## Estimate the change in the density of water in ocean at a depth of 500m below the surface….

Q: Estimate the change in the density of water in ocean at a depth of 500 m below the surface. The density of water at the surface = 1030 kgm-3 and the bulk modulus of water = 2.2×109 Nm-2

(a) 2.363 kg/m³

(b) 1.363 kg/m³

(c) 4.363 kg/m³

(d) 3.363 kg/m³

Ans: (a)

Sol:

Bulk Modulus of Elasticity is $\displaystyle K = \frac{P}{-\Delta V/V}$ $\displaystyle \frac{-\Delta V}{V} = \frac{P}{K}$ $\displaystyle \frac{\Delta \rho}{\rho} = \frac{P}{K}$ $\displaystyle \Delta \rho = \frac{P \rho}{K}$ $\displaystyle \Delta \rho = \frac{\rho g h \times \rho}{K}$

on putting given values ,

= 2.363 kg/m³

## A 30.0 kg hammer, moving with speed 20.0m/s, strikes a steel spike 2.30 cm in diameter….

Q: A 30.0 kg hammer, moving with speed 20.0m/s, strikes a steel spike 2.30 cm in diameter. The hammer rebounds with speed 10.0 m/s after 0.110s. What is the average stress in the spike during the impact?

(a) 1.97 × 107 N/m²

(b) 3.2 ×107 N/m²

(c) 4.6 × 107 N/m²

(d) 8.2 ×107 N/m²

Ans: (a)

Sol:
Force = Change in momentum per sec.

Change in momentum = 2 m u (since hammer rebounds with same speed) $\displaystyle F = \frac{2 m u}{t}$ $\displaystyle Stress = \frac{F}{A} = \frac{F}{\pi r^2}$

## Two rods of different materials having coefficient of thermal expansion α1, α2 and Young’s moduli Y1, Y2 respectively…..

Q: Two rods of different materials having coefficient of thermal expansion α1, α2 and Young’s moduli Y1, Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 : α2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to:

(a) 2 : 3

(b) 1 : 1

(c) 3 : 2

(d) 4 : 9

Ans: (c)

Sol: $\displaystyle Y_1 \alpha_1 \Delta \theta = Y_2 \alpha_2 \Delta \theta$ $\displaystyle Y_1 \alpha_1 = Y_2 \alpha_2$ $\displaystyle \frac{Y_1}{Y_2} = \frac{\alpha_2}{\alpha_1}$ $\displaystyle \frac{Y_1}{Y_2} = \frac{3}{2}$

## A wire of cross section A is stretched horizontally between two clamps located ‘2l’ m apart….

Q: A wire of cross section A is stretched horizontally between two clamps located ‘2l’ m apart. A weight W kg is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance x << l the strain produced is

(a) $\displaystyle \frac{2 x^2}{l^2}$

(b) $\displaystyle \frac{ x^2}{l^2}$

(c) $\displaystyle \frac{ x^2}{2 l^2}$

(d) $\displaystyle \frac{ x }{2 l^2}$

Ans: (c)

Sol: Change in Length $\displaystyle \Delta l = 2\sqrt{l^2 + x^2} - 2 l$ $\displaystyle \Delta l = 2l\sqrt{1 + \frac{x^2}{l^2}} - 2 l$ $\displaystyle \Delta l = [2l(1+ \frac{x^2}{l^2})^{1/2}- 2l]$ $\displaystyle \Delta l = 2l [(1+ \frac{x^2}{l^2})^{1/2}- 1]$ $\displaystyle \Delta l = 2l [(1+ \frac{1}{2}\frac{x^2}{l^2}) - 1]$ $\displaystyle \Delta l = 2l (\frac{x^2}{2l^2})$ $\displaystyle \frac{\Delta l}{2l} = (\frac{x^2}{2l^2})$ $\displaystyle Strain = (\frac{x^2}{2l^2})$

## One end of a long metallic wire of length L, area of cross-section A and Young’s modulus Y is tied to the ceiling…..

Q: One end of a long metallic wire of length L, area of cross-section A and Young’s modulus Y is tied to the ceiling. The other end is tied to a massless spring of force constant K. A mass m hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by

(a) $\displaystyle 2\pi \sqrt{\frac{m}{k}}$

(b) $\displaystyle 2\pi \sqrt{\frac{mYA}{k L}}$

(c) $\displaystyle 2\pi \sqrt{\frac{mY}{k}}$

(d) $\displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$

Ans:(d)

Sol: For wire , $\displaystyle Y = \frac{F L}{A \Delta L}$ $\displaystyle F = (\frac{A Y}{L} )\Delta L$ $\displaystyle F = K' \Delta L$

Where $\displaystyle K' = \frac{A Y}{L}$

For Spring , Force Constant = K

Both Wire & Spring are in Series , $\displaystyle \frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{K'}$ $\displaystyle \frac{1}{K_{eq}} = \frac{1}{K} + \frac{L}{A Y}$

Time Period , $\displaystyle T = 2\pi \sqrt{\frac{m}{K_{eq}}}$ $\displaystyle T = 2\pi \sqrt{m(\frac{1}{K} + \frac{L}{A Y})}$ $\displaystyle T = 2\pi \sqrt{\frac{m(YA + KL)}{kYA}}$