Two blocks, each of mass m, are connected by a massless thread Y and A represent Young's modulus and cross sectional area of wire respectively. The strain developed in the thread is:

\frac{m g (1 + sin θ)}{2 y A}

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\frac{m g}{y A}

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\frac{m g sin θ}{2 y A}

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\frac{2 m g}{y A}

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Solution

The correct option is B $\frac{m g}{y A}$
Now, $y = \frac{F / A}{A l / l}$
$y = \frac{m g / A}{A l / l}$
$\Rightarrow \frac{A l}{l} = \frac{m g}{y A}$
$S t r a i n = \frac{A l}{l} = \frac{m g}{y A}$
Hence option $B$ is correct

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Two blocks, each of mass m, are connected by a massless thread Y and A represent Young's modulus and cross sectional area of wire respectively. The strain developed in the thread is:

The correct option is B mgyANow, y=F/AAl/ly=mg/AAl/l⇒All=mgyAStrain=All=mgyAHence option B is correct

The correct option is B $\frac{m g}{y A}$
Now, $y = \frac{F / A}{A l / l}$
$y = \frac{m g / A}{A l / l}$
$\Rightarrow \frac{A l}{l} = \frac{m g}{y A}$
$S t r a i n = \frac{A l}{l} = \frac{m g}{y A}$
Hence option $B$ is correct