Applied Mechanics and Graphic Statics The time period of a simple pendulum depends on (i) Mass of suspended particle (ii) Length of the pendulum (iii) Acceleration due to gravity Both (ii) and (iii) Both (i) and (iii) Only (i) All are correct Both (ii) and (iii) Both (i) and (iii) Only (i) All are correct ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One end of an elastic string of natural length / and modulus X is kept fixed while to the other end is attached a particle of mass m which is hanging freely under gravity. The particle is pulled down vertically through a distance x, held at rest and then released. The motion is a rectilinear motion with constant speed a damped oscillatory motion None of these a simple harmonic motion a rectilinear motion with constant speed a damped oscillatory motion None of these a simple harmonic motion ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If a projectile is fired with an initial velocity of 10 m/sec at an angle of 60° to the horizontal, its horizontal and vertical velocity at the highest point of trajectory are 5 m/sec and 0 5 and 5√3 m/sec 5√3 m/sec and 0 0 and 5 m/sec 5 m/sec and 0 5 and 5√3 m/sec 5√3 m/sec and 0 0 and 5 m/sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Force polygon method is applicable for Any coplanar force system Concurrent coplanar force system Non-concurrent coplanar force system A system of parallel forces only Any coplanar force system Concurrent coplanar force system Non-concurrent coplanar force system A system of parallel forces only ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A stone is thrown up a slope of inclination 60° to the horizontal. At what angle to the slope must the stone be thrown so as to land as far as possible from the point of projection ? 15° 30° 75° 45° 15° 30° 75° 45° ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The velocity of a body fallen from height ‘h’, on reaching the ground is given by v = √(2gh) v = 2gh v = 2gh² v = 1/√(2gh) v = √(2gh) v = 2gh v = 2gh² v = 1/√(2gh) ANSWER DOWNLOAD EXAMIANS APP
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