JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 25)
A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched
lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constant, k1 and
k2 will be :
$${1 \over {{n^2}}}$$
$${1 \over n}$$
n2
n
Explanation
For a spring, k$$ \times $$$$l$$ = constant.
$$ \therefore $$ k1$${l_1}$$ = k2$${l_2}$$
$$ \Rightarrow $$ $${{{k_1}} \over {{k_2}}} = {{{l_2}} \over {{l_1}}}$$ = $${{{l_2}} \over {n{l_2}}}$$ = $${1 \over n}$$
$$ \therefore $$ k1$${l_1}$$ = k2$${l_2}$$
$$ \Rightarrow $$ $${{{k_1}} \over {{k_2}}} = {{{l_2}} \over {{l_1}}}$$ = $${{{l_2}} \over {n{l_2}}}$$ = $${1 \over n}$$
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