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In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? [2002]
Total student = 100;
for 70 stds. total marks = 75 x 70 = 5250
⇒ Total marks of girls = 7200 – 5250 = 1950 Average of girls = = 65
The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are [2002]
Given P + Q = 18 .......(1)
P^{2} + Q^{2} + 2PQ cosα = 144 ......(2)
⇒ P + Q cosα = 0 ......(3)
From (2) and (3),
Q^{2} – P^{2} = 144 ⇒ (Q – P) (Q + P) = 144
From (1), On solving, we get Q = 13, P = 5
A bead of weight w can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle θ with the vertical then tension of the thread and reaction of the wire on the bead are
∠ TQW = 180 – θ ; ∠ RQW = 2θ ; ∠ RQT = 180 – θ
Applying Lami's theorem at Q,
⇒ R = W and T = 2W cosθ
The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set [2003]
n = 9 then median term term. Last
four observations are increased by 2. The median is 5th observation which is remaining unchanged.
∴ there will be no change in median.
A couple is of moment and the force forming the couple is is turned through a right angle the moment of the couple thus formed is . If instead , the force are turned through an angle α, then the moment of couple becomes [2003]
...........(2)
...........(3)
From (1), (2) & (3),
The resultant of forces is doubled then is d oubled. If the direction of is reversed,then is again doubled. Then P 2 : Q 2 : R 2 is [2003]
.......(1)
.......(2)
.......(3)
.......(4)
.....(5)
.....(6)
....(7)
A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by
Let the body travels from A to B with constant accelerationt and from B to C with constant retardation r.
If AB = x, BC = y, time taken from A to B = t_{1 }
and time taken from B to C = t_{2},
then s = x + y and
t = t_{1} + t_{2}
For the motion from A to B
v^{2} = u^{2} + 2 fs ⇒ v^{2} = 2 fx (∵u=0)
....(1)
and v = u + ft ⇒ v= ft_{1}
...(2)
For the motion from B to C
Adding equations (1) and (3), we get
Adding equations (2) and (4), we get
Two stones are projected from the top of a cliff h metres high , with the same speed u, so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected horizontally and the other is projected at an angle θ to the horizontal then tanq equals [2003]
For the stone projected horizontally, for horizontal motion, using
distance = speed × time ⇒ R = ut and for vertical motion
∴ We get ....(1)
For the stone projected at an angle θ, for horizontal and vertical motions, we have
R = u cosθ x t ....(2)
and ....(3)
From (1) and (2) we get
Substituting this value of t in eq (3) we get
Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity and the other from rest with uniform acceleration . Let α be the angle between their directions of motion.
The relative velocity of the second particle w.r.t. the first is least after a time [2003]
We can consider the two velocities as
and
∴ Relative velocity of second with respect to first
= f ^{2}t^{2} + u^{2}  2uft cosα
Forto be min we should have
Also
∴ v^{2} and hence v is least at the time
The upper th portion of a vertical pole subtends an angle at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is [2003]
or β = θ  α
or
= 40 or 160 metre
∴ Possible height = 40 metre
Let R_{1} and R_{2} respectively be the maximum ranges up and down an inclined plane and R be the maximum range on the horizontal plane. Then R_{1}, R, R_{2} are in [2003]
Let β be the inclination of the plane to the horizontal and u be the velocity of projection of the projectile
and
In an experiment with 15 observations on x, the following results were available : [2003]
∑x^{2 }= 2830, ∑x= 170
One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is [2003]
∑x = 170, ∑x^{2} = 2830 increase in ∑x = 10 , then ∑x' = 170 + 10 = 180
Increase in ∑x^{2} = 900  400 = 500 then ∑x'^{2} = 2830 + 500 = 3330
Variance
= 222  144= 78
Let R = {(1, 3), ( 4, 2), (2, 4), (2, 3), (3,1)} be a relation on the set A = {1,2, 3,4}. . The relation R is [2004]
∵ (1, 1) ∉ R ⇒ R is not reflexive (2,3) ∈ R but (3, 2) ∉ R
∴ R is not symmetric
Consider the following statements :
(A) Mode can be computed from histogram
(B) Median is not independent of change of scale
(C) Variance is independent of change of origin and scale.Which of these is / are correct ? [2004]
Only first (A) and second (B) statements are correct.
In a series of 2 n observations, half of them equal a and remaining half equal –a. If the standard deviation of the observations is 2, then a equals. [2004]
Clearly mean A = 0
Standard deviation
Hence  a  = 2
With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are [2004]
Let forces be P and Q. then P +Q= 4 ....(1)
and P^{2} +Q^{2}= 32 ....(2)
Solving we get the forces
In a right angle ΔABC , ∠A = 90° and sides a, b, c are respectively, 5 cm, 4 cm and 3 cm. If a force has moments 0, 9 and 16 in N cm. units respectively about vertices A, B and C, then magnitude of is [2004]
Since, the moment about A is zero, hence passes through A. Taking A as origin. Let the line of action of force be y = mx.(see figure)
Moment about
Moment about C
Dividing (1) by (2), we get
Three forces acting along IA, IB and IC, where I is the incentre of a ΔABC are in equilibrium. Then [2004]
IA, IB, IC are bisectors of the angles A, B and C as I is incentre of ΔABC.
Now ∠BIC
Applying Lami’s theorem at I
⇒ P : Q : R =
A paticle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5km/hr. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively [2004]
Time taken by the particle in complete journey
∴ Average speed
Average velocity = [using vector addition]
A velocity m / s is resolved into two components along OA and OB making angles 30° and 45° respectively with the given velocity. Then the component along OB is [2004]
If component along OB
If t_{1} and t_{2} are the times of flight of two particles having the same initial velocity u and range R on the horizontal , then t_{1}^{2}_{ }+ t_{2}^{2} is equal to [2004]
For same horizontal range the angles of projection must
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. The relation is [2005]
Reflexive and transitive only. e.g. (3, 3), (6, 6), (9, 9), (12, 12)
[Reflexive] (3, 6), (6, 12), (3, 12)
[Transitive]. (3, 6) ∈ R but (6, 3) ∉ R [ non symmetric]
ABC is a triangle. Forces acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of D ABC. Then P : Q : R is
IA, IB, IC are bisectors of the angles A, B and C as I is incentre of ΔABC.
Now ∠BIC
Applying Lami’s theorem at I
⇒ P : Q : R =
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [2005]
Mode + 2Mean = 3 Median
⇒ Mode = 3 × 22 – 2 × 21 = 66 – 42 = 24.
A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an acceleration of 2 cm /s^{2} and pursues the insect which is crawling uniformly along a straight line at a speed of 20 cm/s. Then the lizard will catch the insect after [2005]
Let the lizard catches the insect after time t then distance covered by lizard = 21cm + distance covered by insect
⇒ t^{2}  20t  21=0 ⇒ t = 21 sec
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