A 60 CP 250 V metal filament lamp has a measured candle power of 17.5 CP at 260 V and 50 C.P. at 240 V.
Ques.41. If the equation for C and V is C = aV b where C = candlepower and V = voltage, the value of constant b is
4.5
0.98 × l0-9
0.98 × 10-6
450
Answer.1.4.5
Explanation:-
The constant for the lamp is
C = aV b
71 .5 = a × (260)b ———(1)
50 = a × (240)b ———–(2)
Dividing 1 by 2 we get
71.5/50 = a × (260)b⁄ a × (240)b
1.43 = (1.083)b
ln1.43 = b ln 1.083
b = ln1.43 ⁄ b ln 1.083
b = 4.5
Ques.42. The value of constant “a” is
0.98 × 10-9
0.98 × 10-7
0.98 × 10-5
0.98 × 10-3
Answer.1.0.98 x 10-9
Explanation:-
Substituting the value of “a” in equation 1 we get
71 .5 = a × (260)4.5
a = 0.97 × 10-9
Ques.43. The change in candle power per volt at 250 V will be
1.1 V
4.4 V
8.8 V
17.6 V
Answer.1.1.1 V
Explanation:-
a = 0.97 × 10-9
Since C = aV b
where
a = 9.7 × 10−1
b = 4.5
V = 250 Volt
∴ C = 9.7 × 10−10 × (250)4.5 candela
Differentiating the above equation
dC/dV = 9.7 × 10−10 × 4.5(250)3.5
= 1.1 candela per volt
Questions 44 to 46 refer to data given below:
The 220 V lamps, one of 60 W and the other of 75 W are connected in series across a 440 V supply.
Ques.44. The potential difference across 60 W lamp will be
195 V
220 V
242 V
440 V
Answer.3.242 V
Explanation:-
Voltage rating 220 V
Power of bulb 1 = 60 W
Power = V2/R
60 = (220)2 ⁄ R
R1 = 806.7 Ω
Power of bulb 2 = 75 W
Power = V2/R
75 = (220)2 ⁄ R
R2 = 645.3Ω
As two bulb are connected in the series their total resistance
R = R1 + R2
R = 806.7 + 645.3 = 1452Ω
Hence current flowing through each resistance (current remans same in series combination of resistances)
I = V/R = 440/1452 = 0.30 Amp
V1 = Potential difference across bulb 1
V1 = IR1 = 806.7 × 0.30 = 242.01 V
Ques.45. The potential difference across 75-watt lamps will be
193 V
220 V
245 V
440 V
Answer.1.193 V
Explanation:-
As solved in above question i.e question No. 44
I = 0.30
R2 = 645.3 Ω
V2 = Potential difference across bulb 2
V2 = IR2 = 645.3 × 0.30 = 193.5 V
V2 = 193.5 V
Ques.46. The essential requirement of good heating elements are
High Specific resistance
Free from oxidation
Low-temperature coefficient of resistance
All of the above
Answer.4.All of the above
Explanation:-
Requirement of good heating elements
The materials used for the heating element should have the following properties:
High-specific resistance:- Material should have high-specific resistance so that a small length of wire may be required to provide a given amount of heat.
High-melting point:- It should have a high-melting point so that it can withstand high temperature, a small increase in temperature will not destroy the element.
Low-temperature coefficient of resistance:- For accurate temperature control, the variation of resistance with the operating temperature should be very low. This can be obtained only if the material has a low-temperature coefficient of resistance.
Free from oxidation The element material should not be oxidized when it is subjected to high temperatures; otherwise, the formation of oxidized layers will shorten its life.
High-mechanical strength:- The material should have high mechanical strength and should withstand mechanical vibrations.
Non-corrosive:- The element should not corrode when exposed to the atmosphere or any other chemical fumes.
Economical:- The cost of material should not be so high.
Questions 47 to 48 refer to data given below:
A 110 V lamp 16 C.P. and a lamp of the same material and worked at the same efficiency develops 25 CP on 220 V.
Ques.47. The ratio of diameters of the filaments will be
0.54
1.0
1.18
1.78
Answer.4.1.78
Explanation:-
For two filaments of the same material working at the same temperature and efficiency, the relation is
(d1 ⁄ d2) = (I1 ⁄ I2)2/3
Given that the two lamps are having the same efficiency their operating temperatures must be the same.
Let d1 and d2 are the diameters of filaments, I1 and I2 are their lengths, and L1 and L2 are currents flowing through them.
Let us assume, the input power taken by the two lamps is proportional to their output.
∴ For Lamp 1 = 16 × 110 I1 ———— 1
For Lamp 2 = 25 × 220 I2———— 2
From equation 1 and 2 we get
I1/I2 = (16/25) × (220/110)
I1/I2 = 32/25
Since their operating temperature are same therefore
(d1 ⁄ d2) = (32 ⁄ 25)2/3
(d1 ⁄ d2) = 1.177
Ques48. The ratio of lengths of the filaments will be
0.54
1.0
1.18
1.78
Answer.1.0.54
Explanation:-
For two filaments working at the same temperature, the flux per unit area is the same. Denoting their lengths by l1 and l2 and their diameters by d1 and d2 respectively, we have, Lumen output ∝ l1d1 ∝ l2d2= constant
From the given data in question 47
Since output ∝ l1d1
16/25 = l1d1 ⁄ l2d2
l1/l2 = (16/25) × (d2/d1)
l1/l2 = (16/25) × (1/1.17)
l1/l2 = 0.54
Ques49. Which of the following filament material has the lowest melting point?
Carbon
Tungsten
Tantalum
Osmium
Answer.1.Tantalum
Explanation:-
The melting point of the carbon is 3550°C
The melting point of the tungsten is 3422°C
The melting point of the tantalum is 3017°C
The melting point of the Osmium is 3033°C
Ques.50. While comparing tungsten filament lamps with fluorescent tubes, all of the following are the advantages in favor of tungsten filament lamp EXCEPT
Longer life
Less costly
More brightness
Simple installation
Answer.1.Longer Life
Explanation:-
Incandescent lamps
Incandescent lamps have been around for a long time and are still commonly used, but they don’t perform as well as some of the other lamps on the market.
Incandescent lamps produce a warm light that is pleasing in homes.
Incandescent lamps are sensitive to vibration.
Incandescent lamps are hot to the touch.
The incandescent lamp produces 15 – 20 lumens/watt
Incandescent lamps are low efficiency. e. Incandescent lamps have a short life expectancy (1200 hours).
Incandescent lamps are inexpensive.
Compact fluorescent (CFL)
Compact fluorescent (CFL) lamps are becoming more popular.
CFLs are available in a wide spectrum of colors (2700 K is the closest to incandescent
CFLs take a few minutes to warm up and produce full intensity.
