A stone of mass 1 kg is tied with a string and it is whirled in a vert...
At highest point, T'+mg=mv^2/r At lowest point, T-mg=mv^2/r again, T-T'=6mg so, 6mg+14-mg=mv^2/r 64=v^2 v=8
A stone of mass 1 kg is tied with a string and it is whirled in a vert...
Solution:
Given:
Mass of the stone, m = 1 kg
Radius of the circle, r = 1 m
Tension at the highest point, T = 14 N
Let the velocity of the stone at the lowest point be v.
Calculating Velocity at the Highest Point:
At the highest point, the tension in the string is equal to the weight of the stone.
Therefore, T = mg
where g is the acceleration due to gravity, which is equal to 9.8 m/s^2.
So, 14 = 1 × 9.8
Thus, g = 14/9.8 = 1.43 m/s^2.
At the highest point, the net force on the stone is given by:
Fnet = T - mg
= 14 - 1 × 9.8
= 4.2 N
Using Newton's second law of motion, we know that:
Fnet = ma
where a is the acceleration of the stone.
So, 4.2 = 1 × a
Thus, a = 4.2 m/s^2.
Therefore, the velocity of the stone at the highest point is given by:
v1 = √(gr)
= √(1.43 × 1)
= 1.2 m/s
Calculating Velocity at the Lowest Point:
At the lowest point, the tension in the string is given by:
T = mv^2/r + mg
where v is the velocity of the stone.
Substituting the given values, we get:
14 = 1 × v^2/1 + 1 × 9.8
Simplifying, we get:
v^2 = 14 × 2 - 9.8
Thus, v = √(27.2) = 5.2 m/s
Therefore, the velocity of the stone at the lowest point is 5.2 m/s.
Hence, the correct option is 3) 6 m/s (closest to 5.2 m/s).
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.