Question
An elastic string of unstretched length and force constant is stretched by a small length . It is further stretched by another small length . The work done in the second stretching is
The correct answer is:
Related Questions to study
A shell of mass moving with velocity suddenly breaks into 2 pieces. The part having mass remains stationary. The velocity of the other shell will be
A shell of mass moving with velocity suddenly breaks into 2 pieces. The part having mass remains stationary. The velocity of the other shell will be
If 
then B=
If then B=
The set S : = { 1, 2, 3 .........12} is to be partitioned into three sets A, B, C of equal size. Thus A 
B C = S, A 
B = B C = A C = 
. The number of ways to partition S is -
×
= 
The set S : = { 1, 2, 3 .........12} is to be partitioned into three sets A, B, C of equal size. Thus A B C = S, A B = B C = A C = . The number of ways to partition S is -
× = 
A one kilowatt motor is used to pump water from a well 10 m deep. The quantity of water pumped out per second is nearly
A one kilowatt motor is used to pump water from a well 10 m deep. The quantity of water pumped out per second is nearly
Assertion (A) : The Remainder obtained when the polynomial 
is divided by 
Is 1
Reason (R): If 
is divided by 
then the remainder is f(a)
Assertion (A) : The Remainder obtained when the polynomial is divided by Is 1
Reason (R): If is divided by then the remainder is f(a)
A ball hits the floor and rebounds after inelastic collision. In this case
By the conservation of momentum in the absence of external force total momentum of the system (ball + earth) remains constant
A ball hits the floor and rebounds after inelastic collision. In this case
By the conservation of momentum in the absence of external force total momentum of the system (ball + earth) remains constant
Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II Principle of conservation of momentum holds true for all kinds of collisions.
Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement II Principle of conservation of momentum holds true for all kinds of collisions.
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
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
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 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.
For selecting r objects from n objects can be done by using the formula as follows
As mentioned in the question, we have to find the total number of intersection points.
For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows =
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows =
For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows =
Hence, the total number of points of intersection is = 28 + 64 + 12 = 104
The greatest possible number of points of intersections of 8 straight line and 4 circles is :
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.
For selecting r objects from n objects can be done by using the formula as follows
As mentioned in the question, we have to find the total number of intersection points.
For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows =
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows =
For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows =
Hence, the total number of points of intersection is = 28 + 64 + 12 = 104
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
As the area above the time axis is numerically equal to area below the time axis therefore net momentum gained by body will be zero because momentum is a vector quantity
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
As the area above the time axis is numerically equal to area below the time axis therefore net momentum gained by body will be zero because momentum is a vector quantity
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 need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
Here, e is for the even places and O is for the odd places of the digit number.
The digits which are even are 2, 2, 8, 8 and 8.
Number of even digits = 5
The digits which are odd are 3, 3, 5 and 5.
Number of odd digits = 4
We have to arrange the odd digits in even places.
Number of ways to arrange the odd digits in 4 even places =
= 6Now, we have to arrange the even digits in odd places.
Number of ways to arrange the even digits in 5 odd places =
Total number of 9 digits number =
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 need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
Here, e is for the even places and O is for the odd places of the digit number.
The digits which are even are 2, 2, 8, 8 and 8.
Number of even digits = 5
The digits which are odd are 3, 3, 5 and 5.
Number of odd digits = 4
We have to arrange the odd digits in even places.
Number of ways to arrange the odd digits in 4 even places =
= 6Now, we have to arrange the even digits in odd places.
Number of ways to arrange the even digits in 5 odd places =
Total number of 9 digits number =
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 :
We have to find the number of ways in which exactly 10 predictions are correct which can be shown by
ways in which his prediction is correct.And in the remaining 10 matches, he makes wrong predictions i.e. out of 3 outcomes (win, lose, tie) he can pick 2 outcomes out of 3 , which can be done in
ways.Thus, 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 :
We have to find the number of ways in which exactly 10 predictions are correct which can be shown by
ways in which his prediction is correct.And in the remaining 10 matches, he makes wrong predictions i.e. out of 3 outcomes (win, lose, tie) he can pick 2 outcomes out of 3 , which can be done in
ways.Thus, 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) :
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, 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) : 93324
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) :
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, 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) : 93324
