A load of 500 kg was lifted through a distance of 13 cm. by an effort of 25 kg which moved through a distance of 650 cm. The efficiency of the lifting machine is
A. 50 %
B. 40 %
C. 55 %
D. 30 %
Answer: Option B
Solution(By Examveda Team)
Mechanical advantage (MA) = Load (W)/Effort (P).MA = W/P = 500/25 = 20.
Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).
VR = Y/X = 650/13 = 50.
Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.
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Comments ( 3 )
In case of S.H.M. the period of oscillation (T), is given by
A. $${\text{T}} = \frac{{2\omega }}{{{\pi ^2}}}$$
B. $${\text{T}} = \frac{{2\pi }}{\omega }$$
C. $${\text{T}} = \frac{2}{\omega }$$
D. $${\text{T}} = \frac{\pi }{{2\omega }}$$
The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be
A. 0.1 rad/sec
B. 1 rad/sec
C. 10 rad/sec
D. 100 rad/sec
A body is said to move with Simple Harmonic Motion if its acceleration, is
A. Always directed away from the centre, the point of reference
B. Proportional to the square of the distance from the point of reference
C. Proportional to the distance from the point of reference and directed towards it
D. Inversely proportion to the distance from the point of reference
The resultant of two forces P and Q acting at an angle $$\theta $$, is
A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$
B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$
C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$
D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$
E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$
50 % is the wrong answer
Correct ans is 40% follow see
Mechanical advantage (MA) = Load (W)/Effort (P).
MA = W/P = 500/25 = 20.
Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)).
VR = Y/X = 650/13 = 50.
Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.
So option B is right
Yes right answer is 40%
This is wrong answer