Engineering Entrance Exams> 7. In an adiabatic process, the pressure ...

7. In an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The value of (which equals Cp/Cv) is a) 5/4b) 4/3c) 5/3d) 3/2Space for rough work AEE15vA48. A mass is moving towards the origin along the x-axis with constant velocity. Its angular momentum with respect to the origin a) remains constantb) is zeroc) increasesd) decreases 9. The rate of cooling of a liquid is 4o C/s, when its temperature is 80o C and is 2o C/s when its temperature is 50o C. The temperature of the surrounding is a) 30o Cb) 20o Cc) 10o Cd) 25o C 10. A Charged sphere of radius 1m carries a charge of 1 x 10-9 C. The electric fields at a point P, which is at a distance d = 3m from the centre of the sphere and at a point Q, at a distance d = 0.3m from the centre of the sphere are respectively a) 1 N/C and 100 N/Cb) 1 N/C and zeroc) zero and 1 N/Cd) 1 N/C and 3 N/C11. An electric dipole lying along X-axis with moment 5 Am2 is subjected to an electric field of magnitude 10j N/C. The torque experienced is a) 2 Nmb) 10 Nmc) 50 Nmd) 25 Nm 12. A parallel plate capacitor with air gap of 5 mm is 2 MFD. If a metallic plate of thickness3 mm is inserted in between the plates, the new capacitance is a) 5 MFDb) 1 MFDc) 2 MFDd) 2.5 MFD1

dheeraj reddy , 7 Years ago

Grade 12th pass

2 Answers

Vikas TU

Last Activity: 5 Years ago

Dear student

Please type one queston at a time , It is very much difficut for us to type 10 queston in a go.

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Abhiram S

Last Activity: 3 Years ago

Answers

7)According to given

$P \to T^3 \Rightarrow P\cdot T^{^{-3}} = C$ (1)

but we have

$PV = nRT \Rightarrow V = \frac{T}{P}$ (the rest are const for a particular gas in a particular case)

and for an adiabatic

$P V^{\gamma }$ is a constant

and

$P^{1-\gamma } T^{\gamma }$ is a constant

$\Rightarrow P T^{\frac{\gamma}{1-\gamma} }$ is a constant

from (1) and above

$\Rightarrow \frac{\gamma }{1-\gamma } = -3 \therefore \gamma = 3/2$

8) Here we use the concept that the angular momentum is defined as the product of the mass of the body, the velocity of the body, and the perpendicular distance from the line of action of momentum with respect to the point. In this case the reference point is the origin, and the body is moving along the x-axis. So here the ‘perpendicular distance from the line of action of momentum with respect to the point’ remians constant, ie 0 here, So the answer may be 0 or since it does not change, it can be said to be constant.

9) Here using Newton's Rate of Cooling equation

$\frac{\mathrm{d} \Theta }{\mathrm{d} t} = k (T- T_{s})$

where $\frac{\mathrm{d} \Theta }{\mathrm{d} t}$ is the rate of cooling, T is the temperature of the body and T_s is the temperature of the surroundings.

now according to given

at $80^{\circ} C$

$\frac{\mathrm{d} \Theta_1 }{\mathrm{d} t} = k (80- T_{s}) = 4$ (1)

at $20^{\circ} C$

$\frac{\mathrm{d} \Theta_2 }{\mathrm{d} t} = k (20- T_{s}) = 2$ (2)

now dividing (1) by (2) we get

$\frac{80 - T_{s}}{50 - T_s} = 2\Rightarrow T_s = 20^{\circ}C$

10) Here we have two cases

At point P at a dist of 3m from the centre of the sphere, ie oustide the sphere

here by applying Gauss’s Law

$\varphi = E \cdot A = \frac{q}{\varepsilon _{0}}$ (1)

where $\varphi$ is the flux through the Gaussian surface, E is the elctric field due to the charge, A is the area of the surface, q is the net charge enclosed by that surface, $\varepsilon _{0}$ is the permittivty of free space

in our case we need E

$E= \frac{q}{A\varepsilon _{0}} = \frac{q}{4\Pi r^{2}x\varepsilon _{0}}=\frac{(9 x 10 ^{9}) x (10^{-9})}{3^2} = 1$

where $\frac{1}{4\Pi\varepsilon0} = 9x10^{9}$ and r = 3

now for the tricky part

At point Q at a dist of 0.3m from the centre of the sphere

here we have a problem, the sphere could be a solid sphere, in which case you would need the charge per unit volume, or it could be a shell in which case you don’t

if it is a solid sphere

let the charge per unit volume be $\rho = \frac{q}{\frac{4}{3}\Pi r^{3}}$

here $q = 10^{-9}$ and r = 1

$\therefore \rho = \frac{10^{-9}}{\frac{4}{3}\Pi}$

now for a Gaussian surface of radius 0.3m inside this sphere the net charge, $q`$ , enclosed by it will be

$q` = \rho v = \frac{10^{-9}}{\frac{4}{3}\Pi} x \frac{4}{3}\Pi x (0.3)^{3} = 0.027 x 10^{-9}$

where v is the volume enclosed by that surface

and so, from (1) and above for this surface the net electric field E will be

$E= \frac{q`}{A`\varepsilon _{0}} = \frac{q`}{4\Pi r`^{2}x\varepsilon _{0}}=\frac{(9 x 10 ^{9}) x (0.027x10^{-9})}{0.3^2} = 2.7$

but, this is not in the options, so it must have been a mettalic shell

in that case, when the Q is at a dist 0.3m fro the centre, Q is inside the shell, and since inside the shell there would not be any net charge(the charge is distributed on the surface), the net flux through the Gaussian surface with this radius and hence the Electric Field at this point, E, would also be 0. And since this second choice is in the ansewer , it must be true

$\therefore$ at P, E = 1

at Q, E = 0

11) Here we use the formula for torque on an electric dipole in a Uniform Elctric Field

$\tau = PE\sin \Theta$

according to given

the dipole moment P is 5 A/m²

the electric field E is 10j N/C

the angle $\Theta$ is $90^{\circ}$

$\therefore \tau = 5 \cdot 10 \cdot \sin 90$

$= 50$

12) Here we use the equayion for a parallel plate capacitor with a dielctric medium in between the plates as

$C =\frac{ \varepsilon _{0} A}{d-t +\frac{t}{k}}$ (1)

where C is the cpacitance, A is the plate area, d is the plate separation, t is the thickness of the inserted material and k is the material’s dielctric constant.

now if a mettalic plate is used

$k = infinty \Rightarrow \frac{1}{k} = 0$