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Consider the gas A.

Let the final pressure be P. For gas A,

$P_1 V^{\frac{3}{2}} = P (2V)^{\frac[3}{2}}$

$\Rightarrow P_1 = 2^{\frac{3}{2}}P$

Consider gas B:

As it expands isobarically, so it's pressure is a constant.

Thus, P_2 = P

For the third gas, Using PV = constant, we get

P_3 = 2P

Thus the ratio is $2\sqrt{2} : 1 : 2$

option b)

is it v/sqrt{2}?

And thus there will be a variation in the growth and decay of the current.

@vyppi - we need the exact problem to solve it.

Kinetic energy = charge x potential through which it is moved.

Thanks to spidey for correcting my misconception about the problem.

Oh... I was thinking the other way. Anyway, here is the solution :

The torque cannot be written straight away as Bhim did. It is wrong. The way he related the MI and the torque itself is wrong first.

Consider an elementary ring between lengths x and x+dx in the ring.

The area of this element = $\frac{2mxdx}{R^2}$ (Assuming areal distribution of mass)

Now, the torque acting on this element, which is at a distance of x from the center (as dx is almost 0)

$d\tau = 2\frac{\mu mg x^2}{R^2} \, dx$

The total torque can be found out by integration. Thus

$\tau = \int^{R}_0 2\frac{\mu mg x^2}{R^2} \, dx$

$\Rightarrow \tau = \frac{2\mu mgR}{3}$

Now, $\frac{2\mu mgR}{3} = \frac{mR^2}{2} \alpha$

giving $\alpha = \frac{4 \mu g}{3R}$

The required time is given by the equation

$\omega - \alpha t = 0 \\ \\ \Rightarrow t = \frac{3R\omega}{4\mu g}$

D)
In the first case, ie, rolling up the incline, the lowermost point has a tendency to slip backwards.

In the second case, the entire disc has a tendency to slip down due to the gsin@ component.

ganesha is right.

@jasbir - Does the question want the time after which it starts pure rolling? Because a rolling body can never come to rest, unless external forces like air resistance, viscosity are present.
@Bhim - You're answer may be right, but the method is wrong mate.
@spidey - that method will get you nowhere, as this is a 2-D object.

x = ut

y = 0.5at²

where a = qE/m.

Now eliminate t.

Let the number of Ferric ions be x. So number of ferrous ions are 93-x. For charge neutrality,
3x + 2(93-x) = 200
giving x = 14.

Percentage = 14/93 x 100 = 15%