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 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

Find the pressure that has to be applied to the ends of a steel wire of length 10 cm, to keep its length constant…

Q: Find the pressure that has to be applied to the ends of a steel wire of length 10 cm, to keep its length constant when its temperature is raised by 100° C is

( Ys = 2 × 1011 N/m2  αwire = 1.1×10-5/K )

Sol: pressure, $ \large \Delta l = l \alpha \Delta T $

$ \large \frac{\Delta l}{l} = \alpha \Delta T $ …(i)

$\large Y = \frac{F/A}{\frac{\Delta l}{l} }$

$\large Y = \frac{P}{\frac{\Delta l}{l} }$

$\large P = Y \times \frac{\Delta l}{l}$

$\large P = Y \alpha \Delta T $

= 2 × 1011 × 1.1×10-5 × 102

=2.2 × 108 Pa

When a wire of length 10 m subjected to a force of 100 N along its length, the lateral stain produced is 0.01 × 10^(-3)…

Q: When a wire of length 10 m subjected to a force of 100 N along its length, the lateral stain produced is 0.01 × 10-3. The Poisson’s ratio was found to be 0.4. If area of cress section of wire is 0.025 m2, its Young’s modulus is

Sol: Poisson’s ratio $\large \sigma = \frac{Lateral \; Strain}{Longitudinal\; Strain}$

$\large \sigma = \frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$

$\large \frac{\Delta l}{l} = \frac{-\frac{\Delta r}{r}}{\sigma}$

$\large \frac{\Delta l}{l} = \frac{10^{-5}}{0.4} = 25 \times 10^{-6}$

$\large Y = \frac{F/A}{\frac{\Delta l}{l}} $

$\large Y = \frac{100}{25 \times 10^{-3}\times 25 \times 10^{-6}} $

= 1.6 × 108 N/m2