Q. The rubber cord catapult has a cross-sectional area 1 $mm^{2}$and total unsaturated length 10.0 cm. It is stretched to 12.0 cm and then released to project a miscible of mass 5.0 g. Taking Young’s modulus for rubber as, the tension in the cord is
A) 1000 N
B) 100 N
C) 10 N
D) 1 N
Explanation
(b) $Y = \frac{F}{A}\ \times \frac{l}{\mathrm{\Delta}l}\ \ or\ \ F = YA\frac{\mathrm{\Delta}l}{l}$
= $\frac{(5.0 \times 10^{8}) \times 10^{6})\ \times (2 \times 10^{- 2})}{(10 \times 10^{- 2})} = 100\ $N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.