JEE Advance - Physics (2019 - Paper 1 Offline - No. 15)
A block of weight 100 N is suspended by copper and steel wires of same cross-sectional area 0.5 cm2 and length $$\sqrt 3 $$ m and 1 m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are 30$$^\circ $$ and 60$$^\circ $$, respectively. If elongation in copper wire is ($$\Delta {l_c}$$) and elongation in steel wire is ($$\Delta {l_s}$$), then the ratio $${{\Delta {l_c}} \over {\Delta {l_s}}}$$ is .............. .
[Young's modulus for copper and steel are 1 $$ \times $$ 1011 N/m2 and 2 $$ \times $$ 1011 N/m2 respectively.]
[Young's modulus for copper and steel are 1 $$ \times $$ 1011 N/m2 and 2 $$ \times $$ 1011 N/m2 respectively.]
Answer
2
Explanation
$${{{T_s}} \over 2} = {T_c}{{\sqrt 3 } \over 2}$$
$${T_s} = \sqrt 3 {T_c}$$
$$\Delta l = {{Tl} \over {Ay}}$$
$$ \therefore $$ $${{\Delta {l_c}} \over {\Delta {l_s}}} = \left( {{{{T_c}} \over {{T_s}}}} \right)\left( {{{{l_c}} \over {{l_s}}}} \right)\left( {{{{Y_s}} \over {{Y_c}}}} \right)$$
$$ = \left( {{1 \over {\sqrt 3 }}} \right)\left( {{{\sqrt 3 } \over 1}} \right)\left( {{{2 \times {{10}^{11}}} \over {1 \times {{10}^{11}}}}} \right) = 2.00$$
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