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:1View AnswerAnswer:
So, correct option is A
Let r be the radius of the circle.then, perimeter of circle=area of circle
2
r=r22=r.
Thus, r=2 units.Thus, the radius of circle is 2 units.
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Q2
In AP, if a=-10, n=6 and an=10, then the value of d is
(A) 0
(B) 4
(C) -4
(D) 10/3Marks:1View AnswerAnswer:
So, correct option is B
Since,
an = a + (n-1)d10 =-10 + (6-1)d
20 = 5d
d=4.Thus, d is 4.
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Q3
Which of the following cannot be the probability of an event?
(A) 1.5
(B) 3/5
(C) 25%
(D) 0.3Marks:1View AnswerAnswer:
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.
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Q4
The mid-point 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:1View AnswerAnswer:
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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:1View AnswerAnswer:
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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) 30
3
(D) 103
Marks:1View AnswerAnswer:
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=1AB=30m. Thus, the height of the tower is 30m.
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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:1View AnswerAnswer:
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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:1View AnswerAnswer:
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=PBAP+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°
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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:1View AnswerAnswer:
Thus, the correct option is C.
It is given thatAOB=100°DAOB is isosceles because OA=OB= radius
OAB=OBAAOB+OAB+OBA=180°100°+
OAB+OAB=180°2
OAB=180° - 100°OAB=80°/2=40°
Now,
OAT=90° [AT is tangent and OA is radius]Thus,
BAT=OAT-OAB=90°-40°
=50°.
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Q10
The roots of the equation x2+x-p(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:1View AnswerAnswer:
