# NEET (NTA)-National Eligibility cum Entrance Test (Medical) Physics: Questions 1515 - 1520 of 2320

Access detailed explanations (illustrated with images and videos) to 2320 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. Unlimited Access for Unlimited Time! View Sample Explanation or View Features.

Rs. 600.00 or

How to register?

## Question number: 1515

Edit
MCQ▾

### Question

The energy released per fission of U 235 is about 200 MeV. A reactor using U 235 as fuel is producing 1000 kilowatt power. The number of U 235 nuclei undergoing fission per sec is, approximately-

### Choices

Choice (4)Response

a.

10 6

b.

3 × l0 16

c.

2 × 10 8

d.

931

## Passage

In a living organism, the quantity of C 14 is the same as in the atmosphere. But in organisms which are dead, no exchange takes place with the atmosphere and by measuring the decay rate of 14 C in the old bones or wood, the time taken for the activity to reduce to this level can be calculated. This gives the age of the wood or bone.

Given: T 1/2 for 14 C is 5370 years and the ratio of is 1.3 ⨯ 10 -12.

## Question number: 1516 (1 of 3 Based on Passage) Show Passage

Edit
MCQ▾

### Question

The decay rate of 14 C in l gm of carbon in a living organism is,

### Choices

Choice (4)Response

a.

25 Bq

b.

5 Bq

c.

2.5 Bq

d.

0.25 Bq

## Question number: 1517 (2 of 3 Based on Passage) Show Passage

Edit
MCQ▾

### Question

If in an old sample of wood of 10g the decay rate is 30 decays per minute, the age of the wood is

### Choices

Choice (4)Response

a.

1000 years

b.

50 years

c.

13310 years

d.

15300 years

## Question number: 1518 (3 of 3 Based on Passage) Show Passage

Edit
MCQ▾

### Question

The decay rate in another piece is found to be 0.30 Bq per gm then we can conclude,

### Choices

Choice (4)Response

a.

he sample is very recent

b.

There is a statistical error

c.

The observed decay is not that of 14 C alone

d.

All a. , b. and c. are correct

## Question number: 1519

Edit
MCQ▾

### Question

For thorium A = 232 and Z = 90, at the end of some radioactive disintegrations we obtain an isotope of lead with A = 208 and Z = 82. Then the number of α and β particles emitted are,

### Choices

Choice (4)Response

a.

b.

c.

d.

## Question number: 1520

» Thermodynamics » Laws of Thermodynamics » Second Law of Thermodynamics

Edit
Assertion-Reason▾

### Assertion (Ꭺ)

A room can be warmed by opening the door of a refrigerator in a closed room

### Reason (Ꭱ)

Head flows from lower temperature (refrigerator) to higher temperature (room).

### Choices

Choice (4)Response

a.

Both Ꭺ and Ꭱ are false

b.

Ꭺ is false but Ꭱ is true

c.

Both Ꭺ and Ꭱ are true and Ꭱ is the correct explanation of Ꭺ

d.

Ꭺ is true but Ꭱ is false

Developed by: