Question

A mass M is lowered with the help of a string by a distance h at a constant acceleration g/2. The work done by the string will be

## The correct answer is:

### Related Questions to study

### A particle begins at the origin and moves successively in the following manner as shown, 1 unit to the right, 1/2 unit up, 1/4 unit to the right, 1/8 unit down, 1/16 unit to the right etc. The length of each move is half the length of the previous move and movement continues in the ‘zigzag’ manner indefinitely. The co-ordinates of the point to which the ‘zigzag’ converges is –

### A particle begins at the origin and moves successively in the following manner as shown, 1 unit to the right, 1/2 unit up, 1/4 unit to the right, 1/8 unit down, 1/16 unit to the right etc. The length of each move is half the length of the previous move and movement continues in the ‘zigzag’ manner indefinitely. The co-ordinates of the point to which the ‘zigzag’ converges is –

### The greatest possible number of points of intersections of 8 straight line and 4 circles is :

The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows

The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.

The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

### The greatest possible number of points of intersections of 8 straight line and 4 circles is :

The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows

The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.

The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.

The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.

### Assertion (A):If ,then A

Reason (R) :

### Assertion (A):If ,then A

Reason (R) :

A force–time graph for a linear motion is shown in figure where the segments are circular. The linear momentum gained between zero and is

A force–time graph for a linear motion is shown in figure where the segments are circular. The linear momentum gained between zero and is

### How many different nine digit numbers can be formed from the number 2,2,3,3,5,5,8,8,8 by rearranging its digits so that the odd digits occupy even position ?

Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.

### How many different nine digit numbers can be formed from the number 2,2,3,3,5,5,8,8,8 by rearranging its digits so that the odd digits occupy even position ?

Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.

### A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

### A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

### The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Alternatively, we can use the formula for the sum of numbers as

We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

### The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Alternatively, we can use the formula for the sum of numbers as

We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

### Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :

We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

### Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :

We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

### If ^{9}P_{5} + 5 ^{9}P_{4} = , then r =

### If ^{9}P_{5} + 5 ^{9}P_{4} = , then r =

### Assertion (A) :If , then

Reason (R) :

### Assertion (A) :If , then

Reason (R) :

A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively