Two springs A and B have force constants $k_{1}$ and $k_{2}$ respectively. The ratio of work done on A to that done on B when they are stretched by the same force is

Q. Two stretched wires

A

and

B

of the same lengths vibrate independently. If the radius, density and tension of wire

A

are respectively twice those of wire

B

, then the fundamental frequency of vibration of

A

relative to that of

B

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A and B are two wires. The radius of A is twice that of B. They are stretched by the same load. Then the stress on B is

The correct option is B four times that on Astress=ForceArea∴stress∝1πr2Thus, SBSA=[rArB]2=(2)2⇒SB=4SA

$A$ and $B$ are two wires. The radius of $A$ is twice that of $B$ . They are stretched by the same load. Then the stress on $B$ is

The correct option is B four times that on $A$
$s t r e s s = \frac{F o r c e}{A r e a}$
$∴ s t r e s s \propto \frac{1}{π r^{2}}$
Thus, $\frac{S_{B}}{S_{A}} = {[\frac{r_{A}}{r_{B}}]}^{2} = (2)^{2} \Rightarrow S_{B} = 4 S_{A}$