Question

A spring is cut into two parts in ratio 1:10 by kength find the stiffness constant of smaller parts

Answer 1

A spring is cut into two parts in ratio 1:10 by kength find the stiffness constant of smaller part is 11 times the stiffness constant of the spring.

Explanation:

If a spring is cut in two or more pieces, the spring constant of the respective pieces are given by

Thus if a spring with spring constant k is cut in the ratio m:n

Then the spring constants of the two pieces respectively will be

$k_1=\frac{(m+n)k}{m}$

And, $k_2=\frac{(m+n)k}{n}$

Here in the question the spring is cut in the ratio 1:10

Therefore, the spring constants will be

Spring constant of the smaller part

$k_1=\frac{(1+10)k}{1}$

$\implies k_1=11k$

And the spring constant of the larger part

$k_2=\frac{(1+10)k}{10}$

$\implies k_2=\frac{11k}{10}$

Hope this helps.

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Answer 2

Answer:

(4) 11K

Explanation:

stiffness constant,

K1=(m+n)/m. =>smaller part

K2=(m+n)/n. =>larger part

according to question

m:n=1:10

so K1=11k. K2=11k/10

hence smallest part =11k

THANK. YOU