## The Inductors of two LR circuits are placed next to each other , as shown in the figure . The values of the self-Inductance ….

Q: The Inductors of two LR circuits are placed next to each other , as shown in the figure . The values of the self-Inductance of the inductors , resistances , mutual-inductance and applied voltage are specified in the given circuit . After both the switches are closed simultaneously , the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is ….. Click to See Solution :
Ans: (55)
Sol: Mutual inductance is producing flux in same direction as self inductance

$\displaystyle U = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 + M I_1 I_2$

$\displaystyle U = \frac{1}{2}(10\times 10^{-3})1^2 + \frac{1}{2}(20\times 10^{-3})2^2 + (5 \times 10^{-3})\times 1 \times 2$

= 55

## A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth .

Q: A cubical solid aluminium (bulk modulus = -VdP/dV = 70 GPa) block has an edge length 1 m on the surface of earth . It is kept on the floor of a 5 km deep ocean . Taking the average density of water and the acceleration due to gravity to be 103 kg/m^3 and 10 m/s^2 respectively , the change in edge length of the block in mm is …..

Click to See Solution :
Ans: (0.24)
Sol: $\displaystyle \frac{\Delta V}{V} = -\frac{\Delta p}{B}$ ; Where B = Bulk modulus

V = l3

$\frac{\Delta V}{V} = 3 \frac{\Delta l}{l}$

$\displaystyle 3\frac{\Delta l}{l} = |-\frac{\Delta p}{B}|=\frac{\rho g h}{B}$

$\displaystyle \Delta l = \frac{\rho g h l}{3 B}$

## Two capacitors with capacitance values C1 = 2000 ±10 pF and C2 = 3000 ±15 pF are connected in series ….

Q: Two capacitors with capacitance values C1 = 2000 ± 10 pF and C2 = 3000 ± 15 pF are connected in series . The voltage applied across this combination is C = 5.00 ±0.02 V . The percentage error in the calculation of the energy stored in this combination is ………….

Click to See Solution :
Ans: (1.3)
Sol: $\displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$

$\displaystyle C = \frac{C_1 C_2}{C_1 + C_2}$

$\displaystyle -\frac{dC}{C^2} = -\frac{dC_1}{C_1^2} – \frac{dC_2}{C_2^2}$

dC = 6 pF

$\displaystyle U = \frac{1}{2}CV^2$

$\displaystyle \frac{dU}{U} \times 100 = (\frac{dC}{C} + \frac{2dV}{V})\times 100$

$\displaystyle \frac{dU}{U} \times 100 = (\frac{6}{1200} + 2\times \frac{0.02}{5})\times 100$

= 13 %

## A beaker of radius r is filled with water (Refractive index = 4/3) up to height H as shown in the figure on the left …..

Q: A beaker of radius r is filled with water (Refractive index = 4/3) up to height H as shown in the figure on the left . The beaker is kept on a horizontal table rotating with angular speed  ω . This makes water surface curved so that the difference in height of water level at the center and at the circumference  of the beaker is h (h<< H , h<< r) as shown in the figure on the right . Take this surface to be approximately spherical with a radius of curvature R . Which of the following is/are correct ? (g is the acceleration due to gravity ) (a) $\displaystyle R = \frac{h^2 + r^2}{2h}$

(b) $\displaystyle R = \frac{3 r^2}{2h}$

(c) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{2}(1 + \frac{\omega^2 H}{2g})^{-1}]$

(d) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{4}(1 + \frac{\omega^2 H}{4g})^{-1}]$

Click to See Solution :
Ans: (a,d)

Sol: $\displaystyle (R-h)^2 + r^2 = R^2$

$\displaystyle R^2 + h^2 – 2 R h + r^2 = R^2$

$\displaystyle h^2 + r^2 = 2Rh$

$\displaystyle R = \frac{h^2 + r^2}{2h}$

Since h << r

$\displaystyle R = \frac{r^2}{2h}$

$\displaystyle h = \frac{\omega^2 r^2}{2 g}$

$\displaystyle R = \frac{r^2 2 g}{2\omega^2 r^2}$

$\displaystyle R = \frac{g}{\omega^2}$

$\displaystyle \frac{\mu_1}{v} – \frac{\mu_2}{u} = \frac{\mu_1 – \mu_2}{R}$

$\displaystyle \frac{1}{v} – \frac{4}{3u} = \frac{1 – 4/3}{R}$

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3(H-h)})$

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3H})$ (Since h << H)

Putting the value of R we get ,

$\displaystyle v = – [\frac{3H}{4}(1 + \frac{\omega^2 H}{4 g})^{-1}]$

## A spherical bubble inside water has radius R . Take the pressure inside the bubble and the water pressure to be po ….

Q: A spherical bubble inside water has radius R . Take the pressure inside the bubble and the water pressure to be Po . The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R-a) . For a << R the magnitude of the work done in the process is given by (4πPoR2 a )X , Where X is a constant and γ = Cp/Cv = 41/30 . The value of X is ……

Click to See Solution :
Ans: (2.05)

Sol: $\displaystyle P_1 V_1^{gamma} = P_2 V_2^{gamma}$

$\displaystyle P_0 (\frac{4}{3}\pi R^3)^{gamma} = P_2 (\frac{4}{3}\pi (R-a)^3)^{gamma}$

$\displaystyle P_2 = P_0 (\frac{R}{R-a})^{3\gamma}$

$\displaystyle W = \frac{P_1 V_1 -P_2 V_2}{\gamma – 1}$

$\displaystyle W = \frac{P_0 (\frac{4}{3}\pi R^3) – P_0 (\frac{R}{R-a})^{3\gamma} (\frac{4}{3}\pi (R-a)^3)}{\frac{41}{30} – 1}$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^{41/10} (R-a)^{-11/10})$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^{41/10}R^{-11/10} (1-a/R)^{-11/10})$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^3 (1-a/R)^{-11/10})$

$\displaystyle W = 4\pi P_0 R^2 a$

Work done is given by (4πPoR2 a )X

On comparing X = 1