An electric lamp is marked 60 W, 220 V. The cost of kilowatt hour of electricity is Rs. 1.25. The cost of using this lamp on 220 V for 8 hours is

Re 0.25
Re 0.60
Re 1.20
Re 4.00

Solution

Energy consumed per day = P × t = 60 × 8 = 480 watt hour = 480/1000 = 0.48 kWh or unit of electricity.Hence the cost = 0.48 × 1.25 = Re 0.60.

Water boils in the electric kettle in 15 minutes after switching on. If the length of heating wire is decreased to 2/3 of its initial value,then the same amount of water will boil with the same supply voltage in

8 minutes
10 minutes
12 minutes
15 minutes

Solution

The internal resistance of a primary cell is 4Ω. It generates a current of 0.2 A in an external reistance of 21 Ω. The rate of chemical energy consumed in providing the current is

0.42 J s^-1
0.84 J s^-1
1 J s^-1
5 J s^-1

Solution

A heater of 220 V heat a volume of water in 5 minutes time.A heater of 110 V heats the same volume of water in

5 minutes
8 minutes
10 minutes
20 minutes

Solution

Heat produced, .H = ^V²t⁄_R.When voltage is halved,the heat produced becomes one fourth. Hence time taken to heat the water becomes four time.

Two identical batteries each of e.m.f. 2 V and internal resistance 1 Ω are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across R using these batteries is

3.2 W
2.0 W
1.28 W
8/9 W

Solution

Two heating wires of equal length are first connected in series and then in parallel to a constant voltage source.The rate of heat produced in two cases is (parallel to series)

1 : 4
4 : 1
1 : 2
2 : 1

Solution

If R is the resistance of each wire, total resistance in series = R + R = 2 R;and total resistance in parallel.

=\(\frac{R\times R}{R+R}=\frac{R}{2}\)

Heat produced per second (= V²/R) will be four times in parallel than in series.

If R₁ and R₂ are the filament resistances of 200 W and a 100W bulb respectively both designed to run at the same voltage,then

R₂ is four times of R₁
R₁ is four times of R₂
R₂ is two times of R₁
R₁ is two times of R₂

Solution

As R= V²/P or R α µ1/P

so R₂/R₁ = P₁/P₂= 200/100 = 2

or R₂ = 2 R₁.

A potentiometer wire, 10m long, has a resistance of 40Ω. It is connected in series with a resistance box and a 2 V storage cell. If the potential gradient along the wire is 0.1 m V/cm,the resistance unplugged in the box is

260 Ω
760 Ω
960 Ω
1060 Ω

Solution

The resistance of a wire at room temperature 30°C is found to be 10 Ω. Now to increase the resistance by 10%, the temperature of the wire must be[ The temperature coefficient of resistance of the material of the wire is 0.002 per °C]

36°C
83°C
63°C
33°C

Solution

Two wires of same metal have the same length but their cross-sections are in the ratio 3: 1. They are joined in series.The resistance of the thicker wire is 10 Ω. The total resistance of the combination is

5/2 Ω
40/3 Ω
40 Ω
100 Ω

Solution

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