23479.Across a metallic conductor of non–uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is
Current density
Current
Drift velocity
Electric field
23480.A resistor R and 2μF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)
3.3 × 107 Ω
1.3 × 104 Ω
1.7 × 105 Ω
2.7 × 106 Ω
23481.In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :
10 A
12 A
14 A
8 A
23482.A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
230 Ω
46 Ω
26 Ω
13 Ω
23483.A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is
32 Ω
40 Ω
44 Ω
48 Ω
23484.The Kirchhoff’s first law (Σi = 0) and second law (ΣiR = ΣE), where the symbols have their usual meanings, are respectively based on
conservation of charge, conservation of energy
conservation of charge, conservation of momentum
conservation of energy, conservation of charge
conservation of momentum, conservation of charge
23485.The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of
200°C
300°C
400°C
500°C
23486.A, B and C are voltmeters of resistance R, 1.5R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively, then
VA = VB = VC
VA ≠ VB = VC
VA = VB ≠ VC
VA ≠ VB ≠ VC
23487.A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be
doubled
four times
one fourth
halved