Two block of masses of 1 kg and 2 kg are connected by a metal wire going over a smooth pulley- The breaking stress of metal is 403- 106N-m2- What should be the minimum radius of wire used if it should not break- g - 10 m-s2

The correct option is C 1.5 mm Breaking stress = FA = 403π× 10^6N/m^2 According to question 403π× 10^6 = 3× 10π r^2 r^2 = 94× 10^6 ...

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Standard XII

Physics

Shear Strain

Two block of ...

Question

Two block of masses of $1 k g$ and $2 k g$ are connected by a metal wire going over a smooth pulley. The breaking stress of metal is $\frac{40}{3 π} \times 10^{6} N / m^{2}$ . What should be the minimum radius of wire used if it should not break?
$(g = 10 m / s^{2})$

0.5 m m

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1 m m

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1.5 m m

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2 m m

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Solution

The correct option is C $1.5 m m$
Breaking stress $= \frac{F}{A}$
$= \frac{40}{3 π} \times 10^{6} N / m^{2}$
According to question
$\frac{40}{3 π} \times 10^{6} = \frac{3 \times 10}{π r^{2}}$
$r^{2} = \frac{9}{4 \times 10^{6}}$
$r = 1.5 \times 10^{- 3} m$
$= 1.5 m m$

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