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 (i) and (iii) Both (ii) and (iii) Only (i) All are correct Both (i) and (iii) Both (ii) and (iii) Only (i) All are correct ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Cartesian form of the equation of catenary is y = c sin x/c y = c sinh x/c y = c cosh x/c y = c tan x/c y = c sin x/c y = c sinh x/c y = c cosh x/c y = c tan x/c ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Three forces which act on a rigid body to keep it in equilibrium. The forces must be coplanar and Parallel None of these Concurrent parallel Concurrent Parallel None of these Concurrent parallel Concurrent ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One Newton is equivalent to 981 dyne 1/9.81 kg. wt 9.81 kg. wt 1 kg. wt 981 dyne 1/9.81 kg. wt 9.81 kg. wt 1 kg. wt ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The resultant of the forces acting on a body will be zero if the body Moves along a curved path Moves with variable velocity in a straight line Rotates Does not move at all Moves along a curved path Moves with variable velocity in a straight line Rotates Does not move at all ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A square hole is punched out of a circular lamina, the diagonal of the square being the radius of the circle. If ‘r’ is the radius of the circle, the C.G. of the remainder from the corner of the square on the circumference will be [r (π + 0.25)]/(π + 0.5) [r (π + 0.25)]/(π - 0.5) [r (π - 0.25)]/(π - 0.5) [r (π - 0.5)]/(π + 0.25) [r (π + 0.25)]/(π + 0.5) [r (π + 0.25)]/(π - 0.5) [r (π - 0.25)]/(π - 0.5) [r (π - 0.5)]/(π + 0.25) ANSWER DOWNLOAD EXAMIANS APP
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