Question

The springs shown are identical. When A=4kg, the elongation of spring is 1 cm. If B=6kg, the elongation produced by it is

A. 4 cm
B. 3 cm
C. 2 cm
D. 1 cm

Answer 1

Hint: As the springs are identical that means constant k will have the same value for all of them, and using spring force formula elongation can be calculated.

Formula used :
$F=Kx$
Here, F = Spring force, K = Spring constant and x = Elongation in spring.

Complete step by step solution:
Two masses A= 4kg and B = 6 kg are suspended by identical springs of constant k and for mass A= 6kg elongation is 1 cm, we have to calculate the elongation for mass B= 6 kg.

As we know spring force for constant k and elongation x is given by,
F= kx
Therefore for mass A = 4 kg we have,
 \[F = k \times 1\,cm \]
\[\Rightarrow mg = \,k\]..........by \[(F = mg)\]

Substituting m = 4 kg and \[g = \,10\,m/{s^2}\] in above equation we get,
\[k = \,4 \times 10\, \Rightarrow k = 40\,N/m\]
Now for mass B, as the mass is suspended by a series combination of two springs, length of the spring combination will be doubled and the effective constant will be \[\dfrac{k}{2}\].

As k = 40 N/m then \[{k_{eff}} = \dfrac{{40}}{2} = 20\,N/m\]
Let for mass B = 6 kg and \[{k_{eff}} = 20N/m\]elongation in spring be x’ then spring force will be,
\[F = 20 \times x' \\
\Rightarrow 6 \times 10 = \,20 \times x'\]
Simplifying above equation we get,
\[x' = \dfrac{{60}}{{30}} = 3\,cm\]
Hence, for mass B = 6 kg elongation is 3 cm.

Therefore, option B is the correct answer.

Note: Elongation in series combination of springs will be equal to sum of elongation due to individual springs, this will also hold true when springs are not identical i.e. when they have different lengths and constants.