## Two spring of spring constants 1500 N/m and 3000 N/m respectively are stretched by the same force…

Q: Two spring of spring constants 1500 N/m and 3000 N/m respectively are stretched by the same force. The potential energy gained by the two spring will be in the ratio

Sol: $\large U = \frac{F^2}{2 K}$

if force F is same on two spring

$\large U \propto \frac{1}{K}$

$\large \frac{U_1}{U_2} = \frac{K_2}{K_1}$

$\large \frac{U_1}{U_2} = \frac{3000}{1500} = \frac{2}{1}$

## A spring of force constant 800 N/m has an extension of 5 cm, The work done in extending it from 5 cm to 15 cm is

Q: A spring of force constant 800 N/m has an extension of 5 cm, The work done in extending it from 5 cm to 15 cm is

Sol: $\large W = \frac{1}{2}K (x_2^2 – x_1^2)$

$\large W = \frac{1}{2} \times 800 (15^2 – 5^2)\times 10^{-4}$

= 400 ×[225 – 25]×10-4

= 4×10-2 ×200 = 8 J

## A uniform cylinder of length Land mass m having cross –section area A is suspended with its length vertical from a fixed point…

Q: A uniform cylinder of length Land mass m having cross –section area A is suspended with its length vertical from a fixed point by a mass less spring, such that it is half submerged in a liquid of density at equilibrium position, the extension x_0 of the spring when it is in equilibrium is

Sol: F= K x0 [restoring force in spring]

Buoyant Force $\large F_B = (\frac{A L}{2}) \sigma g$

At equilibrium Fnet = 0

From free body diagram,

$\large K x_0 + (\frac{A L}{2}) \sigma g = m g$

$\large K x_0 = m g – (\frac{A L}{2}) \sigma g$

$\large x_0 = \frac{m g}{K}[1-\frac{A L \sigma}{2 m}]$

## A copper wire 2 m long is stretched by 1 mm. If the energy stores in the stretched wire is converted into heat…

Q: A copper wire 2 m long is stretched by 1 mm. If the energy stores in the stretched wire is converted into heat, then calculate the rise in temperature of the wire.
(Y= 12.5 × 1010 N/m2;  ρ = 9 × 103 kg /m3 ; s = 385 J/kg – K)

Sol: $\large m s \Delta T = \frac{1}{2} Stress \times Strain \times Volume$

$\large m s \Delta T = \frac{1}{2} (Y \times Strain) \times Strain \times \frac{m}{\rho}$

$\large \Delta T = \frac{Y}{2 \rho s } \times (\frac{\Delta l}{l})^2$

$\large \Delta T = \frac{12.5 \times 10^{10}}{2 \times 9 \times 10^3 } \times (\frac{10^{-3}}{2})^2$

= 0.0045 ° C

So the rise in temperature of wire is 0.0045 °C.

## A 40 kg boy whose legs are 4 cm^2 in area 50 cm long falls through a height of 2 m without breaking his leg bones.

Q: A 40 kg boy whose legs are 4 cm2 in area 50 cm long falls through a height of 2 m without breaking his leg bones. If the bones can withstand a stress of 0.9 × 108 N/m2. Calculate the Young’s modulus of material of the bone.

Sol: Mass = 40 kg, area of each leg = 4 cm2 = 4 × 10-4 m2

breaking stress = 0.9×108 N/m2 ,

length of each leg = 50 cm = 50 × 10-2 m

As $\large U = \frac{1}{2}\times Stress \times Strain \times Volume$

$\large U = \frac{1}{2}\times Stress \times \frac{Stress}{Y} \times Volume$

where elastic energy of bone in the form of potential energy, U = m g h;

For two legs,

$\large m g h = 2 (\frac{1}{2}\times Stress \times \frac{Stress}{Y} \times Volume )$

On Putting the given values , we get

Y = 2.05 × 109 N/m2