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Gist of EPW May Week 1, 2021

The Economic and Political Weekly (EPW) is an important source of study material for IAS, especially for the current affairs segment. In this... View Article

Pressure Swing Adsorption (PSA)

The Pressure Swing Adsorption (PSA) technology is in the news because of the shortage of medical oxygen that India is facing in the wake of the... View Article

06 May 2021: The Hindu Analysis

UPSC Current Affairs Preparation: The Hindu Analysis Watch expert analysis of ‘The Hindu’ dated 6th May 2021. Important news and views to boost... View Article

Difference Between NRI and PIO

The terms Non-Resident Indian (NRI) and Person of Indian Origin (PIO) are used to describe individuals of Indian birth or descent who live... View Article

06 May 2021: UPSC Exam Comprehensive News Analysis

CNA 6th May 2021:- Download PDF Here TABLE OF CONTENTS A. GS 1 Related GEOGRAPHY 1. Unfamiliar lineament among Assam earthquake factors B. GS 2... View Article

Like parallel forces act at the vertices A, B and C of a triangle ABC and are proportional to the lengths BC, AC and AB respectively. The centre of the forces is at the

1) Centroid 2) Circum-centre 3) Incentre 4) None of these Solution: (3) Incentre Let λa, λb, λc be the forces acting at the points A, B, C... View Article

If the resultant of two forces of magnitude P and 2P is perpendicular to P, then the angle between the forces is

1) 2π/ 3 2) 3π/ 4 3) 4π/5 4) 5π/6 Solution: (1) 2π/ 3 Let the angle between the forces P and 2P be ɑ. Since the resultant of P and 2P is... View Article

Forces M and N acting at a point Omake an angle 150°. Their resultant acts at O has magnitude 2 unit and is perpendicular to M. Then, in the same unit, the magnitudes of M and N are

1) 2√3, 4 2) √3 / 2, 2 3) 3, 4 4) 4, 5 Solution: (1) 2√3, 4 22 = M2 + N2 + 2MN cos 150o 4 = M2 + N2 - √3MN --- (i) tan (π / 2) = [N sin 150o] /... View Article

2 forces P and Q are acting parallel to the length and base of an inclined plane respectively would each of them singly support a weight W, on the plane, then (1 / p²) – (1 / q²) =

1) 1 / W2 2) 2 / W2 3) 3 / W2 4) None of these Solution: (1) 1 / W2... View Article

The resultant of two forces P and Q is R. If the direction of P is reversed keeping the direction Q the same, the resultant remains unaltered. The angle between P and Q is

1) 90° 2) 60° 3) 45° 4) 30° Solution: (1) 90°... View Article

If two forces P and Q act on such an angle that their resultant force R is equal to force P, then if P is doubled then the angle between the new resultant force and Q will be

1) 30° 2) 60° 3) 45° 4) 90° Solution: (4) 90° Clearly P2 = P2 + Q2 + 2PQ cos ɑ Q2 + 2PQ cos ɑ = 0 --- (i) Q + 2P cos ɑ = 0 P is changed... View Article

The resultant of two forces acting on a particle is at right angles to one of them and its magnitude is one-third of the magnitude of the other. The ratio of the larger force to the smaller is

1) 3 : 2√2 2) 3 √3 : 2 3) 3 : 2 4) 4 : 3 Solution: (1) 3 : 2√2 Let P and Q are 2 forces and R be the resultant. tan (π / 2) = Q sin θ / [P +... View Article

If the resultant of two forces of magnitudes P and Q acting at a point at an angle of 60° is √7 Q, then P/Q is

1) 1 2) 3 / 2 3) 2 4) 4 Solution: (3) 2 R2 = P2 + Q2 + 2PQ cos θ (√7Q)2 = P2 + Q2 + 2PQ cos 60o 7Q2 = P2 + Q2 + PQ P2 + PQ - 6Q2 = 0 P2 +... View Article

With two forces acting at a point, the maximum effect is obtained when their resultant is 4 N. If they act at right angles, then their resultant is 3 N. Then the forces are

1) (2 + (1 / 2) √3)N and (2 - (1 / 2) √3N) 2) (2 + √3)N and (2 - √3)N 3) (2 + (1 / 2)√2)N and (2 - (1 / 2) √2)N 4) (2 + √2)N and (2 - √2)N... View Article

Let ABC be an equilateral triangle formed by weightless inextensible strings, the side AB is horizontal. A and B are tied to fixed points D and E by equal weightless inextensible strings AD, BE; a weight of W gm is attached at C. The angle DAB, ABE are each 150°. Then which of the following statements is true

1) The tensions in the strings AB, BC and CA are equal 2) The tensions in the strings BE, AD are inversely proportional 3) The tension in BE is... View Article

Forces proportional to AB, BC and 2CA act along the sides of triangle ABC in order, their resultants represented in magnitude and direction is

1) CA 2) AC 3) BC 4) CB Solution: (1) CA Resultant AB + BC + 2 CA = AC + 2CA = - CA + 2 CA = CA... View Article

If the angle ɑ between 2 forces of equal magnitude is reduced to ɑ – (π / 3, then the magnitude of their resultant becomes √3 times the earlier one. The angle ɑ is

1) π /2 2) 2π /3 3) π / 4 4) 4π / 5 Solution: (2) 2π /3 Let R = √F2 + F2 + 2F2 cos ɑ R’ = √F2 + F2 + 2F2 cos (ɑ - (π / 3)) R’ = √3R √3R = √3... View Article

ABC is an isosceles triangle right angled at A. Forces of magnitude 2 √2, 5 and 6 act along BC, CA and AB respectively. The magnitude of their resultant forces is

1) 4 2) 5 3) 11 + 2 √2 4) 30 Solution: (2) 5 R cos θ = 6 cos 0o + 2 √2 cos (180 - B) + 5 cos 270o R cos θ = 6 - 2 √2 cos B --- (i) R sin θ... View Article

Three forces P, Q, R are acting at a point in a plane. The angle between P, Q and Q, R are 150° and 120° respectively, then for equilibrium; forces P, Q, R are in the ratio

1) 1 : 2 : 3 2) 1 : 2 : 31/2 3) 3 : 2 : 1 4) (3)1/2 : 2 : 1 Solution: (4) (3)1/2 : 2 : 1 Clearly, the angle between P and R is 360o - (150o... View Article

If N is resolved in two components such that the first is twice of other, the components are

1) 5N, 5 √2N 2) 10N, 10 √2N 3) (N / √5), (2N / √5) 4) None of these Solution: (3) (N / √5), (2N / √5) Let the 2 components of N be Nx and Ny... View Article