Each of the three blocks P, Q and R shown in figure has a mass of 3 kg. Each of the wires A and B has cross-sectional area 0.005 cm^{2} and Young modulus 2 \times 10^{11} N~m^{2}. Neglect friction. Find the longitudinal strain developed in wire B. Take $g = 10 m s^{- 2} .$

3 \times 10^{- 4}

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4 \times 10^{- 4}

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10^{- 4}

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2 \times 10^{- 4}

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Solution

The correct option is D $2 \times 10^{- 4}$
The block R will descend vertically and the blocks P and Q will move on the frictionless horizontal table. Let the common magnitude of the acceleration be a. Let the tensions in the wires A and B be $T_{A} a n d T_{B}$ respectively.
Writing the equations of motion of the blocks P, Q and R, we get,

$T_{A} = (3 k g) a$

$T_{B} - T_{A} = (3 k g) a$ ...(ii)
and $(3 k g) g - T_{B} = (3 k g) a .$ ...(iii)
By (i) and (ii),

$T_{A} + T_{B} = (3 k g) g = 30 N$

or, $3 T_{A} = 30 N$

or, $T_{A} = 10 N a n d T_{B} = 20 N .$

Logitudinal strain $= \frac{L o n g i t u d i n a l s t r e s s}{Y o u n d m o d u l u s}$

Strain in wire $A = \frac{10 N / 0.005 c m^{2}}{2 \times 10^{11} N m^{- 2}} = 10^{- 4}$ and strain in wire $B = \frac{20 N / 0.005 c m^{2}}{2 \times 10^{11} N m^{- 2}} = 2 \times 10^{- 4} .$

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