A particular force ( F ) applied on a wire increases its length by 2 × 10 - 3 m . To increase the wire's length by 4 × 10 - 3 m, the applied force will be

Y = F A l L = F × L A × l ⇒ F ∝ l F 1 F 2 = l 1 l 2 ⟹ F F 2 = 2 × 10 - 3 4 × 10 - 3 F 2 = 2 F

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Physics - Elasticity Question with Solution | TestHub

PhysicsElasticityMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A particular force $(F)$ applied on a wire increases its length by $2 \times 10^{- 3} m .$ To increase the wire's length by $4 \times 10^{- 3} m,$ the applied force will be

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Answer:

Solution:

$Y = \frac{\frac{F}{A}}{\frac{l}{L}} = \frac{F \times L}{A \times l}$

$\Rightarrow F \propto l$

$\frac{F_{1}}{F_{2}} = \frac{l_{1}}{l_{2}}$

⟹ $\frac{F}{F_{2}} = \frac{2 \times 10^{- 3}}{4 \times 10^{- 3}}$

$F_{2} = 2 F$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElasticitySubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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