Two wires ‘A’ and ‘B’ of the same material have radii in the ratio 2 : 1 and lengths in the ratio 4 : 1.

Q: Two wires $‘A’$ and $‘B’$ of the same material have radii in the ratio $2 : 1$ and lengths in the ratio $4 : 1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

a (a) $F = Y \times A \times \frac{l}{L}$ $⇒$ $F \propto \frac{{{r^2}}}{L}$ $(Y$ and $l$ are constant$)$ $\frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\left( {\frac{{{L_2}}}{{{L_1}}}} \right) = {\left( {\frac{2}{1}} \right)^2}\left( {\frac{1}{4}} \right) = 1$ $⇒$ $\frac{{{F_1}}}{{{F_2}}} = 1:1$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

Two wires $‘A’$ and $‘B’$ of the same material have radii in the ratio $2 : 1$ and lengths in the ratio $4 : 1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

A$1:1$
B$2:1$
C$1:4$
D$1:2$

Medium

STD 11 Science
JEE
STD 11- 8. mechanical properties of solids
Physics

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