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Two springs A and B fixed at the top and are stretched by $8 \mathrm{~cm}$ and $16 \mathrm{~cm}$ respectively, when loads of $20 \mathrm{~N}$ and $10 \mathrm{~N}$ are suspended at the lower ends. The ratio of the spring constants of the springs $A$ and $B$ is
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Verified Answer
The correct answer is:
$4: 1$
Spring constant, $K=\frac{F}{x}$
Given, $\mathrm{F}_{\mathrm{A}}=20 \mathrm{~N}, \mathrm{~F}_{\mathrm{B}}=10 \mathrm{~N}, \mathrm{x}_{\mathrm{A}}=8 \mathrm{~cm}$ and $\mathrm{x}_{\mathrm{B}}=16 \mathrm{~cm}$
$$
\begin{aligned}
& \therefore \frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}}}{\mathrm{x}_{\mathrm{B}}}}=\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}} \times \frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}=\frac{20}{8} \times \frac{16}{10}=4 \\
& \therefore \mathrm{K}_{\mathrm{A}}: \mathrm{K}_{\mathrm{B}}=4: 1
\end{aligned}
$$
Given, $\mathrm{F}_{\mathrm{A}}=20 \mathrm{~N}, \mathrm{~F}_{\mathrm{B}}=10 \mathrm{~N}, \mathrm{x}_{\mathrm{A}}=8 \mathrm{~cm}$ and $\mathrm{x}_{\mathrm{B}}=16 \mathrm{~cm}$
$$
\begin{aligned}
& \therefore \frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}}}{\mathrm{x}_{\mathrm{B}}}}=\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}} \times \frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}=\frac{20}{8} \times \frac{16}{10}=4 \\
& \therefore \mathrm{K}_{\mathrm{A}}: \mathrm{K}_{\mathrm{B}}=4: 1
\end{aligned}
$$
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