JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 12)
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1mm2
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 × 109
N/m2
and 60 × 109
N/m2]
8.0 × 106 N/m2
1.2 × 106 N/m2
0.2 × 106 N/m2
1.8 × 106 N/m2
Explanation
Corresponding to the stress ($$\sigma $$)
Total elongation $$\Delta {I_{net}} = {{\sigma {L_1}} \over {{Y_1}}} + {{\sigma {L_2}} \over {{Y_2}}}$$
$$\sigma = \Delta I\left( {{{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}} \right)$$
$$ = 0.2 \times {10^{ - 3}} \times \left( {{{120 \times 60} \over {180}}} \right) \times {10^9}$$
$$ = 8 \times {10^6}{N \over {{m^2}}}$$
Total elongation $$\Delta {I_{net}} = {{\sigma {L_1}} \over {{Y_1}}} + {{\sigma {L_2}} \over {{Y_2}}}$$
$$\sigma = \Delta I\left( {{{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}} \right)$$
$$ = 0.2 \times {10^{ - 3}} \times \left( {{{120 \times 60} \over {180}}} \right) \times {10^9}$$
$$ = 8 \times {10^6}{N \over {{m^2}}}$$
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