A stone of mass 20, {g} is projected from a rubber catapult of length 0.1, {m} and area of cross sec

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8.Mechanical Properties of Solids

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A stone of mass $20\, {g}$ is projected from a rubber catapult of length $0.1\, {m}$ and area of cross section $10^{-6} \,{m}^{2}$ stretched by an amount $0.04\, {m}$. The velocity of the projected stone is $....\,m\,/s.$ (Young's modulus of rubber $=0.5 \times 10^{9}\, {N} / {m}^{2}$ )

A

$10$

B

$15$

C

$25$

D

$20$

(JEE MAIN-2021)

Solution

$\frac{1}{2} \cdot k \cdot x ^{2}=\frac{1}{2} \cdot \frac{ YA }{ L }\cdot x ^{2}$

By energy conservation

$\frac{1}{2} \cdot \frac{ YA }{ L } \cdot x ^{2}=\frac{1}{2} mv ^{2}$

$\frac{0.5 \times 10^{9} \times 10^{-6} \times(0.04)^{2}}{0.1}=\frac{20}{1000} v ^{2}$

$\therefore \quad v ^{2}=400$

$ v =20 m / s$

Standard 11

Physics

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