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A rubber cord catapult has cross-section...

A rubber cord catapult has cross-sectional area `25 mm^(2)` and initial length of rubber cord is `10 cm`. It is stretched to `5 cm`. And then released to protect a missle of mass `5 gm` Taking `Y_("nibber")=5xx10^(7)N//m^(2)` velocity of projected missle is

A

`20 ms^(-1)`

B

`100 ms^(-1)`

C

`250 ms^(-1)`

D

`200 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`A= 25 xx 10^(-6) m^(2), l= 10 xx 10^(-2) m, Delta l= 5 xx 10^(-2) m`
`m= 5 xx 10^(-3) kg, Y= 5 xx 10^(8) N//m^(2)`
PE energy `= (1)/(2) mv^(2)`
`rArr (1)/(2) (YA (Delta l)^(2))/(l)= (1)/(2) mv^(2)`
`v= sqrt((YA (Delta l)^(2))/(ml))= sqrt((5 xx 10^(8) xx 25 xx 10^(-6) xx (5 xx 10^(-2))^(2))/(5 xx 10^(-3) xx 10 xx 10^(-2)))`
= 250 m/s
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