ 8 }\left( { 1000 }^{ 4 }{ 500 }^{ 4 } \right) }{ \frac { 1 }{ 0.8 } +\frac { 1 }{ 0.7 } 1 } = 31.67 kW/m^{2}
13. For an opaque surface, the absorptivity (α), transmissivity (\tau ) and reflectivity (ρ) are related by the equation
(a) α + ρ = \tau
(b) α + ρ + \tau = 0
(c) α + ρ = 1
(d) α + ρ = 0
(1 Mark, 2012)

Ans: c
Explanation:
We know: \alpha +\rho +\tau =1
For opaque surface, \tau = 0
So, \alpha +\rho =1
14. Consider two infinitely long thin concentric tubes of circular cross section as shown in the figure. If D_{1} and D_{2} are the diameters of the inner and outer tubes respectively, then the view factor F_{22} is given by
(a) \left( \frac { { D }_{ 2 } }{ { D }_{ 1 } } \right) 1
(b) Zero
(c) \left( \frac { { D }_{ 1 } }{ { D }_{ 2 } } \right)
(d) 1\left( \frac { { D }_{ 1 } }{ { D }_{ 2 } } \right)
(2 Mark, 2012)

Ans: d
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1}F_{21} = A_{2 }[As F_{12 }= 1]
=> F_{21 }= \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \pi D_{ 1 }L }{ \pi D_{ 2 }L } =\frac { D_{ 1 } }{ D_{ 2 } }
Now, F_{22 }+ F_{21} = 1
=> F_{22 }= 1 – F_{21}
=> F_{22 }= 1 – \frac { D_{ 1 } }{ D_{ 2 } }
15. Two large diffuse gray parallel plates, separated by a small distance, have surface temperatures of 400 K and 300 K. If the emissivities of the surfaces are 0.8 and the StefanBoltzmann constant is 5.67\times 10^{8} W/m^{2}K^{4}, the net radiation heat exchange rate in kW/m^{2} between the two plates is
(a) 0.66
(b) 0.79
(c) 0.99
(d) 3.96
(2 Mark, 2013)

Ans: a
Explanation:
Q=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1 }{ { \varepsilon }_{ 1 } } +\frac { 1 }{ { \varepsilon }_{ 2 } } 1 }=> Q=\frac { 5.67\times { 10 }^{ 8 }\left( { 400 }^{ 4 }{ 300 }^{ 4 } \right) }{ \frac { 1 }{ 0.8 } +\frac { 1 }{ 0.8 } 1 } = 0.66 kW/m^{2}
16. A hemispherical furnace of 1 m radius has the inner surface (emissivity, ε = 1) of its roof maintained at 800 K, while its floor (ε = 0.5) is kept at 600 K. StefanBoltzmann constant is 5.668×10^{8} W/m^{2}K^{4}. The net radiative heat transfer (in kW) from the roof to the floor is _______.
(2 Mark, 2014[2])

Ans: 24.9
Explanation:
Here, F_{21} = 1
\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 2\pi { r }^{ 2 } }{ \pi { r }^{ 2 } } =2Heat exchange between any two bodies:
{ Q }=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1{ \epsilon }_{ 1 } }{ { A }_{ 1 }{ \epsilon }_{ 1 } } +\frac { 1 }{ { A }_{ 2 }{ F }_{ 21 } } +\frac { 1{ \epsilon }_{ 2 } }{ { A }_{ 2 }{ \epsilon }_{ 2 } } }=> { Q }=\frac { { A }_{ 1 }\sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1{ \epsilon }_{ 1 } }{ { \epsilon }_{ 1 } } +\frac { { A }_{ 1 } }{ { A }_{ 2 }{ F }_{ 21 } } +\left( \frac { { A }_{ 1 } }{ { A }_{ 2 } } \right) \frac { 1{ \epsilon }_{ 2 } }{ { \epsilon }_{ 2 } } }
=> { Q }=\frac { 2\pi { 1 }^{ 2 }\times 5.67\times { 10 }^{ 8 }\left( { 800 }^{ 4 }{ 600 }^{ 4 } \right) }{ \frac { 11 }{ 1 } +\frac { 2 }{ 1 } +\left( \frac { 2 }{ 1 } \right) \frac { 10.5 }{ 0.5 } } = 24.91 kW
17. A solid sphere of radius r_{1} = 20 mm is placed concentrically inside a hollow sphere of radius r_{2} = 30 mm as shown in the figure.
The view factor F_{21} for radiation heat transfer is
(a) 2/3
(b) 4/9
(c) 8/27
(d) 9/4
(2 Mark, 2014[3])

Ans: b
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1} = A_{2 }F_{21 }[As, F_{12}_{ }= 1]
=> F_{21 }= \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\frac { 4\pi \times { 2 }^{ 2 } }{ 4\pi \times { 3 }^{ 2 } } =\frac { 4 }{ 9 }
18. Two infinite parallel plates are placed at a certain distance apart. An infinite radiation shield is inserted between the plates without touching any of them to reduce heat exchange between the plates. Assume that the emissivities of plates and radiation shield are equal. The ratio of the net heat exchange between the plates with and without the shield is
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/8
(2 Mark, 2014[4])

Ans: a
Explanation:
If the shield has the same emissivity as that of the plate and the number of shields kept between two plates = N
Then, Q_{with shield} = \frac { 1 }{ 1+N } Q_{without shield}
If N= 1, \frac { { Q }_{ withshield } }{ { Q }_{ withoutshield } } =\frac { 1 }{ 1+N } =\frac { 1 }{ 2 }
19. The total emissive power of a surface is 500 W/m^{2} at a temperature T_{1} and 1200 W/m^{2} at a temperature T_{2}, where the temperatures are in kelvin. Assuming the emissivity of the surface to be constant, the ratio of the temperatures is
(a) 0.308
(b) 0.416
(c) 0.803
(d) 0.874
(2 Mark, 2015[2])

Ans: c
Explanation:
Emissive power (E) \alpha { T }^{ 4 }
=> \frac { { E }_{ 1 } }{ { E }_{ 2 } } =\frac { { T }_{ 1 }^{ 4 } }{ { T }_{ 2 }^{ 4 } }
=> \frac { { T }_{ 1 } }{ { T }_{ 2 } } ={ \left( \frac { 500 }{ 1200 } \right) }^{ 1/4 }=0.803
20. A solid sphere 1 of radius ‘r’ is placed inside a hollow, closed hemispherical surface 2 of radius ‘4r’. The shape factor F_{21 }is
(a) 1/12
(b) 1/2
(c) 2
(d) 12
(2 Mark, 2015[3])

Ans: a
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1} = A_{2 }F_{21 }[As, F_{12}_{ }= 1]
=> F_{21 }= \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 2\pi { r }_{ 2 }^{ 2 }+\pi { r }_{ 2 }^{ 2 } } =\frac { 4\pi { r }^{ 2 } }{ 2\pi { \left( 4r \right) }^{ 2 }+\pi { \left( 4r \right) }^{ 2 } }
=> F_{21 }= \frac { 4\pi { r }^{ 2 } }{ 48\pi { r }^{ 2 } } =\frac { 1 }{ 12 }
21. An infinitely long furnace of 0.5 m × 0.4 m crosssection is shown in the figure below. Consider all surfaces of the furnace to be black. The top and bottom walls are maintained at temperature T_{1} = T_{3} = 927^{0}C while the side walls are at temperature T_{2} = T_{4} = 527^{0}C. The view factor, F_{12 }is 0.26. The net radiation heat loss or gain on side 1 is _____ W/m.
StefanBoltzmann constant = 5.67 × 10^{8} W/m^{2}K^{4}.

Ans: 24530.68
Explanation:
Heat exchange between 1 and 3 = 0 [As both the side at same temperature]
Heat exchange between 1 and 2 = Heat exchange between 1 and 4
The net radiation heat loss or gain on side 1 (Q)= Heat exchange between 1 and 2 + Heat exchange between 1 and 3 + Heat exchange between 1 and 4
=> Q = 2 * Heat exchange between 1 and 2
=> Q = 2\times { L }_{ 1 }\times { F }_{ 12 }\sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right)
(\varepsilon =1 because all surfaces are black.)
=> Q = 2\times 0.5\times 0.26\times 5.67\times { 10 }^{ 8 }\left( { 1200 }^{ 4 }{ 800 }^{ 4 } \right) =
24530.68 W/m
22. Consider the radiation heat exchange inside an annulus between two very long concentric cylinders. The radius of the outer cylinder is R_{o} and that of the inner cylinder is R_{i}. The radiation view factor of the outer cylinder onto itself is
(a) 1\sqrt { \frac { { R }_{ i } }{ { R }_{ o } } }
(b) \sqrt { 1\frac { { R }_{ i } }{ { R }_{ o } } }
(c) 1{ \left( \frac { { R }_{ i } }{ { R }_{ o } } \right) }^{ 1/3 }
(d) 1\frac { { R }_{ i } }{ { R }_{ o } }
(1 Mark, 2016[2])

Ans: d
Explanation:
We know: A_{1}F_{12} = A_{2}F_{21}
=> A_{1}F_{12} = A_{2 }[As F_{21 }= 1]
=> F_{12 }= \frac { { A }_{ 2 } }{ { A }_{ 1 } } =\frac { 2\pi { R }_{ i }L }{ 2\pi { R }_{ o }L } =\frac { { R }_{ i } }{ { R }_{ o }}
Now, F_{11 }+ F_{12} = 1
=> F_{11 }= 1 – F_{12}
=> F_{11 }= 1 – \frac { { R }_{ i } }{ { R }_{ o } }
23. Two large parallel plates having a gap of 10 mm in between them are maintained at temperatures T_{1} = 1000 K and T_{2} = 400 K. Given emissivity values, { \varepsilon }_{ 1 } = 0.5, { \varepsilon }_{ 2 } = 0.25 and StefanBoltzmann constant σ = 5.67 × 10^{−8} W/m^{2}K^{4}, the heat transfer between the plates (in kW/m^{2}) is ____.
(2 Mark, 2016[3])

Ans: 11.05
Explanation:
Q=\frac { \sigma \left( { T }_{ 1 }^{ 4 }{ T }_{ 2 }^{ 4 } \right) }{ \frac { 1 }{ { \varepsilon }_{ 1 } } \frac { 1 }{ { \varepsilon }_{ 2 } } 1 }=> Q=\frac { 5.67\times { 10 }^{ 8 }\left( { 1000 }^{ 4 }{ 400 }^{ 4 } \right) }{ \frac { 1 }{ 0.5 } \frac { 1 }{ 0.25 } 1 } = 11.05 kW/m^{2}
24. Two black surfaces, AB and BC, of lengths 5 m and 6 m, respectively, are oriented as shown. Both surfaces extend infinitely into the third dimension. Given that view factor F_{12} = 0.5. T_{1} = 800K, T_{2} = 600K, T_{surrounding} = 300K and Stefan Boltzmann constant, σ = 5.67 \times 10^{8} W/(m^{2}K^{4}), the heat transfer rate from surface 2 to the surrounding environment is _______ kW.
(2 Mark, 2017[1])

Ans: 13.778
Explanation:
Instead of “Surface 2 to the surrounding” the question should be modified to “between surface 2 and surrounding”. Now the answer will be given below.
We know:
A_{1 }F_{12 }= A_{2 }F_{21}
=> 6 * 0.5 = 5 * F_{21}
=> F_{21 }= 0.6
Again, F_{21 }+ F_{22 }+ F_{2s }= 1
[F_{2s }= Fraction of radiation energy leaving from section 2 and reaches to the surrounding]
=> 0.6 + 0 + F_{2s }= 1 => F_{2s }= 0.4
Heat transfer from surface 2 to surrounding (Q):
Q = { F }_{ 23 }{ A }_{ 2 }\sigma \left( { T }_{ 2 }^{ 4 }{ T }_{ \infty }^{ 4 } \right)
=> Q = 0.4 * 5 * 5.67 * 10^{8}(600^{4} – 300^{4})
=> Q = 13.778 kW
25. The emissive power of a blackbody is P. If its absolute temperature is doubled, the emissive power becomes
(a) 2P
(b) 4P
(c) 8P
(d) 16P
(1 Mark, 2017[2])

Ans: d
Explanation:
Emissive power (E) \alpha { T }^{ 4 }
=> \frac { { E }_{ 1 } }{ { E }_{ 2 } } =\frac { { T }_{ 1 }^{ 4 } }{ { T }_{ 2 }^{ 4 } }
=> \frac { P }{ { E }_{ 2 } } =\frac { { T }^{ 4 } }{ { 2 }^{ 4 }.{ T }^{ 4 } }
=> E_{2 }= 16P
26. The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is
(a) 500
(b) 1000
(c) 4000
(d) 8000
(1 Mark, 2018[2])

Ans: b
Explanation:
Using Wein’s displacement law:
{ \lambda }_{ max }T=constant=> 1.45 * 2000 = 2.9 * T_{2}
=> T_{2 }= 1000 K
27. A 0.2 m thick infinite black plate having a thermal conductivity of 3.96 W/mK is exposed to two infinite black surfaces at 300 K and 400 K as shown in the figure. At steady state, the surface temperature of the plate facing the cold side is 350 K. The value of Stefan Boltzmann constant, σ, is 5.67 × 10^{8} W/m^{2} K^{4}. Assuming 1D heat conduction, the magnitude of heat flux through the plate (in W/m^{2}) is ________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 391.62
Explanation:
Under steady state condition:
heat transfer between 400K surface to plate = heat transfer via conduction inside plate = Heat transfer between plate to 300K surface
So, magnitude of heat flux (q)= \sigma \left( { 350 }^{ 4 }{ 300 }^{ 4 } \right)
=> q = 5.67\times { 10 }^{ 8 }\times \left( { 350 }^{ 4 }{ 300 }^{ 4 } \right) = 391.62 W/m^{2}
28. Sphere 1 with a diameter of 0.1 m is completely enclosed by another sphere 2 of diameter 0.4 m. The view factor F_{12} is
(a) 0.0625
(b) 0.5
(c) 1.0
(d) 0.25
(1 Mark, 2019[2])

Ans: c
Explanation:
F_{11 }= 0
We know: F_{11 }+ F_{12 }= 1
=> 0 + F_{12 }= 1 => F_{12 }= 1
29. Three sets of parallel plate LM, NR and PQ are given in Figures 1, 2 and 3. The view factor F_{IJ} is defined as the fraction of radiation leaving plate I that is intercepted by plate J. Assume that the values of F_{LM} and F_{NR} are 0.8 and 0.4 respectively. The value of F_{PQ} (round off to one decimal place) is________.
(2 Mark, 2019[2])