In an household electric circuit, which of the following is/are correct?

(A)All electric appliances drawing power are joined in parallel

(B)As witch may be either in series or in parallel with the appliance which it controls

(C)If a switch is in parallel with an appliance, it will draw power when the switch is in the ‘off’ position(open)

(D)If a switch is in parallel with an appliance, the fuse will blow (burn out) when the switch is put‘on’ closed.

A, D
A, C, D
B, C, D
A, B, D

A 100-W bulb and a 25-W bulb are designed for the same voltage. They have filaments of the same length and material. The ratio of the diameter of the 100-W bulb to that of the 25-W bulb is

4 : 1
2 : 1
√2 :1
1 : 2

Solution

In the given circuit, as the sliding contact C is moved from A to B

the readings of both the ammeter and the voltmeter remain constant
the reading of both the ammeter and the voltmeter increase
the reading of the ammeter remains constant but that of the voltmeter increases
the reading of the ammeter remains constant but that of the voltmeter decreases

Solution

As we more from A to B the potential difference across AC increases and hence the reading of voltmeter also increases. But currnet flowing through ammeter remains almost constant.

A torch bulb rated as 4.5 W, 1.5 V is connected as shown in fig. The e.m.f. of the cell, needed to make the bulb glow at full intensity is

4.5 V
1.5 V
2.67 V
13.5 V

Solution

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is

16 : 1
1: 16
1 : 4
4 : 1

Solution

A primary cell has an e.m.f.of 1.5 volt. When short-circuit edit gives a current of 3 ampere. The internal resistance of the cell is

4.5 ohm
2 ohm
0.5 ohm
(1/4.5) ohm

Solution

r = E / I =1.5 / 3 = 0.5 ohm.

A cell when balanced with potentiometer gave a balance length of 50 cm. 4.5 Ωexternal resistance is introduced in the circuit, now it is balanced on 45 cm.The internal resistance of cell is

0.25 Ω
0.5 Ω
1.0 Ω
1.5 Ω

Three copper wires of lengths and cross sectional areas are(l, A), (2 l, A/2) and (l/2, 2A). Resistance is minimum in

wire of cross-sectional area A/2
wire of cross-sectional area A
wire of cross-sectional area 2A
same in all the three cases

Solution

\(R\alpha \frac{\iota }{A};\)

So, the resistance of the wire will be minimum when the area of cross-section is maximum and length is minimum.

If R₁ and R₂ are respectively the filament resistances of a 200 watt bulb and a 100 watt bulb designed to operate on the same voltage

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

Solution

A given resistor has the following colour scheme of the various strips on it : Brown, black, green and silver. Its value in ohm is

1.0 × 10⁴ ± 10%
1.0 × 10⁵ ± 10%
1.0 × 10⁶ ± 10%
1.0 × 10⁷ ± 10%

Solution

Numbers attached for brown,black, green and silver are 1, 0, 5, ± 10%. Therefore the resistance of given resistor

= 10 × 10⁵ Ω ± 10% = 10 × 10⁶ Ω ± 10%.

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