First slide

Work done by Diff type of forces

Question

A load suspended by a massless spring produces an extension of x cm in equilibrium. When it is cut into
two unequal parts, the same load produces an extension of 7.5 cm when suspended by the larger
part of length 60 cm. When it is suspended by the smaller part, the extension is 5.0 cm. Then,

Moderate

A

x = 12.5 cm
B

x = 3.0 cm
C

the length of the original spring is 90 cm
D

the length of the original spring is 80 cm

Solution

Elongation of the wire, $Δ l = \frac{F l}{A Y}$ (Can also be applied for a spring)
$∴$ $Δ l \propto l$
$\frac{7.5}{5.0} = \frac{60}{l_{2}} \Rightarrow l_{2} = 40 cm$
$∴$ Length of original spring is (60+ 40) cm= 100 cm
Now, $\frac{x}{7.5} = \frac{100}{60}$
x = 12.5cm

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