Question

A spiral spring is stretched to 20.5 cm gradation on a metre scale when loaded with a 100 g load and to the 23.3 cm gradation by 200 g load. The spring is used to support a lump of metal in air and the reading now is 24.0 cm. The mass of metal lump is :

• A. 250 gm
• B. 225 gm
• C. 145 gm
• D. 750 gm

Answer 1

Answer: (B)

225 gm

Let original length of spring be $L$.

Thus, $20.5-L=\frac{0.1kg\times g\times L}{AY}$ or $\frac{L}{AY}=20.5-L$

Also, from second loading, we get similarly, $\frac{0.2\times g\times L}{AY}=23.3-L$ or, $\frac{L}{AY}=\frac{23.3-L}{2}$

Solve these to get, $41-2L=23.3-L$ or $L=17.7$

Also, $\frac{L}{AY}=20.5-17.7=2.8$

Thus, for $24cm$, $\frac{mgL}{AY}=24-L$ or $mg=\frac{(24-17.7)}{2.8}=2.25$ or $m=225g$