Physics
General
Easy
Question
Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be



 None of the above
The correct answer is:
or
On waxing the number of beats decreases hence
Related Questions to study
Maths
If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are
a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are
MathsGeneral
a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.
So we have:
maths
Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is 
For 1 i 4, let x_{i} ( 3) be the number of blanks between i^{th} and (i + 1)^{th} letters. Then,
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is 
mathsGeneral
For 1 i 4, let x_{i} ( 3) be the number of blanks between i^{th} and (i + 1)^{th} letters. Then,
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
x_{1} + x_{2} + x_{3} + x_{4} = 15 …..(1)
The number of solutions of (1)
= coefficient of t^{15} in (t^{3} + t^{4} +….)^{4}
= coefficient of t^{3} in (1 – t)^{–4}
= coefficient of t^{3} in [1 + ^{4}C_{1} + ^{5}C_{2} t^{2} + ^{6}C_{3 }t^{3} + …..]
= ^{6}C_{3} = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
maths
The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is 
Note that 7^{r} (r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
Thus, 7^{m} + 7^{n} cannot end in 5 for any values of m, n N. In other words, for 7^{m} + 7^{n} to be divisible by 5, it should end in 0.
For 7^{m} + 7^{n} to end in 0, the forms of m and n should be as follows :
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just 25 values of n for which 7^{m} + 7^{n} ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^{m} + 7^{n} is divisible by 5.
Hence
Thus, 7^{m} + 7^{n} cannot end in 5 for any values of m, n N. In other words, for 7^{m} + 7^{n} to be divisible by 5, it should end in 0.
For 7^{m} + 7^{n} to end in 0, the forms of m and n should be as follows :
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just 25 values of n for which 7^{m} + 7^{n} ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^{m} + 7^{n} is divisible by 5.
Hence
The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is 
mathsGeneral
Note that 7^{r} (r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
Thus, 7^{m} + 7^{n} cannot end in 5 for any values of m, n N. In other words, for 7^{m} + 7^{n} to be divisible by 5, it should end in 0.
For 7^{m} + 7^{n} to end in 0, the forms of m and n should be as follows :
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just 25 values of n for which 7^{m} + 7^{n} ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^{m} + 7^{n} is divisible by 5.
Hence
Thus, 7^{m} + 7^{n} cannot end in 5 for any values of m, n N. In other words, for 7^{m} + 7^{n} to be divisible by 5, it should end in 0.
For 7^{m} + 7^{n} to end in 0, the forms of m and n should be as follows :
m n
1 4r 4s + 2
2 4r + 1 4s + 3
3 4r + 2 4s
4 4r + 3 4s + 1
Thus, for a given value of m there are just 25 values of n for which 7^{m} + 7^{n} ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^{m} + 7^{n} is divisible by 5.
Hence
maths
Consider the following statements:
1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.
2. A pack of 52 cards can be divided equally among four players in order in ways.
Which of these is/are correct?
(1) Total number of ways of arranging m things = m!.
To find the number of ways in which p particular things are together, we consider p particular things as a group.
Number of ways in which p particular things are together = (m – p + 1)! p!
So, number of ways in which p particular things are not together
= m! – (m – p + 1)! p!
Total number of ways =
Hence, both of statements are correct.
To find the number of ways in which p particular things are together, we consider p particular things as a group.
Number of ways in which p particular things are together = (m – p + 1)! p!
So, number of ways in which p particular things are not together
= m! – (m – p + 1)! p!
Total number of ways =
Hence, both of statements are correct.
Consider the following statements:
1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.
2. A pack of 52 cards can be divided equally among four players in order in ways.
Which of these is/are correct?
mathsGeneral
(1) Total number of ways of arranging m things = m!.
To find the number of ways in which p particular things are together, we consider p particular things as a group.
Number of ways in which p particular things are together = (m – p + 1)! p!
So, number of ways in which p particular things are not together
= m! – (m – p + 1)! p!
Total number of ways =
Hence, both of statements are correct.
To find the number of ways in which p particular things are together, we consider p particular things as a group.
Number of ways in which p particular things are together = (m – p + 1)! p!
So, number of ways in which p particular things are not together
= m! – (m – p + 1)! p!
Total number of ways =
Hence, both of statements are correct.
maths
The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to
Let ‘l’ is associated with ‘r’ ,
r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
= =
=
= = 35
Hence (a) is correct answer.
r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
= =
=
= = 35
Hence (a) is correct answer.
The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to
mathsGeneral
Let ‘l’ is associated with ‘r’ ,
r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
= =
=
= = 35
Hence (a) is correct answer.
r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
= =
=
= = 35
Hence (a) is correct answer.
maths
The number of points in the Cartesian plane with integral coordinates satisfying the inequalities x k, y k, x – y k ; is
x k –k x k ….(1)
& y k –k y k ….(2)
& x – y k y – x k ….(3)
– k y – x k x – k y x + k
Number of points having integral coordinates
= (2k + 1)^{2} – 2[k + (k – 1) + …. + 2 + 1]
= (3k^{2} + 3k + 1).
& y k –k y k ….(2)
& x – y k y – x k ….(3)
– k y – x k x – k y x + k
Number of points having integral coordinates
= (2k + 1)^{2} – 2[k + (k – 1) + …. + 2 + 1]
= (3k^{2} + 3k + 1).
The number of points in the Cartesian plane with integral coordinates satisfying the inequalities x k, y k, x – y k ; is
mathsGeneral
x k –k x k ….(1)
& y k –k y k ….(2)
& x – y k y – x k ….(3)
– k y – x k x – k y x + k
Number of points having integral coordinates
= (2k + 1)^{2} – 2[k + (k – 1) + …. + 2 + 1]
= (3k^{2} + 3k + 1).
& y k –k y k ….(2)
& x – y k y – x k ….(3)
– k y – x k x – k y x + k
Number of points having integral coordinates
= (2k + 1)^{2} – 2[k + (k – 1) + …. + 2 + 1]
= (3k^{2} + 3k + 1).
maths
The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is 
The required no. of ways = no. of solution of the equation (x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = K)
Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}
= Coefficient of x^{K} in
= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)
(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)
= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}
= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.
Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}
= Coefficient of x^{K} in
= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)
(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)
= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}
= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.
The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is 
mathsGeneral
The required no. of ways = no. of solution of the equation (x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = K)
Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}
= Coefficient of x^{K} in
= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)
(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)
= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}
= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.
Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}
= Coefficient of x^{K} in
= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)
(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)
= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}
= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.
maths
The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are 
Total number of points = m +n + k. Therefore the total number of triangles formed by these points is ^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2} and k points lie on I_{3} and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
^{m + n +} ^{k}C_{3} – ^{m}C_{3} – ^{n}C_{3} – ^{k}C_{3}.
^{m + n +} ^{k}C_{3} – ^{m}C_{3} – ^{n}C_{3} – ^{k}C_{3}.
The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are 
mathsGeneral
Total number of points = m +n + k. Therefore the total number of triangles formed by these points is ^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2} and k points lie on I_{3} and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
^{m + n +} ^{k}C_{3} – ^{m}C_{3} – ^{n}C_{3} – ^{k}C_{3}.
^{m + n +} ^{k}C_{3} – ^{m}C_{3} – ^{n}C_{3} – ^{k}C_{3}.
Maths
If the line is a normal to the hyperbola then
If the line is a normal to the hyperbola then
MathsGeneral
Maths
If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then
A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.
We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.
We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.
If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then
MathsGeneral
A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.
We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.
We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.
Maths
The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is
The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is
MathsGeneral
Maths
If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –
An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
Now we have given the LCM as: r^{2}t^{4}s^{2}
Consider following cases:
Case 1: if p contains r^{2} then q will have r^{k}, for the value k=0,1.
So 2 ways.
Case 2: if q contains r^{2} then p will have r^{k}, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r^{2}
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.
Now we have given the LCM as: r^{2}t^{4}s^{2}
Consider following cases:
Case 1: if p contains r^{2} then q will have r^{k}, for the value k=0,1.
So 2 ways.
Case 2: if q contains r^{2} then p will have r^{k}, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r^{2}
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.
If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –
MathsGeneral
An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
Now we have given the LCM as: r^{2}t^{4}s^{2}
Consider following cases:
Case 1: if p contains r^{2} then q will have r^{k}, for the value k=0,1.
So 2 ways.
Case 2: if q contains r^{2} then p will have r^{k}, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r^{2}
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.
Now we have given the LCM as: r^{2}t^{4}s^{2}
Consider following cases:
Case 1: if p contains r^{2} then q will have r^{k}, for the value k=0,1.
So 2 ways.
Case 2: if q contains r^{2} then p will have r^{k}, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r^{2}
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.
Maths
A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length
Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m^{2}n^{2}
So it is m^{2}n^{2}.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m^{2}n^{2}
So it is m^{2}n^{2}.
A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length
MathsGeneral
Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m^{2}n^{2}
So it is m^{2}n^{2}.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m^{2}n^{2}
So it is m^{2}n^{2}.
maths
^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)
^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)
mathsGeneral
maths
The coefficient of in is
The coefficient of in is
mathsGeneral