JEE Main & Advanced
JEE Main Online Paper (Held on 9 April 2013)
question_answer
A uniform sphere of weight W and radius 5cm is being held by a string as shown in the figure. The tension in the string will be:
JEE Main Online Paper (Held On 09 April 2013)
A) \[12\frac{W}{5}\]
B) \[5\frac{W}{12}\]
C) \[13\frac{W}{12}\]
D) \[13\frac{W}{12}\]
Correct Answer:
C
Solution :
\[T\cos \theta =w\] \[T=\frac{w}{\cos \theta }\] \[\cos \theta =\frac{AB}{AC}=\frac{AB}{AD+BC}=\frac{AB}{8+5}=\frac{AB}{13}\] From right angle triangle \[AB=12\] So \[T=\frac{w\times 13}{12}\]