﻿ A cylindrical conductor of diameter 0.1 mm carries a current of 90 mA. The current density (in Am-2) is (π ≈ 3) | Current Electricity

## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Current Electricity

#### Multiple Choice Questions

201.

In the following network, potential at O is

• 4 V

• 3 V

• 6 V

• 4.8 V

202.

Effective resistance between A and B in the following circuit

• 10 Ω

• 20 Ω

• 5 Ω

203.

Two heating coils of resistances 10 Ω and 20 Ω are connected in parallel and connected to a battery of emf 12 V and internal resistance 1 Ω. The power consumed by them are in the ratio

• 1 : 4

• 1 : 3

• 2 : 1

• 4 : 1

204.

Of the following graphs, the one that correctly represents the I-V characteristics of a 'Ohmic device' is

205.

The working of magnetic braking of trains is based on

• eddy current

• pulsating current

• alternating current

# 206.A cylindrical conductor of diameter 0.1 mm carries a current of 90 mA. The current density (in Am-2) is (π ≈ 3)12 × 107 3 × 106 6 × 106 2.4 × 107

A.

12 × 107

Given,

Diameter of cylindrical conductor (D) = 0.1 mm

Current (I) = 90 mA = 90 × 10-3 A

We know that,

207.

A straight wire of length 50 cm carrying a current of 2.5 A is suspended in mid-air by a uniform magnetic field of 0.5 T (as shown in figure). The mass of the wire is (g = 10 ms-2)

• 250 gm

• 125 gm

• 62.5 gm

• 100 gm

208.

In meter bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is 'l'. If the temperature of water is increased, the new balancing

• < 1

• > 1

• = 0

• = 1

209.

The power dissipated in 3 Ω resistance in the following circuit is

• 0.25 W

• 0.75 W

• 1 W

• 0.5 W

210.

The value of I in the figure shown below is

• 21 A

• 4 A

• 8 A

• 19 A