# NEET (NTA)-National Eligibility cum Entrance Test (Medical) Physics: Questions 1802 - 1806 of 2184

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## Question number: 1802

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MCQ▾

### Question

Two planets of radii in the ratio 2: 3 are made from the materials of density in the ratio 3: 2. Then the ratio of acceleration due to gravity at the surface of two planets will be,

### Choices

Choice (4) Response

a.

2.25

b.

c.

0.12

d.

1

## Question number: 1803

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MCQ▾

### Question

Two potentiometers A and B having 4 wires and 10 wires, each having 100 cm in length are used to compare e. m. f. of 2 cells. Which one will give a larger balancing length?

### Choices

Choice (4) Response

a.

Potentiometer A

b.

Potentiometer B

c.

Balancing length doesn’t depend on the total length of the wire.

d.

A and B will give same balancing length

## Question number: 1804

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MCQ▾

### Question

If μ 1 and μ 2 are two refractive indices of the materials of core and cladding of an optical fiber, then the loss of light due to its leakage can be minimized by having

### Choices

Choice (4) Response

a.

μ 1 > μ 2

b.

μ 1 = μ 2

c.

μ 1 < μ 2

d.

Question does not provide sufficient data or is vague

## Question number: 1805

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### Question

A body of mass 100 kg falls on the earth from infinity. What will be its energy on reaching the earth? Radius of the earth is 6400 km and g = 9.8 m/s 2. Air friction is negligible.

### Choices

Choice (4) Response

a.

6.27 ⨯ 10 9 J

b.

6.27 ⨯ 10 8 J

c.

6.27 ⨯ 10 10 J

d.

6.27 ⨯ 10 7 J

## Passage

Consider a block of conducting material of resistivity ′ ρ ′ shown in the figure. Current I enter at ‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ∆ V’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:

(i) Take current T entering from ‘A’ and assume it to spread over a hemispherical surface in the block A Block of Conducting Material In figure a block of conducting material of resistivity ρ shown in figure. Current I enter at A and leaves from D.

(ii) Calculate field E (r) at distance ‘r’ from A by using Ohm’s law E = ρ j, where j is the current per unit area at ‘r’.

(iii) From the V dependence of E (r), obtain the potential V (r) at r.

(iv) Repeat (i), (ii) and (iii) for current

## Question number: 1806 (1 of 2 Based on Passage) Show Passage

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MCQ▾

### Question

∆V measured between B and C is,

### Choices

Choice (4) Response

a.

b.

c.

d.