A body having moment of inertia of 30 kg m2 is rotating at 210 RPM and mashes with another body at rest having M.I. of 40 kg m2. The resultant speed after meshing will be
A. 90 RPM
B. 100 RPM
C. 80 RPM
D. None of the mentioned
Answer: Option A
A. 90 RPM
B. 100 RPM
C. 80 RPM
D. None of the mentioned
Answer: Option A
In considering friction of a V-thread, the virtual coefficient of friction (μ1) is given by
A. μ1 = μsinβ
B. μ1 = μcosβ
C. $${\mu _1} = \frac{\mu }{{\sin \beta }}$$
D. $${\mu _1} = \frac{\mu }{{\cos \beta }}$$
The lower pairs are _________ pairs.
A. Self-closed
B. Force-closed
C. Friction closed
D. None of these
In a coupling rod of a locomotive, each of the four pairs is a ________ pair.
A. Sliding
B. Turning
C. Rolling
D. Screw
A kinematic chain is known as a mechanism when
A. None of the links is fixed
B. One of the links is fixed
C. Two of the links are fixed
D. None of these
Solution: Given,
Moment of inertia of first body (I1) = 30 kg m^2
Rotational speed of first body (ω1) = 210 RPM
Moment of inertia of second body at rest (I2) = 40 kg m^2
To find: Resultant speed after meshing Formula:
Conservation of Angular Momentum: I1ω1 = (I1+I2)ω2
where, ω2 = Resultant speed after meshing
Calculation: Substituting the given values in the formula, we get
30 × 210 = (30 + 40) × ω2 ω2 = (30 × 210) / (30 + 40) ω2 = 90 RPM
Therefore, the resultant speed after meshing is 90 RPM.
Wrong answer 157.5 right answer
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