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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
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B
4.0×106N/m2
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C
1.8×106N/m2
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D
1.2×106N/m2
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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.
1247562_1614715_ans_b489eac7989a431185f3e3f87ea8afa6.PNG

Physics

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