CEE Kerala Engineering
CEE Kerala Engineering Solved Paper-2007
question_answer
A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid-point of the rope. The angle made by the rope with the vertical, in equilibrium is
A) \[50{}^\circ \]
B) \[60{}^\circ \]
C) \[30{}^\circ \]
D) \[40{}^\circ \]
E) \[5.5{}^\circ \]
Correct Answer:
D
Solution :
\[\tan \theta =\frac{50}{6g}=0.83\] or \[tan\text{ }\theta =tan\text{ }40{}^\circ \] or \[\theta =40{}^\circ \]