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 

(a) 230$\Omega$          (b) 46$\Omega$

(c) 26$\Omega$            (d) 13$\Omega$

A potentiometer wire is 100 cm long and a constant potential is maintained across it. Two cells are connected in series first to support one another and then in the opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

1. 5:4              2. 3:4

3. 3:2              4. 5:1  

The charge following through a resistance R varies with time $t as Q= at-b{t}^2,$ where a and b are positive constants. The total heat produced in R is

(1) $\frac{a^{3}R}{3b}$               

(2) $\frac{a^{3}R}{2b}$

(3) $\frac{a^{3}R}{b}$               

(4) $\frac{a^{3}R}{6b}$

A potentiometer wire has a length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm of the wire is 

(1)32Ω

(2)40Ω

(3)44Ω

(4)48Ω 

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 V_A, V_B, and V_C respectively. Then,


1. V_A=V_B=V_C

2. V_A≠V_B=V_C

3. V_A=V_B≠V_C

4. V_A≠V_B≠V_C

Across a metallic conductor of non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is :

1. current density
2. current
3. drift velocity
4. electric field

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8Ω all connected in series. If the ammeter has a coil of resistance 480Ω  and a shunt of 20Ω then reading in the ammeter will be :

(1) 0.5A

(2) 0.25A

(3) 2A

(4) 1A

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_o$ and a resistance r₁. An unknown e.m.f. is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by

(1)$\frac{E_{o}rL}{\left(r_1\right)l}$

(2)$\frac{E_{o}rl}{\left(r+r_1\right)L}$

(3)E_ol/L

(4)$\frac{E_{o}rL}{\left(r+r_1\right)l}$

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8V and the  average resistance per km is 0.5 Ω. The power loss in the wire is

(1) 19.2W

(2) 19.2kW

(3) 19.2J

(4) 12.2kW 

The resistances in the two arms of the meter bridge are 5 and R , respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6$l_1$. The resistance R, is


(1)10Ω

(2)15Ω

(3)20Ω

(4)25Ω