CBSE [Delhi]_X_Mathematics_2011_Set II
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Q1
If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) units
(C) 4 units
(D) 7 unitsMarks:1Answer:
So, correct option is A
Let r be the radius of the circle.then, perimeter of circle=area of circle
2r=r^{2}
2=r.
Thus, r=2 units.Thus, the radius of circle is 2 units.

Q2
In AP, if a=10, n=6 and a_{n}=10, then the value of d is
(A) 0
(B) 4
(C) 4
(D) 10/3Marks:1Answer:
So, correct option is B
Since,
a_{n }= a + (n1)d10 =10 + (61)d
20 = 5d d=4.
Thus, d is 4.

Q3
Which of the following cannot be the probability of an event?
(A) 1.5
(B) 3/5
(C) 25%
(D) 0.3Marks:1Answer:
The correct answer is A.
The probability of an event is always greater than or equal to zero and less than or equal to one i.e., 0£P(E)£1.
Here, 3/5=0.6 and 25%=0.25
Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and are less than or equal to 1. But 1.5 is greater than 1. Thus, 1.5 can not be the probability of an event.

Q4
The midpoint of segment AB is the point P (0,4). If the coordinates of B are (2,3) then the coordinates of A are
(A) (2, 5)
(B) (2, 5)
(C) (2, 9)
(D) (2,11)Marks:1Answer:

Q5
The point P which divides the line segment joining the points A(2,5) and B(5,2) in the ratio 2:3 lies in the quadrant.
(A) I
(B) II
(C)III
(D)IVMarks:1Answer:

Q6
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 45°. The height of the tower (in metres) is
(A) 15
(B) 30
(C) 303
(D) 103Marks:1Answer:
The correct option is B
Let AB be the tower and P be the point on the ground. It is given that BP=30 m, P=45°Now, AB/BP=tan45°
AB/30m=1
AB=30m.
Thus, the height of the tower is 30m.

Q7
A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then the water level
rises (in cm) by
(A) 3
(B) 4
(C) 5
(D) 6Marks:1Answer:

Q8
In figure, PA and PB are tangents to the circles with centre O. If APB=60°, then OAB is
(A) 30°
(B) 60°
(C) 90°
(D) 15°Marks:1Answer:
Thus, the correct option is A.
PA and PB are tangents to the circle from an external point O.
Therefore, PA=PB.
PAB is an isosceles triangle where PAB=PBA
P+PAB+PBA=180° [Angle sum property of triangle]
60°+2ÐPAB=180°
2ÐPAB=180°60°
=120°
ÐPAB=120°/2
=60°
It is known that the radius is perpendicular to the tangent at the point of contact.
OAB=90°
PAB+OAB=90°
OAB=90°  60°
=30°

Q9
In figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If AOB=100°, then BAT is equal to
(A) 100°
(B) 40°
(C) 50°
(D) 90°Marks:1Answer:
Thus, the correct option is C.
It is given that AOB=100°DAOB is isosceles because OA=OB= radius
OAB=OBA
AOB+OAB+OBA=180°
100°+OAB+OAB=180°
2OAB=180°  100°
OAB=80°/2
=40°
Now, OAT=90° [AT is tangent and OA is radius]
Thus, BAT=OATOAB
=90°40°
=50°.

Q10
The roots of the equation x^{2}+xp(p+1)=0, where p is a constant, are
(A) p,p+1
(B) p, p+1
(C) p,(p+1)
(D) –p,(p+1)Marks:1Answer: