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Standard XII

Physics

Power, Force and Velocity

When a mass o...

Question

When a mass of 8 kg is suspended from a string, its length is $l_{1}$ . If a mass 10 kg is suspended, its length is $l_{2}$ . Length, when a mass of 16 Kg is suspended from it, is given by

2 l_{2} - l_{1}

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2 l_{2} + l_{1}

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4 l_{2} + 3 l_{1}

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4 l_{2} - 3 l_{1}

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Solution

The correct option is D $4 l_{2} - 3 l_{1}$
$Δ l_{1} = \frac{F_{1} l}{A Y} \Rightarrow l_{1} = l (\frac{F_{1}}{A Y} + 1) = l (\frac{80}{A Y} + 1) a g a i n, Δ l_{2} = \frac{F_{2} l}{A Y} \Rightarrow l_{2} = l (\frac{F_{2}}{A Y} + 1) = l (\frac{100}{A Y} + 1) ∴ l_{3} = 4 l_{2} - 3 l_{1}$ .

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When a mass of 8 kg is suspended from a string, its length is l1. If a mass 10 kg is suspended, its length is l2. Length, when a mass of 16 Kg is suspended from it, is given by

The correct option is D 4l2−3l1Δl1=F1lAY⇒l1=l(F1AY+1)=l(80AY+1)again,Δl2=F2lAY⇒l2=l(F2AY+1)=l(100AY+1)∴l3=4l2−3l1.

When a mass of 8 kg is suspended from a string, its length is $l_{1}$ . If a mass 10 kg is suspended, its length is $l_{2}$ . Length, when a mass of 16 Kg is suspended from it, is given by

The correct option is D $4 l_{2} - 3 l_{1}$
$Δ l_{1} = \frac{F_{1} l}{A Y} \Rightarrow l_{1} = l (\frac{F_{1}}{A Y} + 1) = l (\frac{80}{A Y} + 1) a g a i n, Δ l_{2} = \frac{F_{2} l}{A Y} \Rightarrow l_{2} = l (\frac{F_{2}}{A Y} + 1) = l (\frac{100}{A Y} + 1) ∴ l_{3} = 4 l_{2} - 3 l_{1}$ .