Engineering Mechanics The maximum efficiency of a screw jack is (1 - sinφ)/(1 + sinφ) (1 - tanφ)/(1 + tanφ) (1 + sinφ)/(1 - sinφ) (1 + tanφ)/(1 - tanφ) (1 - sinφ)/(1 + sinφ) (1 - tanφ)/(1 + tanφ) (1 + sinφ)/(1 - sinφ) (1 + tanφ)/(1 - tanφ) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The unit of angular velocity is Revolutions/min rad/s m/min Both (B) and (C) Revolutions/min rad/s m/min Both (B) and (C) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The centre of percussion of the homogeneous rod of length ‘L’ suspended at the top will be 2L/3 L/3 L/2 3L/4 2L/3 L/3 L/2 3L/4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If u₁ and u₂ are the velocities of two moving bodies in the same direction before impact and v₁ and v₂ are their velocities after impact, then coefficient of restitution is given by (u₁ - u₂)/(v₁ - v₂) (u₂ + u₁)/(v₂ + v₁) (v₂ - v₁)/(u₁ - u₂) (v₁ - v₂)/(u₁ - u₂) (u₁ - u₂)/(v₁ - v₂) (u₂ + u₁)/(v₂ + v₁) (v₂ - v₁)/(u₁ - u₂) (v₁ - v₂)/(u₁ - u₂) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics A trolley wire weighs 1.2 kg per meter length. The ends of the wire are attached to two poles 20 meters apart. If the horizontal tension is 1500 kg find the dip in the middle of the span 3.0 cm 4.0 cm 2.5 cm 5.0 cm 3.0 cm 4.0 cm 2.5 cm 5.0 cm ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) IP = IG + Ah2 IP = IG - Ah2 IP = Ah2 / IG IP = IG / Ah2 IP = IG + Ah2 IP = IG - Ah2 IP = Ah2 / IG IP = IG / Ah2 ANSWER DOWNLOAD EXAMIANS APP
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