Question

# In an experiment, bras and steel wires of length lm each with areas of cross section $$1 mm^2$$ are used. the wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of $$0.2 mm$$ is : (Given, the Young's Modulus for steel and brass are respectively, $$120 \times 10^9 N/m^2$$ and $$60 \times 10^9 N/m^2$$)

A
0.2×106N/m2
B
4.0×106N/m2
C
1.8×106N/m2
D
1.2×106N/m2

Solution

## The correct option is A $$4.0 \times 10^6 N/m^2$$$$k_1 = \dfrac{y_1 A_1}{\ell_1} = \dfrac{120 \times 10^9 \times A}{1}$$$$k_2 = \dfrac{y_2 A_2}{\ell_2} = \dfrac{60 \times 10^9 \times A}{1}$$$$k_{eq} = \dfrac{k_1 k_2}{k_1 \times k_2} = \dfrac{120 \times 60}{180} \times 10^9 \times A$$$$k_{eq} = 40 \times 10^9 \times A$$$$F = k_{eq} (x)$$$$F = (40 \times 10^9) A. (0.2 \times 10^{-3})$$$$\dfrac{F}{A} = 8 \times 10^6 N/m^2$$No option is matching.Hence question must be bonus.Physics

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