Laws of MotionMedium
Question
Two blocks, each having mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be :
Options
A.2/3 Mg sin θ
B.3/2 Mg sin θ
C.
D.2 Mg sin θ
Solution
Mg sinθ − T = Ma [Newton′s II law for block 1]
T = Ma [Newton′s II law for block 2]
By dividing both equations
2 T = Mg sinθ T =
Create a free account to view solution
View Solution FreeMore Laws of Motion Questions
An object of mass 150 kg. is to be lowered with the help of a string whose breaking strength is 100 kg.-wt. What should ...When a lift is descending with an acceleration a, the force exerted by an object of mass m on the lift will be:...In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined by a string...In figure shown, both block are released from rest. The time to cross each other is...Tension in the massless rope at rigid support is (g = 10ms-2)...