JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 7)
Two steel wires having same length are
suspended from a ceiling under the same load.
If the ratio of their energy stored per unit
volume is 1 : 4, the ratio of their diameters is:
1 : 2
2 : 1
$$1:\sqrt 2 $$
$$\sqrt 2 :1$$
Explanation
$${{du} \over {dv}}$$ = $${1 \over 2}$$ $$ \times $$ stress × strain
= $${1 \over 2}{F \over A} \times {F \over {AY}}$$ $$ \propto $$ $${1 \over {{A^2}}}$$ $$ \propto $$ $${1 \over {{d^4}}}$$
$${{du} \over {dv}}$$ = $${1 \over 4}$$
$$ \Rightarrow $$ $${\left( {{{{d_1}} \over {{d_2}}}} \right)^4}$$ = 4
$$ \Rightarrow $$ $${{{d_1}} \over {{d_2}}} = {\left( 4 \right)^{{1 \over 4}}}$$ = $$\sqrt 2 :1$$
= $${1 \over 2}{F \over A} \times {F \over {AY}}$$ $$ \propto $$ $${1 \over {{A^2}}}$$ $$ \propto $$ $${1 \over {{d^4}}}$$
$${{du} \over {dv}}$$ = $${1 \over 4}$$
$$ \Rightarrow $$ $${\left( {{{{d_1}} \over {{d_2}}}} \right)^4}$$ = 4
$$ \Rightarrow $$ $${{{d_1}} \over {{d_2}}} = {\left( 4 \right)^{{1 \over 4}}}$$ = $$\sqrt 2 :1$$
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