Applied Mechanics and Graphic Statics When a body falls freely under gravitational force, it possesses No effect on its weight Minimum weight No weight Maximum weight No effect on its weight Minimum weight No weight Maximum weight ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A cable loaded with 10 kN/m of span is stretched between supports in the same horizontal line 100 m apart. If the central dip is 10 m, then the maximum and minimum pull in the cable respectively are 1346.3 kN and 1250 kN 1436.2 kN and 1500 kN 1436.2 kN and 1250 kN 1346.3 kN and 1500 kN 1346.3 kN and 1250 kN 1436.2 kN and 1500 kN 1436.2 kN and 1250 kN 1346.3 kN and 1500 kN ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If two forces each equal to ‘T’ in magnitude act at right angles, their effect may be neutralized by a third force acting along their bisector in opposite direction whose magnitude will be 2 T 1/2 T √2 T 3 T 2 T 1/2 T √2 T 3 T ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The centre of gravity of the trapezium as shown in below figure from the side is at a distance of Applied Mechanics and Graphic Statics mcq question image (h/3) × [(2b + a)/(b + a)] (h/3) × [(b + 2a)/(b + a)] (h/2) × [(2b + a)/(b + a)] (h/2) × [(b + 2a)/(b + a)] (h/3) × [(2b + a)/(b + a)] (h/3) × [(b + 2a)/(b + a)] (h/2) × [(2b + a)/(b + a)] (h/2) × [(b + 2a)/(b + a)] ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If two bodies of masses M1 and M2(M1 > M2) are connected by alight inextensible string passing over a smooth pulley, the tension in the string, will be given by T = g(M1 + M2)/(M1 × M2) T = g(M2 + M1)/(M2 - M1) T = g(M1 - M2)/(M1 + M2) T = g(M2 - M1)/(M1 + M2) T = g(M1 + M2)/(M1 × M2) T = g(M2 + M1)/(M2 - M1) T = g(M1 - M2)/(M1 + M2) T = g(M2 - M1)/(M1 + M2) ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The unit of force in C.G.S. system of units, is called Kg Dyne Newton All of these Kg Dyne Newton All of these ANSWER DOWNLOAD EXAMIANS APP
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