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Two points A and B move from rest along a straight line with constant acceleration f and f ' respectively. If A takes m sec. more than B and describes ‘n’units more than B in acquiring the same speed then [2005]
As per question if point B moves s distance in t time then point A moves (s + n) distance in time (t + m) after which both have same velocity v.
Then using equation v = u + at we get
....(1)
Using equation v^{2} = u^{2}+ 2 , as we get
v^{2} = 2 f (s + n) = 2 f's ....(2)
Also for point B using the eqn we get
Substituting values of t and s from equations (1) and (2) in the above relation, we get
A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance [2005]
Let A and B be displaced by a distance x then Change in moment of (A + B) = applied moments
⇒ (A + B) x x = H ⇒ x
Let x_{1} ,x_{2} , .............. x_{n} be n observations such that = 400 and = 80. Then the possible value of n among the following is [2005]
We know that for positive real numbers x_{1}, x_{2}, ...., x_{n}, A.M. of kth powers of x'_{i}^{s} ≥ kth the power of A.M. of x'_{i}^{s}
⇒ n ≥ 16 . So only possible value for n = 18
A particle is projected from a point O with velocity u at an angle of 60° with the horizontal. When it is moving in a direction at right angles to its direction at O, its velocity then is given by [2005]
u cos 60° = v cos 30° (as horizontal component of velocity remains the same)
The resultant R of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to the smaller one is [2005]
Let F be the larger force
Given R =
Resolving F in horizontal and vertical direction
R =F cosθ ⇒ cosθ =
F' = Fsinθ =
ABC is a triangle, right angled at A. The resultant of the forces acting along with magnitudes respectively is the force along , where D is the foot of the perpdicular from A onto BC. The magnitude of the resultant is [2006]
If we consider unit vectors in the direction AB and AC respectively, then as per quesiton, forces along AB and AC respectively are
∴ Their resultant along AD
∴ Magnitude of resultant is
But from figure ΔABC ~ΔDBA
∴ The required magnitude of resultant becomes
Let W denote the words in the English dictionary. Define the relation R by R = {(x, y) ∈ W × W the words x and y have at least one letter in common.} Then R is
Clearly ( x, x) ∈ R"x ∈ W . So R is relexive.
Let ( x, y) ∈ R , then ( y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric.
But R is not transitive for example Let x = INDIA, y = BOMBAY and z = JOKER then ( x, y) ∈R (A is common) an d ( y, z) ∈ R (O is common) but ( x, z) ∉ R . (as no letter is common)
Suppose a population A has 100 observations 101, 102, ............., 200 and another population B has 100 obsevrations 151, 152, ................ 250. If V_{A} and V_{B} represent the variances of the two populations, respectively then is
(Here deviations are taken from the mean).
Since A and B both have 100 consecutive integers, therefore both have same standard deviation and hence the variance.
(As is same in both the cases)
A particle has two velocities of equal magnitude inclined to each other at an angle θ . If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then q is [2006]
For two velocities u and u at an angle θ to each other the resultant is given by
Now in second case, the new resultant AE (i.e., R") bisects ∠CAB , therefore using angle bisector theorem in ΔABC , we get
A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10m/s ^{2 }, then the height above the point P from where the body began to fall is [2006]
Using and 400 =
Subtracting, we get 400 = 8g + 4gt ⇒ t = 8 sec
= 320
∴ Desired height = 320 + 400 = 720 m
The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be . The value of P is [2007]
Given : Force P = Pn, Q = 3n, resultant R = 7n & P' = Pn,
Q' = (–3)n, R' =
We know that R^{2} = P^{2} + Q^{2 }+ 2PQ cosα
⇒ (7)^{2} = P^{2} + (3)^{2 }+ 2 × P × 3 cosα
⇒ 49 = P^{2} + 9 + 6P cosα
⇒ 40 = P^{2} + 6P cosα .....(i)
And= P^{2} + (–3)^{2 }+ 2P × –3 cosα
⇒ 19 = P^{2} + 9 – 6P cosα
⇒ 10 = P^{2} – 6P cosα.....(ii)
Adding (i) and (ii)
50 = 2P^{2} ⇒ P^{2} = 25 ⇒ P = 5n.
A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection.
The angle of projection is [2007]
Let B be the top of the wall whose coordinates will be (a, b). Range (R) = c
B lies on the trajectory
⇒
⇒
⇒
The angle of projection,
The average marks of boys in class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is [2007]
Let the number of boys be x and that of girls be y.
⇒ 52x + 42y = 50(x + y) ⇒ 52x – 50x = 50y – 42y
⇒ 2x = 8y ⇒ and
Required % of boys =
A body weighing 13 kg is suspended by two strings 5m and 12m long, their other ends being fastened to the extremities of a rod 13m long. If the rod be so held that the body hangs immediately below the middle point, then tensions in the strings are [2007]
∵ 13^{2} = 5^{ 2} + 12^{2} ⇒ AB^{2} = AC^{2} + BC^{2}
⇒ ∠ACB = 90°
Q m is mid point of the hypotenuse AB, therefore MA = MB = MC ⇒ ∠A = ∠ACM =θ
Applying Lami’s theorem at C, we get
⇒ T_{1} = 13 sinθ and T_{2 }= 13 cosθ
⇒ T_{1} = 5 kg and T_{2 }= 12 kg
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b ? [2008]
Mean of a, b, 8, 5, 10 is 6
= ⇒ a + b = 7 ...(i)
Variance of a, b, 8, 5, 10 is 6.80
⇒ a^{2} – 12a + 36 + (1 –a )^{2} + 21= 34 [using eq. (i)]
⇒ 2a^{2} –14a + 24 = 0 ⇒ a^{2} – 7a + 12 = 0
⇒ a = 3 or 4 ⇒ b = 4 or 3
∴ The possible values of a and b are a = 3 and b = 4 or, a = 4 and b = 3
Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “ x is a rational number if f y is a transcendental number”. [2008] Statement1 : r is equivalent to either q or p Statement2 : r is equivalent to ~(p⇿~q).
(None) p : x is an irrational number
q : y is a transcendental number
r : x is a rational number iff y is a transcendental number. clearly r :~p ⇿q
Let us use truth table to check the equivalence of ‘r’ and ‘q or p’; ‘r’ and ~ (p ⇿~ q)
From columns (1), (2) and (3), we observe, none of the these statements are equivalent to each other.
∴ Statement 1as well as statement 2 both are false.
∴ None of the options is correct.
The statement p → (q→p) is equivalent to [2008]
Let us make the truth table for the given statements, as follows :
From table we observe p → (q→p) is equivalent to p→(pνq)
Statement1 : ~ ( p ⇿ ~q) is equivalent to p ⇿ q .
Statement2 : ~ ( p ⇿ ~q) is a tautology [2009]
The truth table for the logical statements, involved in statement 1, is as follows :
We observe the columns for ~ (p ⇿ ~q) and p ⇿ q are identical, therefore ~(p ⇿ ~q) is equivalent to p ⇿ q
But ~ (p ⇿ ~q) is not a tautology as all entries in its column are not T.
∴ Statement1 is true but statement2 is false.
Statement1 : The variance of first n even natural numbers
Statement2 : The sum of first n natural numbers isand the sum of squares of first n natural numbers is [2009]
For the numbers 2, 4, 6, 8, ......., 2n
And Var =
∴ Statement1 is false. Clearly, statement  2 is true.
If A, B and C are three sets such that A ∩ B = A∩C and A ∪ B= A∪C , then [2009]
Let x∈ A and x ∈ B ⇔ x ∈A∪B
⇔ x ∈ A∪C (∵ A ∪ B = A∪C) ⇔ x∈C
∴ B = C.
Let x ∈A and x ∈ B ⇔ x ∈ A∩B
⇔ x ∈ A∩C (∵A ∩ B = A∩C)
⇔ x∈C ∴ B=C
If the mean deviation of the numbers 1, 1 + d, 1 + 2d, .... 1 + 100d from their mean is 255, then d is equal to : [2009]
Mean =
∵ Mean deviation from the mean = 255
Let S be a nonempty subset of R. Consider the following statement : P : There is a rational number x ∈ S such that x > 0.Which of the following statements is the negation of the statement P ? [2010]
P : there is a rational number x ∈S such that x > 0
~ P : Every rational number x ∈S satisfies x ≤ 0
Consider the following relations: R = {(x, y)  x, y are real numbers and x = wy for some rational number w}; and q are integers such that n, q ≠ 0 and qm = pn}.Then [2010]
x Ry need not implies yRx ∴ R is not symmetric and hence not an equivalence relation
Given qm = pn
qm = pn, ps= rq
⇒ ms = rn (transitive).
S is an equivalence relation.
For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is
[2010]
Variance of combined data
Let R be the set of real numbers.Statement1: A = {(x, y) ∈ R × R : y – x is an integer} is an equivalence relation on R.
Statement2: B = {(x, y) ∈ R × R : x = ay for some rational number a} is an equivalence relation on R. [2011]
x – y is an integer.
∵ x – x = 0 is an integer ⇒ A is reflexive.
Let x – y is an integer ⇒ y – x is an integer
⇒ A is symmetric Let x – y, y – z are integers
⇒ x – y + y – z is also an integer
⇒ x – z is an integer ⇒ A is transitive
∴ A is an equivalence relation.
Hence statement 1 is true.
Also B can be considered as
xBy if a rational number
is a rational number
But a rational number need not imply a rational number because
is rational ⇒ is not rational
∴ B is not an equivalence relation.
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