Test: Electrical Energy and Power (December 4) - NEET MCQ

# Test: Electrical Energy and Power (December 4) - NEET MCQ

Test Description

## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Electrical Energy and Power (December 4)

Test: Electrical Energy and Power (December 4) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Electrical Energy and Power (December 4) questions and answers have been prepared according to the NEET exam syllabus.The Test: Electrical Energy and Power (December 4) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electrical Energy and Power (December 4) below.
Solutions of Test: Electrical Energy and Power (December 4) questions in English are available as part of our Daily Test for NEET Preparation for NEET & Test: Electrical Energy and Power (December 4) solutions in Hindi for Daily Test for NEET Preparation course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Electrical Energy and Power (December 4) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Daily Test for NEET Preparation for NEET Exam | Download free PDF with solutions
Test: Electrical Energy and Power (December 4) - Question 1

### In a potentiometer of 10 wires, the balance point is obtained on the 7th wire. To shift the balance point to 9th wire, we should

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 1

To shift the balance point on higher length, the potential gradient of the wire is to be decreased. The same can be obtained by decreasing the current of the main circuit, which is possible by increasing the resistance in series of potentiometer wire.

Test: Electrical Energy and Power (December 4) - Question 2

### 3V poteniometer used for the determination of internal resistance of a 2.4V cell. The balanced point of the cell in open circuit is 75.8cm. When a resistor of 10.2Ω is used in the external circuit of the cell the balance point shifts to 68.3cm length of the potentiometer wire. The internal resistance of the cell is

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 2

Internal resistance of cell

= 3.05 V
l2 = 68.3cm or
= 1.12Ω

 1 Crore+ students have signed up on EduRev. Have you?
Test: Electrical Energy and Power (December 4) - Question 3

### AB is a wire of potentiometer with the increase in the value of resistance R, the shift in the balance point J will be

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 3

Due to increase in resistance R the current through the wire will decrease and hence the potential gradient also decreases, which results in increase in balancing length. So / will shift towards B.

Test: Electrical Energy and Power (December 4) - Question 4

Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference. Their resistivities and lengths are; ρ and L (wire 1), 1.2ρ and 1.2L (wire 2), 0.9ρ and 0.9L (wire 3) and ρ and 1.5L (wire 4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first,

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 4

Resistance of a wire, R = ρl/A
Rate of energy dissipated as heat is
H = V2/R = V2A/ρl
For a wire 1,

For a wire 2,

For a wire 3,

For a wire 4,

∴H3 > H1 > H2 > H4.

Test: Electrical Energy and Power (December 4) - Question 5

If voltage across a bulb rated 220V, 100W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 5

Power, P = V2/R
As the resistance of the bulb is constant
∴ ΔP/P = 2ΔV / V
% decrease in power = ΔP/P × 100
= 2ΔV/V × 100
= 2 × 2.5% = 5%

Test: Electrical Energy and Power (December 4) - Question 6

A heater coil is rated 100 W, 200 V. It is cut into two idential parts. Both parts are connected together in parallel, to the same source of 200 V. The energy liberated per second in the new combination is

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 6

The resistance of heater coil,

= 400Ω
The resistance of either half part = 200 Ω.
Equivalent resistance when both parts are connected in parallel.

The energy liberated per second when the combination is connected to a source of 200 V,

= 400 J

Test: Electrical Energy and Power (December 4) - Question 7

Two 2Ω resistances are connected in parallel in circuit X and in series in circuit Y. The batteries in the two circuits are identical and have zero internal resistance. Assume that the energy transferred to resistor A in circuit X within a certain time is W. The energy transferred to resistor B in circuit Y in the same time will be

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 7

In a circuit X, both the resistance are in parallel, Therefore V is the same and, Power (energy transferred in unit time) is
V2/2 = W
In a circuit Y, both resistance are in series.
Therefore, VB + VB' = V or VB = V/2
In a circuit Y, power supplied to
B  =

Test: Electrical Energy and Power (December 4) - Question 8

In a potentiometer a cell of emf 1.5 V gives a balanced point at 32 cm length of the wire. If the cell is replaced by another cell then the balance point shifts to 65.0 cm then the emf of second cell is

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 8

Here, in the balance condition of potentiometer
ε12 = l1/l2
or  ε= 1.5V , l= 32cm, l2  = 65cm
∴ ε2 = ε1  × l2/l1
= 1.5 × 65/32 = 3.05V

Test: Electrical Energy and Power (December 4) - Question 9

In the circuit shown in figure heat developed across 2Ω, 4Ω and 3Ωresistances are in the ratio

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 9

Current through 2Ω,

Hence produced per second, H1
=
Current through, 4Ω,

Hence produced per second H2

Current through, 3Ω,I
Heat produced, H3 = I2 × 3 = 3I2

∴ H1 : H2 : H3 = 8 : 4 : 27

Test: Electrical Energy and Power (December 4) - Question 10

In a potentiometer the balancing with a cell is at length of 220cm. On shunting the cell with a resistance of 3Ω balance length becomes 130cm. What is the internal resistance of this cell?

Detailed Solution for Test: Electrical Energy and Power (December 4) - Question 10

Here, l1 = 220cm, l2 = 130cm, R = 3Ω
∴ Internal resistance,
r =
= 2.08Ω

## Daily Test for NEET Preparation

12 docs|366 tests
Information about Test: Electrical Energy and Power (December 4) Page
In this test you can find the Exam questions for Test: Electrical Energy and Power (December 4) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Electrical Energy and Power (December 4), EduRev gives you an ample number of Online tests for practice

## Daily Test for NEET Preparation

12 docs|366 tests