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 …..

IIT

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 )

IIT

(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:

IIT

$\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