The wheel of scooter has diameter 140 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 132 km per hour?
A. 1100
B. 500
C. 250
D. 1000
Answer : Option B
Distance travelled by wheel in one revolution = circumference of wheel
= 22/7 × 140 = 440 cm.
Speed of scooter = 132 km/hr = 132 × 1000 × 100/60 cm/min = 220,000 cm/min.
The wheel has therefore got to travel 220,000 cm in 1 min i.e. it has to perform 220,000/440 revolution in 1 min i.e. 500 revolutions.
In the figure AC and BC are radii of circles. The length of AB is 8. If AC = 4, what is BC? (BC is tangent to the circle with center A.)
Answer : Option A
Since PR is tangent to circle with centre O or is perpendicular to PR.
Δ ORP is right angled triangle.
So, BC = √(AB2 – AC2)
= √(82) – 42) = √48 = 4√3
Two circles with same center are drawn with O as the centre as shown is the figure given below. The ratio of the area of the annular ring bounded by these two circles and the quadrilateral EBCH is 3×:2. Find the ratio of the radius of the smaller circle to the radius of the larger circle.
A. 1:4
B. 2:7
C. 1:6
D. 1:7
Answer : Option D
Let the radii of the larger and smaller circle be r1 and r2 respectively.
Area of the annular ring = π(r12 - r22)
Area of the quadrilateral EBC = Area (Δ EBO+Δ BOC+Δ COH+Δ HOE) = 1/2(r1 - r2)2
ATQ.π(r12 + r22)/(1/2)(r1 + r2)2 = r1 : r2 = 1:7
A semicircle having centre at O and radius equal to 4 is drawn with PQ as the diameter as shown is the figure given below. OSRU is a rectangle such that the ratio of area of the semicircle to the area of the rectangle is 2π: 3 or cuts the semicircle at T. Find the length of line segment TQ.
A. (8/5)√5
B. (5/3)√5
C. (17/9)√5
D. (9/2)√5
Answer : Option A
Join AP and draw PQ ⊥ AB
As per question π × 42/2×4×SR = 2π/3
SR = 3 unit = OU
∴ OR =√OS2 + SR2 = 5units
Let OV = x, ∴ QV = OQ – OV = (4 – x) units
ΔOTV ˜ ΔORU
⇒ OT/OR = OV/OU ⇒ 4/5 = x/3 ⇒ x = 12/5 units
VQ = (8/5)units
Also TV/RU = OT/OR ⇒ TV =OT/OR × RU =(4/5) × 4 =16/5 units
TQ= √TV2 + VQ2 = √ (256/25) + (64/25) = 8√5/5 units
A circle of radius 3 cm is drawn inscribed in a right angle triangle ABC, right angled at C. If AC is 10 Find the value of CB
A. 10.5cm
B. (20/7)√58 cm
C. 21/2 cm
D. None of these
Answer : Option A
(x + 7)2 = 102 + (x + 3)2
(Tangents from the external point are equal)
⇒ x = 7.5cm ⇒ OB = x + 3 = 10.5 cm
In the circle shown below, AB is a diameter and CD is a chord. The lengths of AB and CD are integers, such that the length of CD is obtained by reversing the two digits of length of AB. If the length of OE is a rational number, find the length of OE.
A. 16
B. 16.5
C. 33
D. None of these
Answer : Option B
Let AB = 10x + y ⇒ CD = 10y + x
By joining OC, a triangle OCE is formed
⇒ OE2 = OC2 – CE2
= ((1/2)AB)2 - ((1/2)CD)2
⇒OE2= (10x + y/2)2 - (10y + x/2)2
⇒OE2 = (1/2)√99x2 - 99y2
⇒OE2 = (3/2)√11(x + y)(x - y)
Since length of OE is a rational number ⇒ √11\(x2 - y2) must be a perfect square.
The only combination is x = 6, y = 5
(Since x + y is the 0 to 9)
∴ OE = (3/2)× 11 = 16.5
In the given figure, AB is the diameter of the circle with center O. If ∠BOD = 15° & ∠EOA = 85°, then find the value of ∠ECA.
The smallest possible circle, touching two opposite sides of a rectangle, is cut out from a rectangle of area 60 sq units. If the area of the circle is 1.5 times the uncut area left in the rectangle, find the diameter of the circle.
Answer : Option B
Radius = a/2
π(a/2)2 = (3/2) [ 60-π(a/2)2]
⇒ 5π(a/2)2 = 180
⇒ 5πa2/4 = 180 ⇒ a =(12/√π) units
The line √3y = x meets the circle having the centre at origin O at point M(√3,1) . Referring to the diagram below, if PQ is a tangent to the circle at M, find the length of PQ.
A. (5/2)√3
B. 3√3
C. 2√3
D. 8/√3
Answer : Option D
PQ is perpendicular to the line y = x/√3
Therefore, Slope of line = -1/(1/√3) = -√3
Let equation of line PQ be y = - √3x + C
When x = √3 , y = 1 ⇒ C = 4
Therefore, Co-ord. of Q , (4/√3, 0),
Co-ordinates of P (0, 4)
Hence PQ= √ (4/√3)2 + 42 = 4√(1/3) + 1 =8/√3 units .
In the figure O and O’ are the centers of the bigger and smaller circles respectively. The smaller circle touches the square ABCD at the midpoint of side AD. The radius of the bigger circle is 30 cm and the side of the square ABCD is 36 cm. Find the radius of the smaller circle.
A. 9.25 cm
B. 9 cm
C. 9.45 cm
D. 10 cm
Answer : Option B
OE = R – EF = R – [2R– (2r + 2a)]
OE2 + EB2 = OB2
i.e. (2a + 2r – R)2+ a2 = R2, for a = 18 cm
& R = 30 cm
⇒ r = 9 cm
OE2 + EB2 = OB2
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